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Solutions to Section 1

Exercise 1.1Show that |a| ≥ a and |a| ≥ −a.

Solution.This follows from the fact that max{−a, a} ≥ a and max{−a, a} ≥ −a

Exercise 1.2Show that

|a| ={

a if a ≥ 0−a if a < 0

That is, the absolute value function is a piecewise defined function. Graphthis function in the rectangular coordinate system.

Solution.If a ≥ 0 then −a ≤ 0 so that |a| = max{−a, a} = a. If a < 0 then −a > 0 sothat |a| = max{−a, a} = −a. The graph is shown in Figure 1.

Figure 1

Exercise 1.3Show that |a| ≥ 0 with |a| = 0 if and only if a = 0.

Solution.We see from the graph of |a| that |a| ≥ 0 and |a| = 0 if and only if a = 0

Exercise 1.4Show that if |a| = |b| then a = ±b.

Solution.Suppose first that a ≥ 0. Then |a| = a.• If b ≥ 0 then |b| = b. In this case, |a| = |b| implies that a = b.

1

• If b < 0 then |b| = −b. In this case, |a| = |b| implies that a = −b.Suppose now that a < 0. Then |a| = −a.• If b ≥ 0 then |b| = b. In this case, |a| = |b| implies that −a = b which isthe same as a = −b.• If b < 0 then |b| = −b. In this case, |a| = |b| implies that −a = −b whichis equivalent to a = b

Exercise 1.5Solve the equation |3x− 2| = |5x + 4|.

Solution.We have either 3x − 2 = 5x + 4 or 3x − 2 = −(5x + 4). Solving the firstequation we find x = −3. Solving the second equation, we find x = −1

4

Exercise 1.6Show that | − a| = |a|.

Solution.If a ≥ 0 then −a < 0. Thus, |a| = a and | − a| = −(−a) = a. Hence,| − a| = |a|. If a < 0 then −a > 0. In this case, |a| = −a and | − a| = −a.That is, | − a| = |a|

Exercise 1.7Show that |ab| = |a| · |b|.

Solution.Suppose first that a ≥ 0. Then |a| = a.• If b ≥ 0 then |b| = b. Moreover, ab ≥ 0. In this case, |ab| = ab = |a| · |b|.• If b < 0 then |b| = −b. Moreover, ab ≤ 0. In this case, |ab| = −ab =a(−b) = |a| · |b|.Suppose now that a < 0. Then |a| = −a.• If b ≥ 0 then |b| = b. Moreover, ab ≤ 0. In this case, |ab| = −ab = (−a)b =|a| · |b|.• If b < 0 then |b| = −b. Moreover, ab > 0. In this case, |ab| = ab = |a| · |b|


∣∣ 1a

∣∣ = 1|a| , where a 6= 0.

2

Solution.If a > 0 then 1

a> 0. Thus,

∣∣ 1a

∣∣ = 1a

= 1|a| . If a < 0 then 1

a< 0. Thus,∣∣ 1

a

∣∣ = − 1a

= 1−a

= 1|a|


∣∣ab

∣∣ = |a||b| where b 6= 0.

Solution.Using both Exercise 1.7 and Exercise 1.8 we can write the following

∣∣ab

∣∣ =∣∣a · 1b

∣∣ = |a| ·∣∣1

b

∣∣ = |a| · 1|b| = |a|

|b|

Exercise 1.10Show that for any two real numbers a and b we have ab ≤ |a| · |b|.

Solution.From Exercise 1.1, we have ab ≤ |ab| = |a| · |b|, where we used Exercise 1.7

Exercise 1.11Recall that a number b ≥ 0 is the square root of a number a, written√

a = b, if and only if a = b2. Show that√

a2 = |a|.

Solution.Since (±a)2 = a2 we can write

√a2 = −a if a < 0 or

√a2 = a if a ≥ 0. But

this is equivalent to writing√

a2 = |a| by Exercise 1.2

Exercise 1.12suppose that A and B are points on a coordinate line that have coordinatesa and b, respectively. Show that |a− b| is the distance between the points Aand B. Thus, if b = 0, |a| measures the distance from the number a to theorigin.

Solution.Let d be the distance between A and B. If a b (See Figure 2(b)) thend = a− b and |a− b| = a− b = d

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Figure 2

Exercise 1.13Graph the portion of the real line given by the inequality(a) |x− a| < δ(b) 0 < |x− a| < δwhere δ > 0. Represent each graph in interval notation.

Solution.(a) The graph is given in Figure 3(a). In interval notation, we have (a −δ, a + δ).(b) The graph is given in Figure 3(b). In interval notation, we have (a −δ, a) ∪ (δ, a + δ)

Figure 3

Exercise 1.14Show that |x− a| < k if and only if a− k < x < a + k, where k > 0.

Solution.From the previous exercise, we can write |x− a| < k ⇔ a− k < x < a + k

Exercise 1.15Show that |x− a| > k if and only if x < a− k or x > a + k, where k ≥ 0.

Solution.Using Figure 4, we see that |x− a| > k ⇔ x− a < −k or x− a > k. This isequivalent to x < a− k or x > a + k

Figure 4

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Exercise 1.16Solve each of the following inequalities: (a) |2x− 3| < 5 and (b) |x + 4| > 2.

Solution.(a) Using Exercise 1.14 we can write |2x − 3| < 5 ⇔ −5 < 2x − 3 < 5 ⇔−5 + 3 < 2x < 5 + 3 ⇔ −1 < x < 4.(b) Using Exercise 1.15 we can write |x + 4| > 2 ⇔ x + 4 < −2 or x + 4 > 2which is equivalent to x < −6 or x > −2

Exercise 1.17 (Triangle inequality)Use Exercise 1.1, Exercise 1.7, and the expansion of (|a + b|)2 to establishthe inequality

|a + b| ≤ |a|+ |b|,where a and b are arbitrary real numbers.

Solution.We have

(|a + b|)2 = (a + b)(a + b) = a2 + 2ab + b2 ≤ a2 + 2|a| · |b|+ b2 = (|a|+ |b|)2.

Now, the result follows by taking the square root of both sides

Exercise 1.18Show that for any real numbers a and b we have |a| − |b| ≤ |a − b|. Hint:Notice that a = (a− b) + b.

Solution.Using Exercise 1.17 we can write |a| = |(a−b)+b| ≤ |a−b|+ |b|. Subtracting|b| from both sides to obtain the desired inequaltiy

Exercise 1.19Let a ∈ R. Show that max{a, 0} = 1

2(a + |a|) and min{a, 0} = 1

2(a− |a|).

Solution.The results are clear if a = 0. If a > 0 then |a| = a and max{a, 0} = a =12(a + |a|). If a < 0 then |a| = −a max{a, 0} = 0 = 1

2(a + |a|). Likewise for

the minimum

Exercise 1.20Show that |a + b| = |a|+ |b| if and only if ab ≥ 0.

5

Solution.Suppose first that |a + b| = |a|+ |b|. Squaring both sides we find a2 + 2ab +b2 = a2 + 2|a||b| + b2 or equivalently ab = |a||b| = |ab|. But this is trueonly when ab ≥ 0. Conversely, suppose that ab ≥ 0. If a = 0 we have|0+ b| = |b| = |0|+ |b|. Likewise when b = 0. So assume that ab > 0. Supposethat a > 0 and b > 0. Then a+b > 0 and in this case |a+b| = a+b = |a|+|b|.Similar argument when a < 0 and b < 0

Exercise 1.21Suppose 0 < x < 1

2. Simplify x+3

|2x2+5x−3| .

Solution.We have

x + 3

|2x2 + 5x− 3|=

x + 3

|(2x− 1)(x + 3)|=

1

|2x− 1|=

1

1− 2x

Exercise 1.22Write the function f(x) = |x + 2| + |x − 4| as a piecewise defined function.Sketch its graph.

Solution.We have

|x + 2| ={

x + 2 if x ≥ −2−(x + 2) if x < −2

Likewise,

|x− 4| ={

x− 4 if x ≥ 4−(x− 4) if x < 4

Combining we find

f(x) =

−2x + 2 if x < −2

6 if −2 ≤ x < 42x− if x ≥ 4

The graph is given below

6

Exercise 1.23Prove that ||a| − |b|| ≤ |a− b| for any real numbers a and b.

Solution.We have |b| − |a| ≤ |b− a| = |a− b| so that −|a− b| ≤ |a| − |b| ≤ |a− b byExercise 1.18. Now using Exercise 1.14, the result follows

Exercise 1.24Solve the equation 4|x− 3|2 − 3|x− 3| = 1.

Solution.Let u = |x− 3|. Then 4u2 − 3u− 1 = 0 implies u = 1 or u = −1

4. Since u ≥ 0

we must have |x− 3| = u = 1. Hence, x− 3 = −1 or x− 3 = 1. Thus, x = 2or x = 4

Exercise 1.25What is the range of the function f(x) = |x|

xfor all x 6= 0?

Solution.If x > 0 then |x|

x= x

x= 1. If x < 0 then |x||

x= −x

x= −1. Thus, the range is

{−1, 1}

Exercise 1.26Solve 3 ≤ |x− 2| ≤ 7. Write your answer in interval notation.

Solution.Solving the inequality |x−2| ≤ 7 we find −5 ≤ x ≤ 9. Solving the inequality|x − 2| ≥ 3 we find x ≤ −1 or x ≥ 5. Thus, the common intervals are[−5,−1] ∪ [5, 9]

7

Exercise 1.27Simplify

√x2

|x| .

Solution.The answer is

√x2

|x| = |x||x| = 1

Exercise 1.28Solve the inequality

∣∣x+1x−2

∣∣ < 3. Write your answer in interval notation.

Solution.We have −3 < x+1

x−2< 3. The inequality x+1

x−2< 3 implies −2x+7

x−2< 0. Solving

this inequality we find x < 2 or x > 72. Likewise, The inequality x+1

x−2> −3

implies 4x−5x−2

> 0. Solving this inequality we find x < 54

or x > 2. Hence, the

common interval is (−∞, 54) ∪ (7

2,∞)

Exercise 1.29Suppose x and y are real numbers such that |x− y| < |x|. Show that xy > 0.

Solution.Since |x − y| < |x| we have −|x| < x − y < |x|. Multiplying through by −1and adding x we obtain x− |x| < y < x + |x|. If x = 0 then 0 < y < 0 whichis impossible. Therefore either x > 0 or x < 0. If x > 0 then 0 < y < 2x.Hence xy > 0. If x < 0 then 2x < y < 0 and so xy > 0

8

Exercise 2.1Prove that A is bounded if and only if there is a positive constant C suchthat |x| ≤ C for all x ∈ A.

Solution.Suppose that A is bounded. Then there exist real numbers m and M suchthat m ≤ x ≤ M for all x ∈ A. Let C = {|m|, |M | + 1} > 0. Then −C ≤−|m| ≤ m ≤ x ≤ M ≤ |M |+ 1 ≤ C. That is, |x| ≤ C for all x ∈ A.Conversely, suppose that |x| ≤ C for all x ∈ A and for some C > 0. Then, by Exercise 1.4, we have −C ≤ x ≤ C for all x ∈ A. Let m = −C andM = C

Exercise 2.2Let A = [0, 1].(a) Find an upper bound of A. How many upper bounds are there?(b) Find a lower bound of A. How many lower bounds are there?

Solution.(a) Any number greater than 1 is an upper bound.(b) Any number less than 0 is a lower bound

Exercise 2.3Consider the set A = { 1

n: n ∈ N}.

(a) Show that A is bounded from above. Find the supremum. Is this supre-mum a maximum of A?(b) Show that A is bounded from below. Find the infimum. Is this infimuma minimum of A?

Solution.(a) The supremum is 1 which is also the maximum of A.(b) The infimum is 0 which is not a minimum of A

Exercise 2.4Consider the set A = {1− 1

n: n ∈ N}.

(a) Show that 1 is an upper bound of A.(b) Suppose L < 1 is another upper bound of A. Let n be a positive integersuch that n > 1

1−L. Such a number n exist by the Archimedian property

which we will discuss below. Show that this leads to a contradiction. Thus,L ≥ 1. This shows that 1 is the least upper bound of A and hence sup A = 1.

9

Solution.(a) Since 1

n> 0 we have 1− 1

n< 1 for all n ∈ N. Thus, 1 is an upper bound

of A.(b) Since n > 1

1−L, we find L < 1− 1

n. But this contradicts the fact that L is

an upper bound of A. Hence, we must have 1 < L. This shows, that 1 is thesmallest upper bound and so is the supremum of A

Exercise 2.5Let a, b ∈ R with a > 0.(a) Suppose that na ≤ b for all n ∈ N. Show that the set A = {na : n ∈ N}has a supremum. Call it c.(b) Show that na ≤ c− a for all n ∈ N. That is, c− a is an upper bound ofA. Hint: n + 1 ∈ N for all n ∈ N.(c) Conclude from (b) that there must be a positive integer n such thatna > b.

Solution.(a) Since na ≤ b for all n ∈ N, the set A is bounded from above. By thecompleteness axiom of R, A has a supremum, denote it by c = sup{A}.(b) Let n ∈ N. Then n + 1 ∈ N so that (n + 1)a ≤ c or na ≤ c− a.(c) From (b), we have that c − a is an upper bound of A. By the definitionof supremum, we must have c ≤ c− a which is impossible. This shows thatA cannot have an upper bound. That is, there must be a positive integer nsuch that na > b

Exercise 2.6Let a and b be two real numbers such that a < b.(a) Let [a] denote the greatest integer less than or equal to a. Show that[a]− 1 < a < [a] + 1.(b) Let n be a positive integer such that n > 1

b−a. Show that na + 1 < nb.

(c) Let m = [na] + 1. Show that na < m < nb. Thus, a < mn

< b. We seethat between any two distinct real numbers there is a rational number.

Solution.(a) Since |[a] − a| < 1, we have −1 < [a] − a < 1 which is equivalent to[a]− 1 < a < [a] + 1.(b) Since n > 1

b−aand b− a > 0 we can have n(b− a) > 1 or na + 1 < nb.

(c) We have m − 1 = [na] ≤ na < [na] + 1 = m < na + 1 < nb. Thus,na < m < nb. Dividing through by n > 0 we obtain a < m

n< b

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Exercise 2.7Consider the set A = { (−1)n

n: n ∈ N}.

(a) Show that A is bounded from above. Find the supremum. Is this supre-mum a maximum of A?(b) Show that A is bounded from below. Find the infimum. Is this infimuma minimum of A?

Solution.(a) The supremum is 1

2which is also a maximum.

(b) The infimum is −1 which is also a minimum

Exercise 2.8Consider the set A = {x ∈ R : 1 < x < 2}.(a) Show that A is bounded from above. Find the supremum. Is this supre-mum a maximum of A?(b) Show that A is bounded from below. Find the infimum. Is this infimuma minimum of A?

Solution.(a) 2 is a supremum that is not a maximum.(b) 1 is an infimum that is not a minimum

Exercise 2.9Consider the set A = {x ∈ R : x2 > 4}.(a) Show x ∈ A and x < 2 leads to a contradiction. Hence, we must havethat x ≥ 2 for all x ∈ A. That is, 2 is a lower bound of A.(b) Let L be a lower bound of A such that L > 2. Let y = L+2

2. Show that

2 < y < L.(c) Use (a) to show that y ∈ A and L ≤ y. Show that this leads to acontradiction. Hence, we must have L ≤ 2 which means that 2 is the infimumof A.

Solution.(a) If x ∈ A and x < 2 then x2 < 4 which contradicts the fact that x ∈ A.Thus, for all x ∈ A we have x ≥ 2. This shows that 2 is a lower bound of A.(b) Since L > 2 we have L + 2 > 4 and this implies y = L+2

2> 2. Also,

y = L+22

< L+L2

= L.(c) Since y > 2 we have y2 > 4 so that y ∈ A. But L is a lower bound of Aso we must have L ≤ y. But this contradicts y < L from (b). It follows that2 is the least lower bound of A

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Exercise 2.10Show that for any real number x there is a positive integer n such that n > x.

Solution.Let a = 1 and b = x in the Archimedean property

Exercise 2.11Let a and b be any two real numbers such that a < b.(a) Let w be a fixed positive irrational number. Show that there is a rationalnumber r such that a < wr < b.(e) Show that wr is irrational. Hence, between any two distinct real numbersthere is an irrational number.

Solution.(a) Since a < b, we have a

w< b

w. By Exercise ??, there is a rational number

r such that aw

< r < bw

or a < rw < b.(e) If rw = s with s rational then w = s

rwhich is a rational, a contradiction.

Hence, rw is irrational

Exercise 2.12Suppose that α = sup A < ∞. Let ε > 0 be given. Prove that there is anx ∈ A such that α− ε < x.

Solution.Suppose the contrary. That is, α− ε ≥ x for all x ∈ A. In this case, α− ε isan upper bound of A. Thus, we must have α ≤ α− ε which is impossible

Exercise 2.13Suppose that β = inf A < ∞. Let ε > 0 be given. Prove that there is anx ∈ A such that β + ε > x.

Solution.Suppose the contrary. That is, β + ε ≤ x for all x ∈ A. In this case, β + ε isa lower bound of A. Thus, we must have β + ε ≤ β which is impossible

Exercise 2.14For each of the following sets S find sup{S} and inf{S} if they exist.(a) S = {x ∈ R : x2 < 5}.(b) S = {x ∈ R : x2 > 7}.(c) S = {− 1

n: n ∈ N}.

12

Solution.(a) sup{S} =

√5 and inf{S} = −

√5.

(b) sup{S} = ∞ and inf{S} = −∞.(c) sup{S} = 0 and inf{S} = −1

13

Exercise 3.1Find a simple expression for the general term of each sequence.(a) 1,−1

2, 1

3,−1

4, · · ·

(b) 2, 32, 4

3, 5

4, · · ·

(c) 1, 13, 1

5, 1

7, 1

9, · · ·

(d) −1, 1,−1, 1,−1, 1, · · ·

Solution.(a) an = (−1)n−1 1

n.

(b) an = n+1n

.(c) an = 1

2n−1

(d) an = (−1)n

Exercise 3.2Show that the sequence

{1n

}∞n=1

converges to 0.

Solution.Let ε > 0 be given. We want to find a positive integer Nε such that if n ≥ Nε

then∣∣ 1n− 0∣∣ < ε. But this last inequality implies that n > 1

ε. Let Nε be a

positive integer greater than 1ε. If n ≥ Nε > 1

εthen

∣∣ 1n− 0∣∣ = 1

n< ε. This

shows that

limn→∞

1

n= 0

Exercise 3.3Show that the sequence

{1 + C

n

}∞n=1

converges to 1, where C 6= 0 is a con-stant.

Solution.Let ε > 0. We want to find a positive integer Nε such that

∣∣1 + Cn− 1∣∣ < ε

for all n ≥ Nε. But∣∣1 + C

n− 1∣∣ < ε implies n > |C|

ε. By choosing Nε to be a

positive integer greater than |C|ε

the result follows

Exercise 3.4Is there a number L with the property that |(−1)n − L| < 1 for all n ≥ N1,where N1 is some positive integer? Hint: Consider the inequality with aneven integer greater than N1 and an odd integer greater than N1.

14

Solution.If ne ≥ N1 is an even integer than |(−1)ne−L| = |1−L| < 1. If no ≥ N1 is anodd integer than |(−1)no − L| = | − 1− L| < 1. This shows that L is withinone unit of both −1 and 1 which is impossible. Thus, L does not exist

Exercise 3.5Use the previous exercise to show that the sequence {(−1)n}∞n=1 is divergent.

Solution.Assume the contrary. That is, suppose there is an L such that limn→∞(−1)n =L. Let ε = 1. Then, there is a positive integer N1 such that n ≥ N1 implies|(−1)n −L| < 1. But by the previous exercise, this is impossible. Hence, thegiven sequence is divergent

Exercise 3.6Suppose that limn→∞ an = a and limn→∞ an = b with a < b. Show that bychoosing ε = b−a

2> 0 we end up with the impossible inequality b−a < b−a.

A similar result holds if b < a. Thus, we must have a = b. Hint: Exercise 1.6and Exercise 1.17.

Solution.Let ε = b−a

2> 0. Since the sequence converges to a, we can find a positive

integer N1 such that

n ≥ N1 =⇒ |an − a| < b− a

2.

Similarly, since the sequence converges to b we can find a positive integer N2

such that

n ≥ N2 =⇒ |an − b| < b− a

2.

Let N = max{N1, N2}. Then for n ≥ N we have n ≥ N1 and n ≥ N2.Moreover, by using Exercise 1.6 and Exercise 1.17 we have

b− a =|b− a| = |(b− an) + (an − a)|≤|b− an|+ |an − a|

=|an − b|+ |an − a| < b− a

2+

b− a

2= b− a

Thus, we conclude that b− a < b− a which is impossible. Likewise, if b < awe end up with a − b < a − b which is impossible. That is, either a < b orb < a leads to a contradiction. Hence, a = b

15

Exercise 3.7Show that each of the following sequences is bounded. Identify M in eachcase.(a) an = (−1)n.(b) an = 1√

n ln (n+1).

Solution.(a) We have |an| = |(−1)n| = 1 ≤ 1 so that M = 1.(b) n ≥ 1 ⇒

√n ≥ 1 ⇒ 1√

n≤ 1. Also, ln (n + 1) ≥ ln 2 ⇒ 1

ln (n+1)≤ 1

ln 2.

Hence, |an| ≤ 1ln 2

= M

Exercise 3.8Let {an}∞n=1 be a sequence such that |an| ≤ K for all n ≥ N. Show that thissequence is bounded. Identify your M.

Solution.Let M = |a1| + |a2| + · · · + |aN−1| + K. Then |an| ≤ M for all n ≥ 1. Thatis, the sequence is bounded

Exercise 3.9Show that a convergent sequence is bounded. Hint: use the definition ofconvergence with ε = 1.

Solution.Let {an}∞n=1 be a convergent sequence with limit L. Let ε = 1. There is apositive integer N1 such that |an − L| < 1 for all n ≥ N1. By Exercise 1.18,we obtain |an| − |L| < 1 or |an| < 1 + |L| for all n ≥ N1. By the previousproblem with K = 1 + |L|, the sequence is bounded

Exercise 3.10Give an example of a bounded sequence that is divergent.

Solution.Let an = (−1)n. We know that this sequence is bounded (Exercise 3.7(a)).We also know that this sequence is divergent (Exercise )

Exercise 3.11Let {an}∞n=1 , {bn}∞n=1 , {cn}∞n=1 be three sequences with the following condi-tions:

16

(1) bn ≤ an ≤ cn for all n ≥ K, where K is some positive integer.(2) limn→∞ bn = limn→∞ cn = L.Show that limn→∞ an = L. Hint: Use the definition of convergence alongwith Exercise 1.14.

Solution.Let ε > 0. By hypothesis, there exist positive integers N1 and N2 such that

|bn − L| < ε for all n ≥ N1 or equivalently L− ε < bn < L + ε for all n ≥ N1

and

|cn −L| < ε for all n ≥ N2 or equivalently L− ε < cn < L + ε for all n ≥ N2.

Let N = N1 + N2 + K. Suppose n ≥ N. Then L− ε < bn ≤ an ≤ cn < L + ε.That is L− ε < an < L+ ε for all n ≥ N. This implies that limn→∞ an = L

Exercise 3.12An expansion of (a+b)n, where n is a positive integer is given by the Binomialformula

(a + b)n =n∑

k=0

C(n, k)akbn−k

where C(n, k) = n!k!(n−k)!

.

(a) Use the Binomial formula to establish the inequality

(1 + x)1n ≤ 1 +

x

n, x ≥ 0

(b) Show that if a ≥ 1 then limn→∞ a1n = 1. Hint: Use Exercise 3.3.

Solution.(a) Using the Binomial formula with a = 1 and b = x

nwe find(

1 + xn

)n= 1 + nx

n+(other positive terms) ≥ 1 + x.

or1 + x ≤

(1 +

x

n

)n

.

Taking the nth root of both sides we find

(1 + x)1n ≤ 1 +

x

n.

17

(b) If a ≥ 1 then a1n ≥ 1. By letting x = a− 1 ≥ 0 in (a) we find

1 ≤ a1n ≤ 1 +

a− 1

n.

By Exercise 3.3 we know that limn→∞ 1+ a−1n

= 1. Thus, by the squeeze rulewe obtain

limn→∞

a1n = 1, a ≥ 1

Exercise 3.13Prove that the sequence {cos (nπ)}∞n=1 is divergent.

Solution.Note that {cos (nπ)}∞n=1 = {(−1)n}∞n=1 and by Exercise , this sequence isdivergent

Exercise 3.14Let {an}∞n=1 be the sequence defined by an = n for all n ∈ N. Explain whythe sequence {an}∞n=1 does not converge to any limit.

Solution.The sequence is unbounded

Exercise 3.15(a) Show that for all n ∈ N we have

n!

nn≤ 1

n.

(b) Show that the sequence {an}∞n=1 where an = n!nn is convergent and find

its limit.

Solution.(a) We know that n−i

n≤ 1 for all 0 ≤ i ≤ n− 1. Thus, n!

nn = n(n−1)(n−2)···2·1n·n···n =

nn· n−1

n· n−2

n· · · 2

n· 1

n≤ 1

n.

(b) By the Squeeze rule we find that limn→∞n!nn = 0

Exercise 3.16Using only the definition of convergence show that

limn→∞

3√

n− 50013√

n− 1001= 1.

18

Solution.Let ε > 0. We want to find a positive integer N such that if n ≥ N then∣∣∣∣ 3

√n− 5001

3√

n− 1001− 1

∣∣∣∣ < ε

or ∣∣∣∣ −40013√

n− 1001

∣∣∣∣ < ε.

Let n > 10013. Then 3√

n−1001 > 0 so that the previous inequality becomes

40013√

n− 1001< ε.

Solving this for n we find

n >

(4001

ε+ 1001

)3

.

Let N be a positive integer greater than(

4001ε

+ 1001)3

. Then for n ≥ N wehave ∣∣∣∣ 3

√n− 5001

3√

n− 1001− 1

∣∣∣∣ < ε

Exercise 3.17Consider the sequence defined recursively by a1 = 1 and an+1 =

√2 + an for

all n ∈ N. Show that an ≤ 2 for all n ∈ N.

Solution.The proof is by induction on n. For n = 1 we have a1 = 1 ≤ 2. Suppose thatan ≤ 2. Then an+1 =

√2 + an ≤

√2 + 2 = 2

Exercise 3.18Calculate limn→∞

(n2+1) cos nn3 .

Solution.We have

−n2 + 1

n3≤ (n2 + 1) cos n

n3≤ n2 + 1

n3.

By the Squeeze rule we conclude that the limit is 0

19

Exercise 3.19Calculate limn→∞

2(−1)n+3√

n.

Solution.We have

− 2√n≤ 2(−1)n+3

√n

≤ 2√n

.

By the Squeeze rule the limit is 0

Exercise 3.20Suppose that limn→∞ an = L with L > 0. Show that there is a positiveinteger N such that 2xN > x.

Solution.Let ε = L

2. Then there is a positive integer N such that if n ≥ N we have

|an − L| < L2. Thus, |aN − L| < L

2or −L

2< aN − L. Hence, aN > L

2or

2aN > L

Exercise 3.21Let a ∈ R and n ∈ N. Clearly, a < a + 1

n.

(a) Show that there is a1 ∈ Q such that a < a1 < a + 1n. Hint: Exercise

2.6(c).(b) Show that there is a2 ∈ Q such that a < a2 < a1.(c) Continuiung the above process we can find a sequence {an}∞n=1 such thata < an < a + 1

nfor all n ∈ N. Show that this sequence converges to a.

We have proved that if a is a real number then there is a sequence of rationalnumbers converging a. We say that the set Q is dense in R.

Solution.(a) This follows from Exercise 2.6(c).(b) Similar to (a).(c) Applying the Squeeze rule, we obtain limn→∞ an = a

20

Exercise 4.1Suppose that limn→∞ an = A and limn→∞ bn = B. Show that

limn→∞

an ± bn = A±B.

Solution.We will prove the result for addition. The difference case is similar. Letε > 0. Since the two sequences are convergent, there exist positive integersN1 and N2 such that

|an − A| < ε2

for all n ≥ N1

and

|bn −B| < ε2

for all n ≥ N2.

Let N = N1 + N2. Then for all n ≥ N we have n ≥ N1 and n ≥ N2. Hence,

|(an +bn)−(A+B)| = |(an−A)+(bn−B)| ≤ |an−A|+ |bn−B| < ε2+ ε

2= ε.

This establishes the desired result

Exercise 4.2Suppose that limn→∞ an = A and limn→∞ bn = B.(a) Show that |bn| ≤ M for all n ∈ N, where M is a positive constant.(b) Show that anbn − AB = (an − A)bn + A(bn −B).(c) Let ε > 0 be arbitrary and K = M +|A|. Show that there exists a positiveinteger N1 such that |an − A| < ε

2Kfor all n ≥ N1.

(d) Let ε > 0 and K be as in (c). Show that there exists a positive integerN2 such that |bn −B| < ε

2Kfor all n ≥ N2.

(e) Show that limn→∞ anbn = AB.

Solution.(a) Since {bn}∞n=1 is convergent, the sequence is bounded. Thus, there is apositive constant M such that |bn| ≤ M for all n ≥ 1.(b) We have (an−A)bn +A(bn−B) = anbn−Abn +Abn−AB = anbn−AB.(c) Let ε1 = ε

2K. Since limn→∞ an = A, we can find a positive integer N1 such

that |an − A| < ε1 = ε2K

for all n ≥ N1.(d) Let ε2 = ε

2K. Since limn→∞ bn = B, we can find a positive integer N2 such

21

that |bn −B| < ε2 = ε2K

for all n ≥ N2.(e) Let N = N1 + N2. Then n ≥ N implies n ≥ N1 and n ≥ N2. In this case,

|anbn − AB| =|(an − A)bn + A(bn −B)| ≤ |an − A||bn|+ |A||bn −B|

<ε

2K·M +

ε

2K· |A|

<ε

2+

ε

2= ε

where we have used the fact that MK

= MM+|A| < 1 and |A|

K= |A|

M+|A| < 1

Exercise 4.3Give an example of two divergent sequences {an}∞n=1 and {bn}∞n=1 such that{anbn} and {an + bn} are convergent.

Solution.Let an = (−1)n and bn = (−1)n+1. Both sequences are divergent. Moreover,anbn = (−1)2n+1 = −1 for all n ≥ 1. Hence, limn→∞ anbn = −1. Finally,an + bn = (−1)n− (−1)n = 0 for all n ≥ 1. Therefore, limn→∞(an + bn) = 0

Exercise 4.4Let k 6= 0 be an arbitrary constant and limn→∞ an = A. Show that limn→∞ kan =kA.

Solution.Let ε > 0 be arbitrary. There is a positive integer N such that |an−A| < ε

|k|for all n ≥ N. Moreover, for n ≥ N we have

|kan − kA| = |k(an − A)| = |k||an − A| < |k| · ε

|k|= ε.

This establishes the desired result

Exercise 4.5Suppose that limn→∞ an = 0 and {bn}∞n=1 is bounded. Show that limn→∞ anbn =0.

Solution.Since {bn}∞n=1 is bounded, we can find a positive constant M such that |bn| ≤M for all n ≥ 1. Let ε > 0 be arbitrary. Since limn→∞ an = 0, there is a

22

positive integer N such that |an − 0| = |an| < εM

for all n ≥ N. Thus, forn ≥ N, we have

|anbn − 0| = |anbn| = |an||bn| <ε

M·M = ε.

This shows that limn→∞ anbn = 0

Exercise 4.6(a) Use the previous exercise to show that limn→∞

sin nn

= 0.(b) Show that limn→∞

sin nn

= 0 using the squeeze rule.

Solution.(a) Let an = 1

nand bn = sin n. Then limn→∞ an = limn→∞

1n

= 0 and|bn| = | sin n| ≤ 1. Now the result follows from the previous exercise.(b) Since −1 ≤ sin n ≤ 1, we obtain − 1

n≤ sin n

n≤ 1

n. But limn→∞

1n

=limn→∞− 1

n= 0 so that by the squeeze rule

limn→∞

sin n

n= 0

Exercise 4.7Suppose that limn→∞ an = A, with A 6= 0. Show that there is a positiveinteger N such that |an| > |A|

2for all n ≥ N. Hint: Use Exercise 1.18.

Solution.Let ε = |A|

2> 0. Then there is a positive integer N such that |an − A| < |A|

2

for all n ≥ N. By Exercise 1.18, we have |A| − |an| < |A|2

or |an| > |A|2

forn ≥ N

Exercise 4.8Let {an}∞n=1 be a sequence with the following conditions:(1) an 6= 0 for all n ≥ 1.(2) limn→∞ an = A, with A 6= 0.(a) Show that there is a positive integer N1 such that for all n ≥ N1 we have∣∣∣∣ 1

an

− 1

A

∣∣∣∣ < 2

|A|2|an − A|.

(b) Let ε > 0 be arbitrary. Show that there is a positive integer N2 such thatfor all n ≥ N2 we have

|an − A| < |A|2

2ε.

23

(c) Using (a) and (b), show that

limn→∞

1

an

=1

A.

Solution.(a) From the previous exercise, we can find a positive integer N1 such that

|an| > |A|2

for n ≥ N1. Thus, for n ≥ N1 we obtain∣∣∣∣ 1

an

− 1

A

∣∣∣∣ =|an − A||an||A|

<2

|A|2|an − A|.

(b) Let ε > 0 be arbitrary. Since limn→∞ an = A, we can find a positiveinteger N2 such that for all n ≥ N2 we have

|an − A| < |A|2

2ε.

(c) Let ε > 0 be arbitrary. Let N = N1 + N2. Then n ≥ N implies thatn ≥ N1 and n ≥ N2. Moreover,∣∣∣∣ 1

an

− 1

A

∣∣∣∣ =|an − A||an||A|

<2

|A|2|an − A| < 2

|A|2|A|2

2ε = ε.

This establishes the required result

Exercise 4.9Let 0 < a < 1. Show that limn→∞ a

1n = 1. Hint: Use Exercise 3.12 (b).

Solution.Since 0 < a < 1, we have 1

a> 1. By Exercise 3.12(b), we find

limn→∞

(1

a

) 1n

= limn→∞

1

a1n

= 1.

By the previous exercise, we can write

limn→∞

a1n = lim

n→∞

11

a1n

=1

1= 1

24

Exercise 4.10Show that if limn→∞ an = A and limn→∞ bn = B with bn 6= 0 for all n ≥ 1and B 6= 0, then

limn→∞

an

bn

=A

B.

Solution.Using Exercise ?? and Exercise ?? we can write

limn→∞

an

bn

= limn→∞

an ·1

bn

= A · 1

B=

A

B

Exercise 4.11Given that limn→∞ an = A and limn→∞ bn = B with an ≤ bn for all n ≥ 1.(a) Suppose that B < A. Let ε = A−B

2> 0. Show that there exist positive

integers N1 and N2 such that A − ε < an < A + ε for n ≥ N1 and B − ε <bn < B + ε for n ≥ N2.(b) Let N = N1 + N2. Show that for n ≥ N we obtain the contradictionbn < an. Thus, we must have A ≤ B.

Solution.(a) Since limn→∞ an = A, there exists a positive integer N1 such that |an −A| < ε for all n ≥ N1. By Exercise 1.14, this is equivalent to A−ε < an < A+εfor n ≥ N1. Similarly, since limn→∞ bn = B, there exists a positive integerN2 such that |bn−B| < ε for all n ≥ N2. By Exercise 1.14, this is equivalentto B − ε < bn < B + ε for n ≥ N2.(b) If n ≥ N then n ≥ N1 and n ≥ N2. Thus,

bn <B + ε = B +A−B

2=

A + B

2

=A− A−B

2= A− ε < an

This contradicts the fact that an ≤ bn for all n ≥ 1. Hence, we conclude thatA ≤ B

Exercise 4.12Suppose that limn→∞

an

bn= L and limn→∞ bn = 0 where bn 6= 0 for all n ∈ N.

Find limn→∞ an.

25

Solution.We have

limn→∞

an = limn→∞

an

bn

· bn =

(lim

n→∞

an

bn

)(lim

n→∞bn

)= L · 0 = 0

Exercise 4.13The Fibonacci numbers are defined recursively as follows:

a1 = a2 = 1 and an+2 = an+1 + an for all n ∈ N.

Suppose that limn→∞an+1

an= L. Find the value of L.

Solution.We have

an+2 = an+1 + an

Divide through by an+1 to obtain

an+2

an+1

= 1 +an

an+1

.

Take the limit as n →∞ to obtain

L = 1 +1

Lor

L2 − L− 1 = 0.

Solving this quadratic equation for L > 0 we find L = 1+√

52

Exercise 4.14Show that the sequence defined by

an =n

n + 1+ (−1)n n2 + 3

n2 + 7

have two limits by finding limn→∞ a2n and limn→∞ a2n+1.

Solution. We have

limn→∞

a2n = limn→∞

2n

2n + 1+

4n2 + 3

4n2 + 7= 2

and

limn→∞

a2n+1 = limn→∞

2n + 1

2n + 2− (2n + 1)2 + 3

(2n + 1)2 + 7= 0

26

Exercise 4.15Use the properties of this section to find

limn→∞

√2n2 + 5n

n + 4.

Solution.We have

limn→∞

√2n2 + 5n

n + 4= lim

n→∞

√2 + 5

n

1 + 4n

= limn→∞

limn→∞

√2 + 5

n

limn→∞(1 + 4n)

= limn→∞

√limn→∞(2 + 5

n)

1 + limn→∞4n

=√

2

Exercise 4.16Find the limit of the sequence defined by

an = n1

2 ln n .

Solution.We have limn→∞ ln an = limn→∞

12 ln n

· ln n = 12. Thus, limn→∞ an = e

12

Exercise 4.17Consider the sequence defined by

an =1√1

+1√2

+ · · ·+ 1√n

.

(a) Show that an ≥√

n for all n ∈ N.(b) Show that the sequence {an}∞n=1 is divergent. Hint: Exercise 4.11.

Solution.(a) By induction on n. If n = 1 we have a1 = 1√

1=√

1. Suppose that

an ≥√

n. Then an+1 ≥√

n+ 1√n+1

=√

n√

n+1+1√n+1

≥√

n√

n+1√n+1

= n+1√n+1

=√

n + 1.

(b) Since an ≥√

n for all n ∈ N and limn→∞√

n = ∞, we conclude thatlimn→∞ an = ∞

27

an = ln (2n +√

n)− ln n.

Solution.We have limn→∞[ln (2n +

√n)−ln n] = limn→∞ ln

(2n+

√n

n

)= limn→∞ ln

(2 + 1√

n

)=

ln 2

Exercise 4.19Consider the sequence defined by an = n

√3n + 1.

(a) Show that 3 < an < 3 n√

2 for all n ≥ 2.(b) Find the limit of an as n →∞.

Solution.(a) We have

3 =n√

3n < n√

3n + 1 < n√

3n + 3n < 3n√

2.

(b) Since limn→∞n√

2 = 1 we conclude by the Squeeze rule that limn→∞ an =3

Exercise 4.20Let {an}∞n=1 be a convergent sequence of nonnegative terms with limit L.Suppose that the terms of sequence satisfy the recursive relation anan+1 =an + 2 for all N ∈ N. Find L.

Solution.We have limn→∞ anan+1 = limn→∞(an + 2) → limn→∞ an · limn→∞ an+1 =limn→∞ an + limn→∞ 2 → L · L = L + 2 → L2 − L − 2 = 0. Solving thisequation we find L = −1 and L = 2. But an ≥ 0 for all n ∈ N so that L ≥ 0.Thus, L = 2


an = cos1

n+

sin n

n.

Solution.We have

limn→∞

an = limn→∞

cos 1n + limn→∞

sin n

n= 1 + 0 = 1

28

Exercise 4.22Suppose that an+1 = a2

n+1an

. Show that the sequence {an}∞n=1 is divergent.

Solution.Suppose the contrary and let L = limn→∞ an. Then L = L2+1

Lor L2 = L2 +1

which leads to the contradiction 0 = 1. Hence, the given sequence must bedivergent

29


Exercise 5.1Show that the sequence { 1

n}∞n=1 is decreasing.

Solution.We have n + 1 ≥ n ⇒ 1

n+1≤ 1

n⇒ an ≥ an+1 for all n ≥ 1. This shows that

the sequence is decreasing


1+e−n}∞n=1 is increasing.

Solution.We have n ≤ n+1 ⇒ en ≤ en+1 ⇒ e−(n+1) ≤ e−n ⇒ 1+ e−(n+1) ≤ 1+ e−n ⇒

11+e−n ≤ 1

1+e−(n+1) ⇒ an ≤ an+1 for all n ≥ 1. This shows that the sequenceis increasing


n}∞n=1 is bounded from below. What is a lower

bound? Are there more than one lower bound?

Solution.Since 1

n≥ 0 for all n ≥ 1, the given sequence is bounded from below with a

lower bound 0. Any negative number is a lower bound


1+e−n}∞n=1 is bounded from above. What is an upperbound? Are there more than one upper bound?

Solution.We have 1 + e−n ≥ 1 ⇒ 1

1+e−n ≤ 1. Thus, the given sequence is boundedfrom above with an upper bound equals to 1. Any number larger than 1 isan upper bound

Exercise 5.5Let {an}∞n=1 be an increasing sequence that is bounded from above.(a) Show that there is a finite number M such that M = sup{an : n ≥ 1}.(b) Let ε > 0 be arbitrary. Show that M − ε cannot be an upper bound ofthe sequence.

30

(c) Show that there is a positive integer N such that M − ε < aN .(d) Show that M − ε < an for all n ≥ N.(e) Show that M − ε < an < M + ε for all n ≥ N.(f) Show that limn→∞ an = M. That is, the given sequence is convergent.

Solution.(a) Since the sequence {an}∞n=1 is bounded from above, by the CompletenessAxiom there is a finite number M such that M = sup{an : n ≥ 1}. We willshow that {an}∞n=1 converges to M.(b) Let ε > 0 be arbitrary. Consider the number M − ε. This number iseither an upper bound of the sequence or not. If it is an upper bound thenwe must have M ≤ M − ε which is an impossible inequality.(c) If such an N does not exist, then we will have an ≤ M − ε for all n ≥ 1.This means that M − ε is an upper bound which is impossible by (b). Thus,there is a positive integer N such that

M − ε < aN ≤ M.

(d) Since the sequence is increasing, for all n ≥ N we can write aN ≤ an.Hence, for all n ≥ N we have M − ε < an.(e) For all n ≥ N we can write

M − ε < an ≤ M < M + ε

or equivalently |an −M | < ε for all n ≥ N.(f) This follows from the definition of convergence and (e)

Exercise 5.6Consider the sequence {an}∞n=1 defined recursively by a1 = 3

2and an+1 =

12an + 1 for n ≥ 2.

(a) Show by induction on n ≥ 2, that an+1 = an + 12n+1 .

(b) Show that this sequence is increasing.(c) Show that {an}∞n=1 is bounded from above. What is an upper bound?(d) Show that {an}∞n=1 is convergent. What is its limit? Hint: In finding thelimit, use the arithmetic operations of sequences.

Solution.(a) We have a2 = 1

2a1 + 1 = 3

4+ 1 = 7

4= 3

2+ 1

4= a1 + 1

22 . Suppose thatan = an−1+

12n . Then an+1 = 1

2an+1 = 1

2an−1+

12n+1 +1 =

(12an−1 + 1

)+ 1

2n+1 =

31

an + 12n+1 .

(b) Since an+1 − an = 12n+1 > 0, the given sequence is increasing.

(c) By (b), we have an+1 = 12an +1 > an. Solving this equality for an we find

an < 2 so that the sequence {an}∞n=1 is bounded from above with an upperbound equals to 2.(d) By the previous exercise, the sequence is convergent say to A. Using thearithmetic operations of sequences, we can write

limn→∞

an+1 = limn→∞

(1

2an + 1

)=

1

2lim

n→∞an + lim

n→∞1.

Thus,

A =1

2A + 1.

Solving this equation for A we find A = 2

Exercise 5.7Let {an}∞n=1 be a decreasing sequence such that M ≤ an for all n ≥ 1. Showthat {an}∞n=1 is convergent. Hint: Let bn = −an and use Exercise 5.5 andExercise 4.4.

Solution.Since {an}∞n=1 is decreasing, the sequence {bn}∞n=1 is increasing. Since thesequence {an}∞n=1 is bounded from below, the sequence {bn}∞n=1 is boundedfrom above. By Exercise 5.5, the sequence {bn}∞n=1 is convergent. By Exercise4.4, the sequence {an}∞n=1 is also convergent

Exercise 5.8Show that a monotone sequence is convergent if and only if it is bounded.

Solution.Suppose first that a sequence {an}∞n=1 is a monotone convergent sequence.By Exercise 3.9, the sequence {an}∞n=1 is bounded.Conversely, suppose that {an}∞n=1 is a bounded sequence. Then there is apositive constant M such that |an| ≤ M for all n ≥ 1. By Exercise 1.14,we have −M ≤ an ≤ M for all n ≥ 1. This shows that the sequence isbounded from below as well from above. If the sequence is either increasingor decreasing, then it is convergent by Exercise 5.5 and Exercise 5.7

32

Exercise 5.9Let an be defined by a1 =

√2 and an+1 =

√2 + an for n ∈ N.

(a) Show that an ≤ 2 for all n ∈ N. That is, {an}∞n=1 is bounded from above.(b) Show that an+1 ≥ an for all n ∈ N. That is, {an}∞n=1 is increasing.(c) Conclude that {an}∞n=1 is convergent. Find its limit.

Solution.(a) By induction on n. For n = 1, we have a1 =

√2 ≤ 2. Suppose that

an ≤ 2. Then an+1 =√

2 + an ≤√

2 + 2 = 2. Thus, an ≤ 2 for all n ∈ N.

(b) By induction on n. For n = 1 we have a2 − a1 =√

2 +√

2 −√

2 ≥ 0.Suppose that an − an−1 ≥ 0. Then

an+1 − an =√

2 + an −√

2 + an−1 =an − an−1√

2 + an +√

2 + an−1

≥ 0.

(c) Since the sequence is increasing and bounded from above, it is convergent,say with limit a. Thus, we have a =

√2 + a. Solving this equation we obtain

a = −1 or a = 2. Since a1 =√

2 and the sequence is increasing we concludethat a = 2

Exercise 5.10Let an =

∑nk=1

1k2 .

(a) Show that an < 2 for all n ∈ N. Hint: Recall that∑n

k=11

(n+1)n= 1− 1

n+1.

(b) Show that {an}∞n=1 is increasing.(c) Conclude that {an}∞n=1 is convergent.

Solution.(a) We have an =

∑nk=1

1k2 = 1 +

∑nk=2

1k2 ≤ 1 + 1− 1

n< 2.

(b) Since an+1 = an + 1(n+1)2

> an, the given sequence is increasing.

(c) This follows from the fact that an increasing sequence that is boundedfrom above is convergent

Exercise 5.11Consider the sequence {an}∞n=1 defined recursively as follows

a1 = 2 and 7an+1 = 2a2n + 3 for all n ∈ N.

(a) show that 12

< an < 3 for all n ∈ N.(b) Show that an+1 ≤ an for all n ∈ N.(c) Deduce that {an}∞n=1 is convergent and find its limit.

33

Solution.(a) We prove this by induction on n. If n = 1 then 1

2< 2 = a1 < 3. Suppose

that 12

< an < 3. Then 72

< 7an+1 < 21 → 12

< an+1 < 3.(b) We have an+1 − an = 1

7(2a2

n − 7an + 3) = 17(2an − 1)(an − 3) < 0. Thus,

an+1 ≤ an for all n ∈ N.(c) We deduce that the given sequence is convergent, say with limit L. Thus,7L = 2L2 + 3. Solving this equation for L we find L = 1

2and L = 3. Since

the sequence is decreasing and 12

< an < 3 we must have L = 12

Exercise 5.12Let {an}∞n=1 be an increasing sequence. Define bn = a1+a2+···+an

n. Show that

the sequence {bn}∞n=1 is increasing.

Solution.note that s1 ≤ s2 ≤ · · · ≤ sn ≤ sn+1, hence s1 +s2 + · · ·+sn ≤ nsn+1. Addingn(s1 + · · ·+ sn) to both sides, we obtain

n(s1 + · · ·+ sn) + (s1 + · · ·+ sn) ≤ n(s1 + · · ·+ sn) + nsn+1,

or, in other words,

(n + 1)(s1 + · · ·+ sn) ≤ n(s1 + · · ·+ sn + sn+1),

Dividing both sides by n(n + 1), we obtain

s1 + · · ·+ sn

n≤ s1 + · · ·+ sn + sn+1

n + 1,

or, in other words, bn ≤ bn+1 for all n. This proves our claim

Exercise 5.13Give an example of a monotone sequence that is divergent.

Solution.One example is the sequence defined by an = n

Exercise 5.14Consider the sequence defined recursively by a1 = 1 and an+1 = 3 + an

2for

all n ∈ N.(a) Show that an ≤ 6 for all n ∈ N.(b) Show that {an}∞n=1 is increasing.(c) Conclude that the sequence is convergent. Find its limit.

34

Solution.(a) If n = 1 we have a1 = 1 < 6. Suppose that an ≤ 6. Then an+1 = 3+ an

2≤

3 + 62

= 6.(b) We have an+1 = 3 + an

2= 6+an

2≥ an+an

2= an

(c) Since the sequence is increasing and bounded from above it is convergentwith limit say equals to L. Thus, L = 3+ L

2and solving for L we find L = 6

Exercise 5.15Give an example of two monotone sequences whose sum is not monotone.

Solution.Let {an}∞n=1 = {1, 1, 2, 2, 3, 3, · · · } and {bn}∞n=1 = {−1,−2,−2,−3,−3, · · · }.The first sequence is increasing and the second sequence is decreasing. How-ever the sum is the sequence {0,−1, 0,−1, 0,−1, · · · } which is not monotone

35

Exercise 6.1Let {ank

}∞k=1 be a subsequence of a sequence {an}∞n=1. Use induction on k toshow that nk ≥ k for all k ∈ N.

Solution. For k = 1 we have n1 ∈ N so that n1 ≥ 1. Suppose that nk ≥ k.Then nk+1 > nk ≥ k. Thus, nk+1 ≥ k + 1

Exercise 6.2Let {an}∞n=1 be a sequence of real numbers that converges to a number L.Let {ank

}∞k=1 be any subsequence of {an}∞n=1.(a) Let ε > 0 be given. Show that there is a positive integer N ′ such that ifn ≥ N ′ then |an − L| < ε.(b) Let N be the first positive integer such that nN ≥ N ′. Show that if k ≥ Nthen |ank

−L| < ε. That is, the subsequence {ank}∞k=1 converges to L. Hence,

every subsequence of a convergent sequence is convergent to the same limitof the original sequence.

Solution.(a) This follows from the fact that the sequence {an}∞n=1 converges to L.(b) Let N be such that nN ≥ N ′. Then if k ≥ N we have nk > nN ≥ N ′.Hence, |ank

−L| < ε. This shows that the subsequence {ank}∞k=1 converges to

L

Exercise 6.3Let {an}∞n=1 be a sequence of real numbers. Let S = {n ∈ N : an > amfor allm > n}.(a) Suppose that S is infinite. Then there is a sequence n1 < n2 < n3 < · · ·such nk ∈ S. Show that ank+1

< ank. Thus, the subsequence {ank

}∞k=1 isdecreasing.(b) Suppose that S is finite. Let n1 be the first positive integer such thatn1 6∈ S. Show that the subsequence {ank

}∞k=1 is increasing.

Solution.(a) By the definition of S we have ank+1

< anksince nk+1 > nk.

(b) Let n1 be the first positive integer such that n1 6∈ S. This means thatthere is a positive integer n2 > n1 such that an1 < an2 . But n2 6∈ S so thatthere is a positive integer n3 > n2 such that an2 < an3 . Continuing thisprocess we find an increasing subsequence {ank

}∞k=1

36

Exercise 6.4 (Bolzano-Weierstrass)Every bounded sequence has a convergent subsequence. Hint: Exercise 5.8

Solution.By the previous exercise, the sequence has a bounded monotone subsequence.By Exercise 5.8 this subsequence is convergent

Exercise 6.5Show that the sequence {esin n}∞n=1 has a convergent subsequence.

Solution.Since | sin n| ≤ 1 we must have 1

e≤ esin n ≤ e for all n ∈ N. Thus, the given

sequence is bounded so that by the Bolzano-Weierstrass theorem it has aconvergent subsequence

Exercise 6.6Prove that the sequence {an}∞n=1 where an = cos nπ

2is divergent.

Solution.Consider the two subsequences {a2n+1}∞n=1 that converges to 0 and {a4}∞n=1

that converges to 1. Thus, the original sequence must be divergent

Exercise 6.7Prove that the sequence {an}∞n=1 where

an =(n2 + 20n + 35) sin n3

n2 + n + 1

has a convergent subsequence. Hint: Show that {an}∞n=1 is bounded.

Solution.We have

|an| =∣∣∣∣(n2 + 20n + 35) sin n3

n2 + n + 1

∣∣∣∣=

(n2 + 20n + 35)| sin n3|n2 + n + 1

≤n2 + 20n + 35

n2 + n + 1

≤n2 + 20n + 35

n2= 1 +

20

n+

35

n2≤ 56

Hence, by the Bolzano Weierstraa theorem the given sequence has a conver-gent subsequence

37

Exercise 6.8Show that the sequence defined by an = 2 cos n − sin n has a convergentsubsequence.

Solution.We have −2 ≤ 2 cos n ≤ 2 and −1 ≤ − sin n ≤ 1. Adding we obtain −3 ≤an ≤ 3 so that the given sequence is bounded. By the Bolzano-Weierstrasstheorem, the given sequence has a convergent subsequence

Exercise 6.9True or false: There is a sequence that converges to 6 but contains a subse-quence converging to 0. Justify your answer.

Solution.By Exercise 6.2, this cannot happen

Exercise 6.10Give an example of an unbounded sequence with a bounded subsequence.

Solution.Let

an =

{0 if n is oddn if n is even

Then {an}∞n=1 is unbounded. However, the subsequence {a2n+1}∞n=1 = {0, 0, 0, · · · }is bounded

Exercise 6.11Show that the sequence {(−1)n}∞n=1 is divergent by using subsequences.

Solution.This sequence has two subsequences {a2n}∞n=1 that converges to 1 and {a2n+1}∞n=1

that converges to −1. Thus, the original sequence cannot be convergent

38

Exercise 7.1Consider the sequence whose n−th term is given by an = 1

n. Let ε > 0 be

arbitrary and choose N > 2ε. Show that for m,n ≥ N we have |am − an| < ε.

That is, the above sequence is a Cauchy sequence. Hint: Exercise 1.17.

Solution.Let n,m ≥ N. Then by Exercise 1.17 we have∣∣∣∣ 1n − 1

m

∣∣∣∣ ≤ 1

n+

1

m<

ε

2+

ε

2== ε

Exercise 7.2Show that any Cauchy sequence is bounded. Hint: Let ε = 1 and use Exercise1.18.

Solution.Let {an}∞n=1 be a Cauchy sequence. Let ε = 1. There is a positive integer Nsuch that whenever n,m ≥ N we have |an − am| < 1. In particular, lettingm = N we can write |an − aN | < 1 for all n ≥ N. By Exercise1.18 we canwrite |an| − |aN | < 1 for all n ≥ N or |an| < 1 + |aN | for all n ≥ N. LetM = max{|a1|, |a2|, · · · , |aN−1|, |aN |+1}. Then |an| ≤ M for all n ≥ 1. Thatis, {an}∞n=1 is bounded

Exercise 7.3Show that if limn→∞ an = A then {an}∞n=1 is a Cauchy sequence. Thus, everyconvergent sequence is a Cauchy sequence.

Solution.Let ε > 0 be arbitrary. Then there is a positive integer N such that |an−A| <ε2

for all n ≥ N. Thus, for all n, m ≥ N we have

|an − am| =|(an − A) + (A− am)| ≤ |an − A|+ |am − A|

<ε

2+

ε

2= ε.

This shows that {an}∞n=1 is a Cauchy sequence

39

Exercise 7.4(a) Using Exercise 7.2, show that for each n ≥ 1, the sequence {an, an+1, · · · }is bounded.(b) Show that for each n ≥ 1 the infimum of {an, an+1, · · · } exists. Call itbn.

Solution.(a) By Exercise 7.2, there is a positive constant M such that |an| ≤ M forall n ≥ 1. In particular, |an| ≤ M, |an+1| ≤ M, |an+2| ≤ M, · · · . That is,{an, an+1, · · · } is bounded.(b) Since {an, an+1, · · · } is bounded, it is bounded from below. By the Com-pleteness Axiom, bn = inf{am : m ≥ n} exists

Exercise 7.5(a) Show that the sequence {bn}∞n=1 is bounded from above.(b) Show that the sequence {bn}∞n=1 is increasing. Hint: Show that bn is alower bound of the sequence {an+1, an+2, · · · }.

Solution.(a) By the definition of infimum, we have bn ≤ an for all n ≥ 1. Thus,bn ≤ an ≤ |an| ≤ M. This, shows that {bn}∞n=1 is bounded from above.(b) We know that bn ≤ am for all m ≥ n. In particular bn ≤ am for allm ≥ n + 1. This shows that bn is a lower bound of {an+1, an+2, · · · }. Butbn+1 is the greatest lower bound of {an+1, an+2, · · · }. Hence, bn ≤ bn+1 andthe sequence {bn}∞n=1 is increasing

Exercise 7.6Show that the sequence {bn}∞n=1 is convergent. Call the limit B.

Solution.This follows from Exercise 5.5

Exercise 7.7(a) Let ε > 0 be arbitrary. Using the definition of Cauchy sequences andExercise ??, show that there is a positive integer N such that aN − ε

2< an <

aN + ε2

for all n ≥ N.(b) Using (a), show that aN− ε

2is a lower bound of the sequence {aN , aN+1, · · · }.

Thus, aN − ε2≤ bn for all n ≥ N.

(c) Again, using (a) show that bn ≤ aN + ε2

for all n ≥ N. Thus, combining

40

(b) and (c), we obtain aN − ε2≤ bn < aN + ε

2.

(d) Using Exercise 4.11, show that aN − ε2≤ B ≤ aN + ε

2.

(e) Using (a), (d), and Exercise 1.17, show that limn→∞ an = B. Thus, everyCauchy sequence is convergent.

Solution.(a) Let ε > 0 be arbitrary. Since {an}∞n=1 is Cauchy, there is a positive integerN such that |an − am| < ε

2for all n, m ≥ N. In particular, for all n ≥ N we

have |an − aN | < ε2

which, by Exercise 1.14, we have aN − ε2

< an < aN + ε2

for all n ≥ N.(b) From part (a), we have aN − ε

2< an for all n ≥ N. This shows, that

aN − ε2

is a lower bound of the sequence {an, an+1, · · · }∞n=N for n ≥ N. Butbn is the greatest lower bound of the sequence {an, an+1, · · · }∞n=N for n ≥ N.We conclude that aN − ε

2≤ bn for all n ≥ N.

(c) For n ≥ N we have bn ≤ an < aN + ε2.

(d) Taking the limit as n →∞ we obtain aN − ε2≤ B ≤ aN + ε

2.

(e) Using (a), (d) and Exercise 1.17, we have that for n ≥ N

|an −B| =|(an − aN) + (aN −B)|| ≤ |an − aN |+ |aN −B|

<ε

2+

ε

2= ε

Exercise 7.8(a) Show that if {an}∞n=1 is Cauchy then {a2

n}∞n=1 is also Cauchy.(b) Give an example of Cauchy sequence {a2

n}∞n=1 such that {an}∞n=1 is notCauchy.

Solution.(a) Since {an}∞n=1 is Cauchy then it is convergent. Since the product of twoconvergent sequences is convergentm the sequence {a2

n}∞n=1 is convergent andtherefore is Cauchy.(a) Let an = (−1)n for all n ∈ N. The sequence {an}∞n=1 is not Cauchy sinceit is divergent. However, the sequence {a2

n}∞n=1 = {1, 1, · · · } converges to 1so it is Cauchy

Exercise 7.9Let {an}∞n=1 be a Cauchy sequence such that an is an integer for all n ∈ N.Show that there is a positive integer N such that an = C for all n ≥ N,where C is a constant.

41

Solution.Let ε = 1

2. Since {an}∞n=1 is Cauchy, there is a positive integer N such that if

m,n ≥ N we have |am − an| < 12. But am − an is an integer so we must have

an = aN for all n ≥ N

Exercise 7.10Let {an}∞n=1 be a sequence that satisfies

|an+2 − an+1| < c2|an+1 − an| for all n ∈ N

where 0 < c < 1.(a) Show that |an+1 − an| < cn|a2 − a1| for all n ≥ 2.(b) Show that {an}∞n=1 is a Cauchy sequence.

Solution.(a) The proof is by induction on n. For n = 2 we have |a3− a2| < c2|a2− a1|.Suppose that |an+2 − an+1| < cn+1|a2 − a1|. Then |an+3 − an+2| < c2|an+2 −an+1| < cn+2|a2 − a1|.(b) Let ε > 0 be given. Since limn→∞ cn = 0 we can find a positive integerN such that if n ≥ N then |c|n < (1− c)ε. Thus, for n > m ≥ N we have

|an − am| ≤|am+1 − am|+ |am+2 − am+1|+ · · ·+ |an − an−1|<cm|a2 − a1|+ cm+1|a2 − a1|+ · · ·+ cn−1|a2 − a1|<cm(1 + c + c2 + · · · )|a2 − a1|

=cm

1− c|a2 − a1| < ε

It follows that {an}∞n=1 is a Cauchy sequence

Exercise 7.11What does it mean for a sequence {an}∞n=1 to not be Cauchy?

Solution.A sequence {an}∞n=1 is not a Cauchy sequence if there is a real number ε > 0such that for all positive integer N there exist n, m ∈ N such that n,m ≥ Nand |an − am| ≥ ε

Exercise 7.12Let {an}∞n=1 and {bn}∞n=1 be two Cauchy sequences. Define cn = |an − bn|.Show that {cn}∞n=1 is a Cauchy sequence.

42

Solution.Let ε > 0 be given. There exist positive integers N1 and N2 such that ifn,m ≥ N1 and n, m ≥ N2 we have |an − am| < ε

2and |bn − bm| < ε

2. Let

N = N1 + N2. If n, m ≥ N then |cn − cm| = ||an − bn| − |am − bm|| ≤|(an − bn) + (am − bm)| ≤ |an − am| + |bn − bm| < ε. Hence, {cn}∞n=1 is aCauchy sequence

Exercise 7.13Suppose {an}∞n=1 is a Cauchy sequence. Suppose an ≥ 0 for infinitely manyn and an ≤ 0 for infinitely many n. Prove that limn→∞ an = 0.

Solution.Let ε > 0 be given. Since {an}∞n=1 is Cauchy, there is a positive integer Nsuch that if n, m ≥ N then |an − am| < ε. Let n ≥ N. Then the element an

is positive, negative or zero.Case 1: Suppose an ≥ 0. Since am ≤ 0 for infinitely many m, there ism ≥ N such that am ≤ 0 (else, there would be less than N nonpositiveterms in the sequence, which contradicts the assumption). Since n,m ≥ N,an ≤ an − am = |an − am| < ε.Case 2: Suppose an < 0. Since amgeq0 for infinitely many m, there is m ≥ Nsuch that am ≥ 0 (else, there would be less than N non-negative terms inthe sequence, which contradicts the assumption). But then, since n,m ≥N, an < −an ≤ −an +am ≤ |an−am| < ε. So, in any case, for any given ε wecan find N ∈ N such that if n ≥ N then |an| < ε, that is, limn→∞ an = 0

Exercise 7.14Explain why the sequence defined by an = (−1)n is not a Cauchy sequence.

Solution.We know that every Cauchy sequence is convergent. We also know that thegiven sequence is divergent. Thus, it can not be Cauchy

43

Exercise 8.1Show that limx→1

x2−1x−1

= 2.

Solution.Let ε > 0 be arbitrary. We note first that∣∣∣∣x2 − 1

x− 1− 2

∣∣∣∣ =

∣∣∣∣(x− 1)(x + 1)

x− 1− 2

∣∣∣∣ = |x + 1− 2| = |x− 1|.

Thus, choose δ ≤ ε. If 0 < |x− 1| < δ (which is ≤ ε) we obtain∣∣∣∣x2 − 1

x− 1− 2

∣∣∣∣ = |x− 1| < ε

which is the required result

Exercise 8.2Let f(x) = |x|

x. Suppose that limx→0 f(x) = L.

(a) Show that there is a positive number δ such that if 0 < |x| < δ then∣∣∣ |x|x− L

∣∣∣ < 14.

(b) Let x1 = δ4

and x2 = − δ4. Compute the value of |f(x1)− f(x2)|.

(c) Use (a) to show that |f(x1)− f(x2)| < 12.

(d) Conclude that L does not exist. That is, limx→0|x|x

does not exist.

Solution.(a) Since limx→0 f(x) = L, for ε = 1

4we can find a δ > 0 such that

∣∣∣ |x|x− L

∣∣∣ <14

whenever 0 < |x− 0| < δ.(b) We have |f(x1)− f(x2)| = |1− (−1)| = 2.(c) Since both x1 and x2 satisfy 0 < |x1| < δ and 0 < |x2| < δ, we have|f(x1)− L| < 1

4and |f(x2)− L| < 1

4. Thus,

|f(x1)− f(x2)| =|(f(x1)− L)− (f(x2)− L)| (1)

≤|f(x1)− L|+ |f(x2)− L|

<1

4+

1

4=

1

2

But by (b), we have that |f(x1)− f(x2)| = 2. We conclude that 2 < 12

whichin impossible.(d) The contradiction obtained in (c) shows that limx→0

|x|x

does not exist

44

Exercise 8.3Let f(x) = sin

(1x

). Suppose that limx→0 f(x) = L.

(a) Show that there is a positive number δ such that if 0 < |x| < δ then∣∣sin ( 1x

)− L

∣∣ < 14.

(b) Let n be a positive integer such that x1 = 2(2n+1)π

< δ and x2 = 1(2n+1)π

<

δ. Compute the value of |f(x1)− f(x2)|.(c) Use (a) to show that |f(x1)− f(x2)| < 1

2.

(d) Conclude that L does not exist. That is, limx→0 sin(

1x

)does not exist.

Solution.(a) Since limx→0 f(x) = L, for ε = 1

4we can find a δ > 0 such that∣∣sin ( 1

x

)− L

∣∣ < 14

whenever 0 < |x− 0| < δ.(b) We have |f(x1)− f(x2)| = | − 1− 0| = 1.(c) Since both x1 and x2 satisfy 0 < |x1| < δ and 0 < |x2| < δ, we have|f(x1)− L| < 1

4and |f(x2)− L| < 1

4. Thus,

|f(x1)− f(x2)| =|(f(x1)− L)− (f(x2)− L)| (2)

≤|f(x1)− L|+ |f(x2)− L|

<1

4+

1

4=

1

2

But by (b), we have that |f(x1)− f(x2)| = 1. We conclude that 1 < 12

whichin impossible.(d) The contradiction obtained in (c) shows that limx→0 sin

(1x

)does not

exist

Exercise 8.4Suppose that limx→a f(x) exists. Also, suppose that limx→a f(x) = L1 andlimx→a f(x) = L2. So either L1 = L2 or L1 6= L2.(a) Suppose that L1 6= L2. Show that there exist positive constants δ1 and δ2

such that if 0 < |x−a| < δ1 then |f(x)−L1| < |L1−L2|2

and if 0 < |x−a| < δ2

then |f(x)− L2| < |L1−L2|2

.(b) Let δ = min{δ1, δ2} so that δ < δ1 and δ < δ2. Show that if 0 < |x−a| < δthen |L1 − L2| < |L1 − L2| which is impossible.(c) Conclude that L1 = L2. That is, whenever a function has a limit, thatlimit is unique.

Solution.(a) Since L1 6= L2, we have ε = |L1−L2|

2> 0. Since limx→a f(x) = L1, there

45

is δ1 > 0 such that if 0 < |x − a| < δ1 then |f(x) − L1| < |L1−L2|2

. Similarly,since limx→a f(x) = L2 there is δ2 > 0 such that if 0 < |x − a| < δ2 then

|f(x)− L2| < |L1−L2|2

.(b) Suppose 0 < |x− a| < δ. Then

|L1 − L2| =|f(x)− L2 + L1 − f(x)|≤|f(x)− L1|+ |f(x)− L2|

<|L1 − L2|

2+|L1 − L2|

2= |L1 − L2|

(c) The contradiction obtained in (b) implies L1 = L2

Exercise 8.5Using the εδ definition of limit show that

limx→−1

(2x2 + x + 1) = 2.

Solution.Let ε > 0 be given. Note first that |2x2 + x + 1 − 2| = |(2x − 1)(x + 1)| =|2x − 1||x + 1| ≤ (2|x| + 1)|x + 1|. If 0 < |x + 1| < 1 then |x| − 1 < 1 andthis implies |x| < 2. So choose δ = min{1, ε

5}. Clearly, δ ≤ 1 and δ ≤ ε

5. So if

0 < |x+1| < δ we have |2x2+x+1−2| ≤ (2|x|+1)|x+1| < 5|x+1| ≤ 5· ε5

= ε

Exercise 8.6Prove directly from the definition that limx→1

xx+3

= 14.

Solution.First note that ∣∣∣∣ x

x + 3− 1

4

∣∣∣∣ =3

4

|x− 1||x + 3|

.

If |x− 1| < 1 then 0 < x < 2. Thus, |x + 3| = x + 3 > 3. Let ε > 0 be given.

Choose δ = min{1, 4ε}. Then if 0 < |x− 1| < δ we have∣∣ xx+3

− 14

∣∣ = 34|x−1||x+3| <

34|x−1|

3= |x−1|

4< 4ε

4= ε

Exercise 8.7In this exercise we discuss the concept of sided limits.(a) We say that L is the left side limit of f as x approaches a from the leftif and only if

46

∀ε > 0,∃δ > 0 such that 0 < a− x < δ ⇒ |f(x)− L| < ε

and we write limx→a− f(x) = L. Show that limx→0−|x|x

= −1.(b) We say that L is the right side limit of f as x approaches a from theright if and only if

∀ε > 0,∃δ > 0 such that 0 < x− a < δ ⇒ |f(x)− L| < ε

and we write limx→a+ f(x) = L. Show that limx→0+|x|x

= 1.

Exercise 8.8Prove that L = limx→a f(x) if and only if limx→a− f(x) = limx→a+ f(x) = L.

Solution. Suppose first that limx→a f(x) = L. Let ε > 0 be given. Thenthere is δ > 0 such that 0 < |x − a| < δ ⇒ |f(x) − L| < ε. Suppose that0 < a− x < δ. Then 0 < a− x = |x− a| < δ ⇒ |f(x)− L| < ε. This showsthat L = limx→a− f(x). Likewise, one can show that L = limx→a+ f(x).Conversely, suppose limx→a− f(x) = limx→a+ f(x) = L. Let ε > 0 be given.Then there exist δ1 > 0 and δ2 > 0 such that 0 < a−x < δ1 ⇒ |f(x)−L| < εand 0 < x − a < δ2 ⇒ |f(x) − L| < ε. Let δ = min{δ1, δ2}. Suppose0 < |x− a| < δ. Since |x− a| = ±(x− a) we must have |f(x)− L| < ε. Thisshows that limx→a f(x) = L

Exercise 8.9Using ε and δ, what does it mean that limx→a f(x) 6= L?

Solution. If lim x → af(x) 6= L then there is an ε > 0 such that for all δ > 0there is an x in the domain of f such that 0 < |x−a| < δ but |f(x)−L| ≥ ε

47

Exercise 9.1Suppose that limx→a f(x) = L1 and limx→a g(x) = L2. Show that

limx→a

[f(x)± g(x)] = L1 ± L2.

Solution.Let ε > 0 be arbitrary. Since limx→a f(x) = L1, we can find δ1 > 0 such that

0 < |x− a| < δ1 =⇒ |f(x)− L1| <ε

2.

Similarly, since limx→a g(x) = L2, we can find δ2 > 0 such that

0 < |x− a| < δ2 =⇒ |g(x)− L2| <ε

2.

Let δ = min{δ1, δ2}. Notice that δ ≤ δ1 and that δ ≤ δ2. Thus, if 0 < |x−a| <δ then

|(f(x) + g(x))− (L1 + L2)| =|(f(x)− L1) + (g(x)− L2)|≤|f(x)− L1|+ |g(x)− L2|

<ε

2+

ε

2= ε

A similar argument holds for f(x)− g(x)

Exercise 9.2Suppose that limx→a f(x) = L1 and limx→a g(x) = L2. Show the following:(a) There is a δ1 > 0 such that

0 < |x− a| < δ1 =⇒ |f(x)| < 1 + |L1|.

Hint: Notice that f(x) = (f(x)− L1) + L1.(b) Given ε > 0, there is a δ2 > 0 such that

0 < |x− a| < δ2 =⇒ |f(x)− L1| <ε

2(1 + |L2|).

48

Solution.(a) Let ε = 1. Since limx→a f(x) = L1, we can find δ1 > 0 such that

0 < |x− a| < δ1 =⇒ |f(x)− L1| < 1.

But then

|f(x)| = |(f(x)− L1) + L1| ≤ |f(x)− L1|+ |L1| < 1 + |L1|.

(b) Since ε2(1+|L2|) > 0 and limx→a f(x) = L1, we can find δ2 > 0 such that

0 < |x− a| < δ2 =⇒ |f(x)− L1| <ε

2(1 + |L2|)

Exercise 9.3Suppose that limx→a f(x) = L1 and limx→a g(x) = L2.(a) Show that f(x)g(x)− L1L2 = f(x)(g(x)− L2) + L2(f(x)− L1).(b) Show that |f(x)g(x)− L1L2| ≤ |f(x)||g(x)− L2|+ |L2||f(x)− L1|.(c) Show that limx→a f(x)g(x) = L1L2. Hint: Use the previous exercise.


f(x)(g(x)− L2) + L2(f(x)− L1) =f(x)g(x)− L2f(x) + L2f(x)− L1L2

=f(x)g(x)− L1L2.

(b) By (a) and Exercise 1.7/Exercise1.17, we have

|f(x)g(x)− L1L2| =|f(x)(g(x)− L2) + L2(f(x)− L1)|≤|f(x)(g(x)− L2)|+ |L2(f(x)− L1)|=|f(x)||(g(x)− L2)|+ |L2||(f(x)− L1)|

(c) Let ε > 0 be arbitrary. By Exercise 9.2(a), there is δ1 > 0 such that

0 < |x− a| < δ1 =⇒ |f(x)| < 1 + |L1|.

Also, by Exercise 9.2(b), there exist δ2 > 0 and δ3 > 0 such that

0 < |x− a| < δ2 =⇒ |f(x)− L1| <ε

2(1 + |L2|)

49

and0 < |x− a| < δ3 =⇒ |g(x)− L2| <

ε

2(1 + |L1|).

Let δ = min{δ1, δ2, δ3}. Notice that δ ≤ δ1, δ ≤ δ2, and δ ≤ δ3. Suppose that0 < |x− a| < δ. Using (b) and the above inequalities we find

|f(x)g(x)− L1L2| ≤|f(x)||g(x)− L2|+ |L2||f(x)− L1|

<(1 + |L1|) ·ε

2(1 + |L1|)+ |L2| ·

ε

2(1 + |L2|)<(1 + |L1|) ·

ε

2(1 + |L1|)+ (1 + |L2|)| ·

ε

2(1 + |L2|)=

ε

2+

ε

2= ε


Exercise 9.4(a) Suppose that |f(x)| ≤ M for all x in its domain and limx→a g(x) = 0.Show that

limx→a

f(x)g(x) = 0.

Hint: Recall Exercise 4.5(b) Show that limx→0 x sin

(1x

)= 0.

Solution.(a) Let ε > 0 be arbitrary. Since limx→a g(x) = 0, there is a positive numberδ such that if 0 < |x − a| < δ then |g(x) − 0| = |g(x)| < ε

M. Thus, for

0 < |x− a| < δ we have

|f(x)g(x)− 0| = |f(x)g(x)| = |f(x)||g(x)| < M · ε

M= ε.

This shows that limx→a f(x)g(x) = 0.(b) Let f(x) = x and g(x) = sin

(1x

). Note that limx→0 f(x) = 0 and |g(x)| =∣∣sin ( 1

x

)∣∣ ≤ 1. It follows from (a) that

limx→0

x sin

(1

x

)= lim

x→0f(x)g(x) = 0

50

Exercise 9.5Suppose that limx→a f(x) = L with L 6= 0. Show that there exists a δ > 0such that

0 < |x− a| < δ =⇒ |f(x)| > |L|2

> 0.

Hint: Recall the solution to Exercise 4.7

Solution.Let ε = |L|

2> 0. Then there exists δ > 0 such that |f(x)−L| < |L|

2whenever

0 < |x−a| < δ. But by Exercise 1.18, we have |L|−|f(x)| < |L|2

or |f(x)| > |L|2

whenever 0 < |x− a| < δ

Exercise 9.6Let g(x) be a function with the following conditions:(1) g(x) 6= 0 for all x in the domain of g.(2) limx→a g(x) = L2, with L2 6= 0.(a) Show that there is a δ1 > 0 such that if 0 < |x− a| < δ1 then∣∣∣∣ 1

g(x)− 1

L2

∣∣∣∣ < 2

|L2|2|g(x)− L2|.

(b) Let ε > 0 be arbitrary. Show that there is δ2 > 0 such that if 0 < |x−a| <δ2 then

|g(x)− L2| <|L2|2

2ε.

(c) Using (a) and (b), show that

limx→a

1

g(x)=

1

L2

.

Hint: Recall Exercise 4.8

Solution.(a) From Exercise 9.5, we can find δ1 > 0 such that |g(x)| > |L2|

2whenever

0 < |x− a| < δ1. Thus, for 0 < |x− a| < δ1 we obtain∣∣∣∣ 1

g(x)− 1

L2

∣∣∣∣ =|g(x)− L2||g(x)||L2|

<2

|L2|2|g(x)− L2|.

51

(b) Let ε > 0 be arbitrary. Since limx→a g(x) = L2, we can find a positivenumber δ2 such that if 0 < |x− a| < δ2 then

|g(x)− L2| <|L2|2

2ε.

(c) Let ε > 0 be arbitrary. Let δ = min{δ1, δ2}. Notice that δ ≤ δ1 andδ ≤ δ2. Now, if 0 < |x− a| < δ then∣∣∣∣ 1

g(x)− 1

L2

∣∣∣∣ =|g(x)− L2||g(x)||L2|

<2

|L2|2|g(x)− L2| <

2

|L2|2|L2|2

2ε = ε.


Exercise 9.7Show that if limx→a f(x) = L1 and limx→a g(x) = L2 where g(x) 6= 0 in itsdomain and L2 6= 0 then

limx→a

f(x)

g(x)=

L1

L2

.

Hint: Recall Exercise 4.10.

Solution.Using Exercise 9.6 and Exercise 9.3 we can write

limx→a

f(x)

g(x)= lim

x→a

[f(x) · 1

g(x)

]= L1 ·

1

L2

=L1

L2

Exercise 9.8Let f(x) and g(x) be two functions with a common domain D and a a pointin D. Suppose that f(x) ≤ g(x) for all x in D. Show that if limx→a f(x) = L1

and limx→a g(x) = L2 then L1 ≤ L2. Hint: Recall Exercise 4.11

Solution.Since limx→a f(x) = L1, there exists a positive number δ1 such that |f(x)−L1| < ε whenever 0 < |x − a| < δ1. By Exercise 1.14, this is equivalentto L1 − ε < f(x) < L1 + ε whenever 0 < |x − a| < δ1. Similarly, sincelimx→a g(x) = L2, there exists a positive number δ2 such that |g(x)−L2| < εwhenever 0 < |x − a| < δ2. By Exercise 1.14, this is equivalent to L2 − ε <

52

g(x) < L2 + ε whenever 0 < |x− a| < δ2.Suppose that L1 > L2. Let ε = L1−L2

2. We have

g(x) <L2 + ε = L2 +L1 − L2

2=

L1 + L2

2

=L1 −L1 − L2

2= L1 − ε < f(x)

This contradicts the fact that f(x) ≤ g(x) for all x in D. Hence, we concludethat L1 ≤ L2

Exercise 9.9Let D be the domain of a function f(x). Suppose that f(x) ≥ 0 for all x inD and limx→a f(x) = L with L > 0.(a) Show that √

f(x)−√

L =f(x)− L√f(x) +

√L

.

(b) Let ε > 0. Show that there exists δ > 0 such that |f(x) − L| < ε√

Lwhenever 0 < |x− a| < δ.(c) Show that

limx→a

√f(x) =

√L.


√f(x)−

√L =

(√

f(x)−√

L)(√

f(x) +√

L)√f(x) +

√L

=f(x)− L√f(x) +

√L

.

(b) Since limx→a f(x) = L and ε√

L > 0 there exists δ > 0 such that |f(x)−L| < ε

√L whenever 0 < |x− a| < δ.

(c) For 0 < |x− a| < δ we have

|√

f(x)−√

L| =

∣∣∣∣∣ f(x)− L√f(x) +

√L

∣∣∣∣∣ ≤ |f(x)− L|√L

<ε√

L√L

= ε.

This establishes a proof of the required result

53

Exercise 9.10 (Squeeze Rule)Let f(x), g(x) and h(x) be three functions with common domain D and a bea point in D. Suppose that(1) g(x) ≤ f(x) ≤ h(x) for all x in D.(2) limx→a g(x) = limx→a h(x) = L.Show that limx→a f(x) = L. Hint: Recall Exercise 3.11

Solution.Let ε > 0. From (2) we can find δ1 > 0 such that if 0 < |x− a| < δ1 then

|g(x)− L| < ε ⇔ L− ε < g(x) < L + ε.

Similarly, we can find δ2 > 0 such that if 0 < |x− a| < δ2 then

|h(x)− L| < ε ⇔ L− ε < h(x) < L + ε.

Let δ = min{δ1, δ2}. Suppose 0 < |x − a| < δ. Then L − ε < g(x) ≤ f(x) ≤h(x) < L + ε. That is L− ε < f(x) < L + ε whenever 0 < |x− a| < δ. Thisimplies that limx→a f(x) = L

Exercise 9.11Consider the following figure.

where 0 < x < π2.

(a) Using geometry, establish the inequality

0 < sin x < x.

Hint: The area of a circular sector with radius r and central angle θ is givenby the formula 1

2r2θ.

54

(b) Show that limx→0+ sin x = 0.(c) Show that limx→0− sin x = 0. Thus, we conclude that limx→0 sin x = 0.Hint: Recall that the sine function is an odd function.(d) Show that limx→0 cos x = 1. Hint: cos2 x + sin2 x = 1.(e) Using geometry, establish the double inequality

sin x cos x

2<

x

2<

tan x

2.

(f) Using (a) show that

cos x <sin x

x<

1

cos x.

(g) Show that

limx→0+

sin x

x= 1.

(h) Show that for −π2

< x < 0 we have also

limx→0−

sin x

x= 1.

Solution.(a) The area of the triangle OBC is smaller than the circular area. Thus

0 <sin x

2<

x

2

or0 < sin x < x.

(b) This follows from (a) and the squeeze rule.(c) For x < 0 we can write

limx→0−

sin x = limx→0−

− sin (−x) = − lim−x→0+

sin (−x) = 0.

(d) Since cos x ≥ 0 for −π2≤ x ≤ π

2we have cos x =

√1− sin2 x. Thus,

using Exercise 9.9, we find

limx→0

cos x = limx→0

√1− sin2 x =

√limx→0

(1− sin2 x) =√

1− 0 = 1.

(e) The area of the triangle OAB is sin x cos x2

. The area of the triangle OCDis tan x

2. From the graph we see that

55

Area of triangle OAB < Area of circular sector < Area of triangle OCD.

Thus,sin x cos x

2<

x

2<

tan x

2.

(f) Since 0 < x < π2, sin x > 0. Simple algebra leads to

cos x <sin x

x<

1

cos x.

(g) Notice that limx→0 cos x = 1 and limx→01

cos x= 1. Now the result follows

by applying the squeeze rule.(h) Since sin x = − sin (−x) and 0 < −x < π

2for −π

2< x < 0, we can use

(g) to obtain

limx→0−

sin x

x= lim

−x→0+

sin (−x)

−x= 1

Exercise 9.12Find each of the following limits:

(1) limx→1

√x2+3−2

√x

x2−1.

(2) limx→2−x−2

|x2−5x+6| .


limx→1

√x2 + 3− 2

√x

x2 − 1= lim

x→1

(√

x2 + 3− 2√

x)(√

x2 + 3 + 2√

x

(√

x2 + 3 + 2√

x)(x2 − 1)

= limx→1

x2 − 4x + 3

(√

x2 + 3 + 2√

x)(x2 − 1)

= limx→1

(x− 1)(x− 3)

(√

x2 + 3 + 2√

x)(x− 1)(x + 1)

= limx→1

x− 3

(√

x2 + 3 + 2√

x)(x + 1)= −1

4

(2) We have

limx→2−

x− 2

|x2 − 5x + 6|= lim

x→2−

x− 2

|(x− 2)(x− 3)|

= limx→2−

x− 2

(x− 2)(x− 3)

= limx→2−

1

x− 3= −1

56

Exercise 9.13Find limx→∞

x2+xx2−x

by using the change of variable u = 1x.

Solution.By letting u = 1

xwe find

limx→∞

x2 + x

x2 − x= lim

u→0

1 + u

1− u= 1

Exercise 9.14Find limx→0

3√

x sin 1x.

Solution.We have for x > 0

− 3√

x ≤ 3√

x sin1

x≤ 3√

x

and for x < 0

− 3√

x ≥ 3√

x sin1

x≥ 3√

x

By the squeeze rule we conclude that limx→03√

x sin 1x

= 0

Exercise 9.15Find limx→0 x2 tan x.

Solution.For −π

4≤ x ≤ π

4we have −1 ≤ tan x ≤ 1. Thus, −x2 ≤ x2 tan x ≤ x2. By

the squeeze rule we conclude that limx→0 x2 tan x = 0

Exercise 9.16Let n be a positive integer. Prove that limx→a[f(x)]n = [limx→a f(x)]n .

Solution.Follows by a simple induction on n in Exercise 9.3

57

Exercise 10.1Suppose that limx→a f(x) = L, where a is in the domain of f. Let {an}∞n=1

be a sequence whose terms belong to the domain of f and are different froma and suppose that limn→∞ an = a.(a) Let ε > 0 be arbitrary. Show that there exist a positive integer N anda positive number δ such that for n ≥ N we have |an − a| < δ and for0 < |x− a| < δ we have |f(x)− L| < ε.(b) Use (a) to conclude that for a given ε > 0 there is a positive integer Nsuch that if n ≥ N then |f(an)− L| < ε. That is, limn→∞ f(an) = L.

Solution.(a) Let ε > 0 be given. Since limx→a f(x) = L, we can find a positive constantδ such that if 0 < |x − a| < δ then |f(x) − L| < ε. But for δ > 0 we canfind a positive integer N such that for n ≥ N we have |an − a| < δ due tothe fact that limn→∞ an = a. Since an 6= a for all n ≥ 1 we conclude that0 < |an − a| < δ for all n ≥ N.(b) Using (a), for a given ε > 0 we can find a positive integer N such thatif n ≥ N then 0 < |an − a| < δ which implies that |f(an) − L| < ε. Thisestablishes that

limn→∞

f(an) = L

Exercise 10.2Let {an}∞n=1 be an arbitrary sequence of terms in the domain of f with an 6= afor all n ≥ 1. Suppose that if limn→∞ an = a then limn→∞ f(an) = L. Clearly,either

limx→a f(x) = L or limx→a f(x) 6= L.

(a) Suppose first that limx→a f(x) 6= L. Show that there is an ε > 0 and asequence {an}∞n=1 of terms in the domain of f such that 0 < |an−a| < 1

nand

|f(an)− L| ≥ ε.(b) Use the squeeze rule to show that limn→∞ |an − a| = 0.(c) Use the fact that −|a| ≤ a ≤ |a| for any number a and the squeeze ruleto show that limn→∞(an − a) = 0.(d) Use Exercise 4.1 to show that limn→∞ an = a.(e) Using (a), (d), the given hypothesis and Exercise 4.11, show that ε ≤ 0.

58

Thus, this contradiction shows that limx→a f(x) 6= L cannot happen. Weconclude that

limx→a

f(x) = L

Solution.(a) Since limx→a f(x) 6= L, there is ε > 0 such that for every n ≥ 1, there isan in the domain of f such that 0 < |an − a| < 1

nbut |f(an)− L| ≥ ε.

(b) Let bn = 0, cn = |an − a|, and dn = 1n. Then we have bn < cn < dn. Since

limn→∞ bn = limn→∞1n

= 0. By the squeeze rule we obtain limn→∞ |an−a| =0.(c) Since−|an−a| ≤ an−a ≤ |an−a|, by the squeeze rule we find limn→∞(an−a) = 0.(d) Since an = (an − a) + a, we can use Exercise 4.1 to obtain limn→∞ an =limn→∞(an − a) + limn→∞ a = 0 + a = a.(e) According to (a) and (d), we can find a sequence {an}∞n=1 of terms in thedomain of f such that an 6= a for all n ≥ 1, |f(an)− L| ≥ ε. By hypothesis,limn→∞ f(an) = L. By Exercise 4.11, ε ≤ 0, a contradiction. Hence, we musthave

limx→a

f(x) = L

Exercise 10.3Let f be a function with domain D and a be a point in D. Suppose that fsatisfies the following Property:

(P) If {an}∞n=1, with an in D, an 6= a for all n ≥ 1 and limn→∞ an = a then{f(an)}∞n=1 is a Cauchy sequence.

(a) Let {an}∞n=1 be a sequence of elements of D such that an 6= a for all n ≥ 1and limn→∞ an = a. Show that the sequence {f(an)}∞n=1 is convergent. Callthe limit L. Hint: See Exercise 7.7(b) Let {bn}∞n=1 be a sequence of elements of D such that bn 6= a for all n ≥ 1and limn→∞ bn = a. Show that the sequence {f(bn)}∞n=1 converges to L′.

Solution.(a) By Property (P) the sequence {f(an)}∞n=1 is a Cauchy sequence. By Ex-erccise 7.7, the sequence {f(an)}∞n=1 converges, say with limit L.(b) By an argument similar to (a), there is a number L′ such that limn→∞ f(bn) =L′

59

Exercise 10.4Let {an}∞n=1 and {bn}∞n=1 be the two sequences of the previous exercise. Definethe sequence

{cn} = {b1, a1, b2, a2, b3, a3, · · · }.

That is, cn = ak if n = 2k and cn = bk if n = 2k + 1 where k ≥ 0.(a) Show that for all n ≥ we have cn ∈ D and cn 6= a.(b) Let ε > 0. Show that there exist positive integers N1 and N2 such that ifn ≥ N1 then |an − a| < ε and if n ≥ N2 then |bn − a| < ε.(c) Let N = 2N1 + 2N2 + 1. Show that if n ≥ N then |cn − a| < ε. Hence,limn→∞ cn = a. Hint: Consider the cases n = 2k or n = 2k + 1.(d) Show that limn→∞ f(cn) = L′′ for some number L′′.

Solution.(a) Since cn is either an or bn and both belong to D and are different froma, we conclude that cn belongs to D and cn 6= a for all n ≥ 1.(b) Let ε > 0 be arbitrary. Since limn→∞ an = a and limn→∞ bn = a we canfind positive integers N1 and N2 such that if n ≥ N1 we have |an − a| < εand if n ≥ N2 we have |bn − a| < ε.(c) Let N = 2N1 + 2N2 + 1. Let n ≥ N. If n = 2k + 1 then 2k + 1 ≥ N ≥2N2 + 1 → k ≥ N2 so that |cn− a| = |bk − a| < ε. If n = 2k ≥ N ≥ 2N1 thenk ≥ N1 and in this case |cn − a| = |ak − a| < ε. Thus, whether n is even orodd we have |cn − a| < ε. It follows that

limn→∞

cn = a.

(d) With an argument similar to Exercise 10.2, we conclude that limn→∞ f(cn) =L′′ for some number L′′

Exercise 10.5Let {an}∞n=1, {bn}∞n=1, and {cn}∞n=1 be as in the previous exercise.(a) Compare {an}∞n=1 and {cn}∞n=1.(b) Let ε > 0 be arbitrary. Show that there is a positive integer N such thatif n ≥ N then |f(cn)− L′′| < ε.(c) Let N1 be a positive integer such that N1 ≥ N

2. Show that if n ≥ N1 then

|f(an)− L′′| < ε. Hence, limn→∞ f(an) = L′′.(d) Show that limn→∞ f(bn) = L′′. Thus, by Exercise 3.6, we must haveL = L′ = L′′.

60

Solution.(a) First note that {an}∞n=1 = {c2n}∞n=1.(b) Let ε > 0 be arbitrary. Since limn→∞ f(cn) = L′′, there is a positiveinteger N such that if n ≥ N we have |f(cn)− L′′| < ε.(c) Let N1 be a positive integer such that N1 ≥ N

2. For n ≥ N1 → 2n ≥ N →

|f(an)−L′′| = |f(c2n)−L′′| < ε. Hence, limn→∞ f(an) = L′′. by Exercise 3.6,we obtain L = L′′.(d) First note that {bn}∞n=1 = {c2n−1}∞n=1. Let ε > 0 be arbitrary. Sincelimn→∞ f(cn) = L′′, there is a positive integer N such that if n ≥ N wehave |f(cn) − L′′| < ε. Let N2 be a positive integer such that N2 ≥ N+1

2.

For n ≥ N2 → 2n − 1 ≥ N → |f(bn) − L′′| = |f(c2n−1) − L′′| < ε. Hence,limn→∞ f(bn) = L′′. by Exercise 3.6, we obtain L = L′′

Exercise 10.6Prove that if a function f satisfies property (P) then limx→a f(x) exists. Hint:Use Exercise 10.2.

Solution.This follows from Exercise 10.2 and Exercise 10.5

Exercise 10.7Consider the function f : R → R defined by

f(x) =

{sin 1

xif x 6= 0

1 if x = 0

Let {an}∞n=1 and {bn}∞n=1 be the two sequences defined by an = 12nπ

andbn = 1

(2n+ 12)π

. Clearly, an, bn 6= 0 for all n ∈ N, an → 0 and bn → 0. Show

that limx→0 f(x) does not exist.

Solution.Since limn→∞ f(an) = 0 and limn→∞ f(bn) = 1, by Exercise 10.1, limx→0 f(x)does not exist

Exercise 10.8Let {an}∞n=1 be a sequence such that an 6= 2 for all n ∈ N and limn→∞ an = 2.

(a) Find limn→∞a2

n−4an−2

= 4.

(b) Find limx→2x2−4x−2

.

61


limn→∞

a2n − 4

an − 2= lim

n→∞(an + 2) = 4.

(b) By Exercise 10.2, we have

limx→2

x2 − 4

x− 2= 4

Exercise 10.9Consider the floor function f : [0, 1] → R given by f(x) = bxc, where bxcdenote the largest integer less than or equal to x. Find limx→1bxc usingsequences.

Solution.Let {an}∞n=1 ⊂ [0, 1] such that an 6= 1 for all n ∈ N and limn→∞ an = 1. Sincean ∈ (0, 1) we have f(an) = 0 for all n ∈ N. Thus, limn→∞ f(an) = 0. Weconclude that limx→1 f(x) = 0

Exercise 10.10Consider the floor function f : R → R given by f(x) = bxc, where bxc denotethe largest integer less than or equal to x.(a) Let an = 1 − 1

nand bn = 1 + 1

nfor all n ∈ N. Find limn→∞ f(an) and

limn→∞ f(bn).(b) Does limx→1bxc exist?

Solution.(a) bf(an)c = 0 for all n ∈ N so limn→∞ f(an) = 0. Likewise, bf(bn)c = 1 forall n ∈ N so limn→∞ f(bn) = 1.(b) By Exercise ??, limx→1bxc does not exist

62

Exercise 11.1Show that the function f(x) = x2 is continuous at x = 0.

Solution.Let ε > 0 be arbitrary. Let 0 < δ <

√ε. Then for any x such that |x− 0| < δ

we have |x2 − 0| = |x|2 < δ2 < ε. This shows that f(x) = x2 is continuous atx = 0

Exercise 11.2Show that f is continuous at x = a if and only if limx→a f(x) = f(a).

Solution.Suppose first that f is continuous at x = a. Let ε > 0 be arbitrary. Then,there is a δ > 0 such that for all x in D, if |x− a| < δ then |f(x)− f(a)| < ε.In particular, for all x in D such that x 6= a if 0 < |x − a| < δ then|f(x)− f(a)| < ε. But this is the same as

limx→a

f(x) = f(a).

Conversely, suppose that limx→a f(x) = f(a). Let ε > 0 be given. There isa δ > 0 such that for all x in D, if 0 < |x − a| < δ then |f(x) − f(a)| < ε.Since |a − a| = 0 < δ and |f(a) − f(a)| = 0 < ε then for all |x − a| < δ wehave |f(x)− f(a)| < ε. This shows that f is continuous at x = a

Exercise 11.3Consider the function

f(x) =

{x2−4x−2

if x 6= 2

0 if x = 2

Show that f is discontinuous at x = 2.

Solution.We have limx→2

x2−4x−2

= limx→2(x−2)(x+2)

x−2= limx→2(x + 2) = 4 6= f(2). By

Exercise 11.2, the function is discontinuous at 2

63

Exercise 11.4Suppose that f is discontinuous at x = a.(a) Show that there is and ε > 0 and a sequence {an}∞n=1 of elements in Dsuch that 0 ≤ |an − a| < 1

nand |f(an)− f(a)| ≥ ε.

(b) Show that limn→∞ |an − a| = 0.(c) Show that limn→∞ an = a

Solution.(a) By Definition 11, there is an ε > 0 such that for each δn = 1

nwe can find

an in D with 0 ≤ |an − a| < 1n

and |f(an)− f(a)| ≥ ε.(b) Using the Squeeze rule we conclude that limn→∞ |an − a| = 0.(c) Since −|an−a| ≤ an−a ≤ |an−a|, by the Squeeze rule we conclude thatlimn→∞(an − a) = 0. But limn→∞ an = limn→∞(an − a) + a = limn→∞(an −a) + limn→∞ a = 0 + a = a

Exercise 11.5Suppose that f is continuous at x = a. Let {an}∞n=1 be a sequence of elementsin D converging to a.(a) Let ε > 0 be given. Show that there is a δ > 0 such that for any x in Dsuch that |x− a| < δ we have |f(x)− f(a)| < ε.(b) With the ε and δ as in (a), show that there is a positive integer N suchthat if n ≥ N then |an − a| < δ.(c) Conclude that limn→∞ f(an) = f(a).

Solution.(a) Let ε > 0 be given. Since f is continuous at x = a, there is a δ > 0 suchthat for any x in D with |x− a| < δ we have |f(x)− f(a)| < ε.(b) Since limn→∞ an = a, for the δ > 0 we can find a positive integer N suchthat if n ≥ N then |an − a| < δ.(c) Using (a) and (b) we see that for any ε > 0 we can find a positive integer Nsuch that if n ≥ N we have |an−a| < δ which implies that |f(an)−f(a)| < ε.This establishes the result

limn→∞

f(an) = f(a)

Exercise 11.6Suppose that for any sequence {an}∞n=1 of elements in D that converges to a,the sequence {f(an)}∞n=1 converges to f(a). Then either f is continuous at a

64

or f is discontinuous at a.(a) Suppose that f is discontinuous at a. Show that there is an ε > 0 anda sequence {an}∞n=1 of elements in D such that limn→∞ an = a and |f(an)−f(a)| ≥ ε for all n ≥ 1.(b) Show that limn→∞ f(an) = f(a).(c) Show that by (a) and (b) we conclude that ε ≤ 0, a contradiction. Thus,f must be continuous at x = a.

Solution.(a) This follows from Exercise ??.(b) By hypothesis, since limn→∞ an = a we must have limn→∞ f(an) = f(a)or limn→∞(f(an)− f(a)) = limn→∞ f(an)− limn→∞ f(a) = f(a)− f(a) = 0.(c) Since 0 ≤ −(f(an)− f(a)) ≤ |f(an)− f(a)| ≤ (f(an)− f(a)) so that bythe Squeeze rule limn→∞ |f(an)− f(a)| = 0. Since, 0 < ε ≤ |f(an)− f(a)| forall n ≥ 1, we can apply the Squeeze rule and get ε ≤ 0, a contradiction. Weconclude that f must be continuous at x = a


f(x) =

{1 if x ≥ 00 if x < 0

(a) Let an = − 1n. Find limn→∞ an and limn→∞ f(an).

(b) Is f continuous at x = 0?

Solution.(a) We have limn→∞ an = 0 and limn→∞ f(an) = 0.(b) From (a) and Exercise 11.5, the function is discontinuous at x = 0

Exercise 11.8Give an example of a continuous function f : R → R and a sequence {an}∞n=1

such that limn→∞ f(an) exists, but limn→∞ an does not exist.

Solution.Let an = (−1)n. We know that this sequence is divergent so that limn→∞ an

does not exist. However, f(an) = 1 for all n ∈ N so that limn→∞ f(an) = 1

65

Exercise 11.9Determine the values of a and b that makes the function f continuous every-where.

f(x) =

2 sin x

xif x < 0

a if x = 0b cos x if x > 0

Solution.We must have limx→0− f(x) = limx→0+ f(x) = f(0) = a. But limx→0− f(x) =2 so that a = 2. Also, limx→0+ f(x) = b = 2

Exercise 11.10Using the ε-δ definition of continuity show that f(x) = x3 is continuous atx = 1. Hint: x3 − 1 = (x− 1)(x2 + x + 1).

Solution.We first note that |x3−1| = |(x−1)(x2+x+1)| = |x−1||x2+x+1| ≤ 3|x−1|.Let ε > 0 be given. Let δ < ε

5. Then |x−1| < δ ⇒ |f(x)−1| < ε. This shows

that f(x) is continuous at x = 1

Exercise 11.11Consider the function f(x) = cos

(1x

).

(a) Let an = 12nπ

and bn = 1(n+ 1

2)π

. Find limn→∞ an, limn→∞ bn, limn→∞ f(an),

and limn→∞ f(bn).(b) Is f continuous at x = 0?

Solution.(a) we have limn→∞ an = limn→∞ bn = 0, limn→∞ f(an) = 1, and limn→∞ f(bn) =0. Thus, limx→0 cos

(1x

)does not exist and therefore the function is not con-

tinuous at x = 0


f(x) =

{x sin

(1x

)if x 6= 0

0 if x = 0

Show that this function is continuous at x = 0 by using the ε-δ definition.

Solution.Let ε > 0 be given. Choose δ = ε. Then |x| < δ ⇒

∣∣x sin(

1x

)− 0∣∣ ≤ |x| < ε

66

Exercise 11.13Prove that if f is continuous at x = a so does |f |. Hint: Exercise 1.23.

Solution.Let ε > 0 be given. Then there is a δ > 0 such that |x − a| < delta ⇒|f(x)− f(a)| < ε. But ||f(x)| − |f(a)|| ≤ |f(x)− f(a)| < ε. This shows that|f | is continuous at x = a

Exercise 11.14Suppose f, g : R → R are continuous on R. Suppose h : R → R satisfiesf(x) ≤ h(x) ≤ g(x) for all x ∈ R. If f(c) = g(c), prove that h is continuousat c.

Solution.Let {an}∞n=1 be any sequence in R that converges to c. We must show that thesequence {h(an)}∞n=1 converges to h(c). By the continuity of f at c, we knowthat {f(an)}∞n=1 converges to f(c), and similarly we know that {g(an)}∞n=1

converges to g(c). We also have the inequality f(an) ≤ h(an) ≤ g(an). Sincelimn→∞ f(an) = f(c) = g(c) = limn→∞ g(an), it follows from the Squeezetheorem that {h(an)}∞n=1 converges, and limn→∞ h(an) = f(c) = h(c). Thush is continuous at c

Exercise 11.15Let f : [0,∞) → R be defined by f(x) =

√x. Show that f is continuous on

[0,∞).

Solution.First we show that f is right continuous at x = 0. Let ε > 0 be given. Chooseδ = ε2. Then if 0 < x < δ we have |

√x− 0| =

√x < ε.

Now let c > 0. Let ε > 0 be given. Let δ < ε√

c. Then

|f(x)− f(c)| =∣∣∣∣(√x−

√c)(√

x−√

c)√x +

√c

∣∣∣∣=

|x− c|√x +

√c

<√

cε · 1√c

= ε

67

Exercise 12.1Let f(x) and g(x) be two functions with common domain D. Suppose thatf and g are continuous at a point a in D. Show the following properties:(i) f ± g is continuous at a.(ii) f · g is continuous at a.(iii) f

gis continuous at a provided that g(a) 6= 0.

Solution.(i) By Exercise 9.1, we have

limx→a

(f ± g)(x) = limx→a

f(x)± limx→a

g(x) = f(a)± g(a) = (f ± g)(a).

By Exercise 11.2, (f ± g) is continuous at a.(ii) By Exercise 9.3, we have

limx→a

(f · g)(x) = limx→a

f(x) · limx→a

g(x) = f(a) · g(a) = (f · g)(a).

By Exercise 11.2, (f · g) is contiuous at a.(iii) By Exercise 9.7, we have

limx→a

(f

g

)(x) =

limx→a f(x)

limx→a g(x)=

f(a)

g(a)=

(f

g

)(a).

By Exercise 11.2, fg

is contiuous at a

Exercise 12.2Let f be continuous at a point a in its domain with f(a) 6= 0. Show thatthere exists a δ > 0 such that

|x− a| < δ =⇒ |f(x)| > |f(a)|2

.

That is, there is an open interval centered at a where the function is alwaysdifferent from zero there. Hint: Look at Exercise 4.7

Solution.Let ε = |f(a)|

2> 0. By the definition of continuity, there is a δ > 0 such that

|x− a| < δ =⇒ |f(x)− f(a)| < |f(a)|2

.

68

By Exercise 1.18, we have |f(a)| − |f(x)| < |f(a)|2

or |f(x)| > |f(a)|2

Exercise 12.3Let f : D → R and g : D′ → R with the range of f contained in D′. Thus,g ◦ f : D → R is a function with domain D. Suppose that f is continuous ata and g is continuous at f(a).(a) Let ε > 0 be given. Show that there is a δ′ > 0 such that for all y in D′

satisfying |y − f(a)| < δ′ we have |g(y)− g(f(a))| < ε.(b) Show that there is a δ′′ > 0 such that if |x−a| < δ′′ then |f(x)−f(a)| < δ′.(c) Show that there is a δ > 0 such that if |x−a| < δ then |g(f(x))−g(f(a))| <ε. In other words, the composite function g(f(x)) is continuous at a. Hence,the composition of two continuous functions is a continuous function.

Solution.(a) Let ε > 0 be given. Since g is continuous at f(a), that is

limy→f(a)

g(y) = g(f(a))

there is a δ′ > 0 such that if |y − f(a)| < δ′ then |g(y)− g(f(a))| < ε, wherey is in D′.(b) Now, since f is continuous at a, there is a δ′′ > 0 such that if |x−a| < δ′′

then |f(x)− f(a)| < δ′.(c) Let δ = min{δ′, δ′′}. If |x − a| < δ then |x − a| < δ′′ which implies that|f(x) − f(a)| < δ′. Letting y = f(x) we find that |y − f(a)| < δ′. But then|g(y) − g(f(a))| = |g(f(x)) − g(f(a))| < ε. This establishes the fact thatg(f(x)) is continuous at a

Exercise 12.4In Exercise 9.11, we established that limx→0 sin x = 0 = sin 0. That is, thesine function is continuous at 0.(a) Using the trigonometric identity

sin (a + b) = sin a cos b + cos a sin b

show that the sine function is continuous at every number a. Hint: Use thesubstitution u = x− a and note that u → 0 as x → a.(b) Show that the cosine function is continuous for every number a. Hint:Note that cos x = sin

(π2− x)

and use Exercise 12.3.

69

Solution.(a) Letting u = x− a we have

limx→a

sin x = limu→0

sin (u + a)

= limu→0

(sin u cos a + cos u sin a)

= limu→0

sin u cos a + limu→0

cos u sin a

=0 + (1) sin a = sin a

(b) Let f(x) = π2− x, g(x) = sin x, and h(x) = cos x. Then h(x) =

sin(

π2− x)

= g(f(x)). Since f and g are continuous for every real number,by the previous exercise, h is continuous for every real number

Exercise 12.5Suppose that f : R → R is continuous such that f(x) = 0 for all x ∈ Q.Prove that f(x) = 0 for all x ∈ R. Hint: Exercise 3.21

Solution.Let a be an irrational number. By Exercise 3.21, there is a sequence ofrational numbers {an}∞n=1 that converges to a. By continuity, we must havef(a) = limn→∞ f(an) = 0. Thus, f(x) = 0 for all x ∈ R


f(x) =

{x if x ∈ Q0 if x 6∈ Q

(a) Prove that f is continuous at x = 0.(b) Let a 6= 0. Prove that f is discontinuous at x = a.

Solution.(a) Let ε > 0 be given. Choose δ = ε. If x is rational such that |x| < δ then|f(x)−f(0)| = |x| < ε. If x is irrational such that |x| < δ then |f(x)−f(0)| =0 < ε. Hence, f is continuous at x = 0.(b) Let a 6= 0. We can find a sequence of rationals {an}∞n=1 that converges toa. In this case, limn→∞ f(an) = limn→∞ an = a. Also, we can find a sequenceof irrationals {bn}∞n=1 that converges to a. In this case, limn→∞ f(bn) = 0.Thus, limx→af(x) does not exist and so the function is discontinuous atx = a

70

Exercise 12.7Suppose f, g : R → R are continuous functions and f(x) = g(x) for everyx ∈ Q. Show that f(x) = g(x) for every x ∈ R.

Solution.Apply Exercise 12.5 to the function h(x) = f(x)− g(x)

Exercise 12.8Use continuity to evaluate limx→π sin (x + sin x).

Solution.We have limx→π sin (x + sin x) = sin (π + 0) = 0

Exercise 12.9Give an example of two functions f and g that are not continuous on theinterval (0, 1) but their sum f + g is continuous on (0, 1).

Solution.Let

f(x) =

{1 if 0 < x < 1

2

2 if 12≤ x < 1

and

g(x) =

{2 if 0 < x < 1

2

1 if 12≤ x < 1

Thus, f and g are discontinuous at x = 12. However, (f + g)(x) = 3 for all

x ∈ (0, 1) which is a continuous function

Exercise 12.10Let f : R → R be a continuous function that satisfies f(x+ y) = f(x)+ f(y)for all x, y ∈ R.(a) Show that f(0) = 0 and f(n) = an and f(−n) = a(−n) for all n ∈ Nwhere a = f(1).(b) Show f

(mn

)= a · m

nwhere m and n are integers with n 6= 0. That is,

f(x) = ax for all x ∈ Q.(c) Show that f(x) = ax for all x ∈ R. Hint: Exercise 12.5 applied to thefunction g(x) = f(x)− ax.

71

Solution.(a) We have f(0) + f(0) = f(0 + 0) = f(0). Subtract f(0) from both sidesto obtain f(0) = 0. Let a = f(1) and n ∈ N. Write n = 1 + 1 + · · ·+ 1. Thenf(n) = f(1 + 1 + · · · + 1) = f(1) + f(1) + · · · + f(1) = nf(1) = an. Also,f(−n) + f(n) = f(n− n) = f(0) = 0 so that f(−n) = −f(n) = a(−n).(b) Let m and n be integers such that n 6= 0. Then f

(mn

)= f

(m · 1

n

)=

mf(

1n

). Since nf

(1n

)= f

(nn

)= f(1) = a we obtain f

(mn

)= am

n. Thus,

f(x) = ax for all x ∈ Q.(c) Let g(x) = f(x)−ax. g is continuous on R and g(x) = 0 for all x ∈ Q. byExercise 12.5, we conclude that g(x) = 0 for all x ∈ R. That is, f(x) = axfor all x ∈ R

Exercise 12.11Prove that if f is continuous on [a, b], then either f(x) = 0 for some x ∈ [a, b],or there is a number ε > 0 such that |f(x)| ≥ ε for all x ∈ [a, b].

Solution.Suppose that for every ε > 0 there is an x ∈ [a, b] such that |f(x)| < ε. By let-ting ε = 1

nwe find a sequence {an}∞n=1 such that |f(an)| < 1

n. Since {an}∞n=1 is

bounded (since all the terms are in [a, b]) we can find a subsequence {ank}∞k=1

of {an}∞n=1 that converges to some number x ∈ [a, b]. Since |f(ank)| < 1

nkand

f is continuous we conclude that f(x) = limn→∞ f(ank) = 0. Thus either

there is some ε > 0 with |f(x)| ≥ ε for all x ∈ [a, b], or there is some x ∈ [a, b]with f(x) = 0

72

Exercise 13.1Show that the function f(x) = x is uniformly continuous.

Solution.Let ε > 0 be given. Let δ = ε. Then for all x1 and x2, if |x1 − x2| < δ then|f(x1) − f(x2)| = |x1 − x2| < δ = ε. Hence, the given function is uniformlycontinuous

Exercise 13.2Consider the function f(x) = 1

xon the set x > 0. Let δ > 0 be any number

and define α = min{2, δ}. Then α ≤ 2 and α ≤ δ. Let x1 = α3

> 0 andx2 = α

6> 0.

(a) Show that |x1 − x2| < δ but |f(x1)− f(x2)| ≥ 32.

(b) Conclude from (a) that f is not uniformly continuous on the interval0 < x < ∞.

Solution.(a) We have |x1 − x2| =

∣∣α3− α

6

∣∣ = α6≤ δ

6< δ. Moreover, |f(x1) − f(x2)| =∣∣ 3

α− 6

α

∣∣ = 3α≥ 3

2.

(b) By letting ε = 32

there is no δ > 0 where the definition of uniformcontinuity is satisfied. Hence, f(x) = 1

xis not uniformly continuous on

(0,∞)

Exercise 13.3(a) Show that if f is uniformly continuous on D then f is continuous at everypoint in D.(b) Using properties of continuous functions, show that the function f(x) = 1

x

is continuous on the interval 0 < x < ∞.(c) Is the converse of (a) always true? That is, every continuous function isuniformly continuous.

Solution.(a) Suppose f is uniformly continuous on D. Let a be a point in D. Let ε > 0be given. Then by uniform continuity, there is a δ > 0 such that for all x, uin D,

if |x− u| < δ =⇒ |f(x)− f(u)| < ε.

73

In particular, the above is true if u = a. Hence, f is continuous at a.(b) Since g(x) = 1 and h(x) = x are both continuous in the interval 0 < x <∞, the ratio function f(x) = 1

xis also continuous there.

(c) No. The function f(x) = 1x

is continuous in the interval 0 < x < ∞ (by(b)) but not uniformly continuous there (by Exercise 13.2(a))

Exercise 13.4Show that if f, g : D → R are uniformly continuous then f + g : D → R isalso uniformly continuous.

Solution.Let ε > 0. Since f is uniformly continuous, there is a δ1 > 0 such that

if |x− u| < δ1 =⇒ |f(x)− f(u)| < ε2

for all x, u in D.

Likewise, since g is uniformly continuous, there is a δ2 > 0 such that

if |x− u| < δ2 =⇒ |g(x)− g(u)| < ε2

for all x, u in D.

Let δ = min{δ1, δ2}. If x, u are in D such that |x− u| < δ then |x− u| < δ1

and |x−u| < δ2. But then |f(x)−f(u)| < ε2

and |g(x)−g(u)| < ε2. Therefore,

|(f + g)(x)− (f + g)(u)| =|(f(x)− f(u)) + (g(x)− g(u))|≤|f(x)− f(u)|+ |g(x)− g(u)|

<ε

2+

ε

2= ε.

It follows that f + g is also uniformly continuous

Exercise 13.5Let f(x) = x2. Suppose that there is a δ > 0 such that |x1 − x2| < δ for allreal numbers x1 and x2. In addition, suppose we want |x2

1− x22| = 1. That is,

|x1− x2||x1 + x2| = 1. One way to achieve that is by setting x1− x2 = δ2

andx1 + x2 = 2

δ.

(a) Find x1 and x2 in terms of δ.(b) Show that f is not uniformly continuous. Hint: Let ε = 1

2in Definition

12.

Solution.(a) Solving the system of equations by the method of elimination we findx1 = δ

4+ 1

δand x2 = 1

δ− δ

4.

(b) By (a), letting ε = 12

we see that |x1 − x2| = δ2

< δ but |f(x1)− f(x2)| =1 > 1

2= ε. This shows that the given function is not uniformly continuous

74

Exercise 13.6Give an example of two functions f, g : D → R that are uniformly continuousbut the product function f · g is not.

Solution.Let f(x) = g(x) = x. Both functions are uniformly continuous by Exercise??. However, the product (f · g)(x) = x2 is not uniformly continuous by theprevious exercise

Exercise 13.7Let f, g : D → R be uniformly continuous and bounded, say |f(x)| ≤ M1

and |g(x)| ≤ M2 for all x in D. Let ε > 0 be arbitrary.(a) Show that there is a δ1 > 0 such that

if |x− u| < δ1 =⇒ |f(x)− f(u)| < ε2M2

for all x, u in D.

(b) Show that there is a δ2 > 0 such that

if |x− u| < δ2 =⇒ |g(x)− g(u)| < ε2M1

for all x, u in D.

(c) Show that f · g : D → R is also uniformly continuous. Note that bound-edness is crucial in this result. Hint: Note that f(x)g(x) − f(u)g(u) =(f(x)− f(u)g(x) + f(u)(g(x)− g(u)).

Solution.(a) Since f is uniformly continuous, there is a δ1 > 0 such that

if |x− u| < δ1 =⇒ |f(x)− f(u)| < ε2M2

for all x, u in D.

(b) Likewise, since g is uniformly continuous, there is a δ2 > 0 such that

if |x− u| < δ2 =⇒ |g(x)− g(u)| < ε2M1

for all x, u in D.

(c)Let δ = min{δ1, δ2}. If x, u are in D such that |x−u| < δ then |x−u| < δ1

and |x − u| < δ2. But then |f(x) − f(u)| < ε2M2

and |g(x) − g(u)| < ε2M1

.Therefore,

|f(x)g(x)− f(u)g(u)| =|(f(x)− f(u))g(x) + f(u)(g(x)− g(u))|≤|f(x)− f(u)||g(x)|+ |f(u)||g(x)− g(u)|

<ε

2M2

M2 + M1ε

2M1

= ε.

It follows that f · g is also uniformly continuous

75

Exercise 13.8Suppose that f : D → R is uniformly continuous. Let {an}∞n=1 be a Cauchysequence of terms in D.(a) Let ε > 0 be arbitrary. Show that there is a δ > 0 such that

If |x1 − x2| < δ =⇒ |f(x1)− f(x2)| < ε for all x1, x2 in D.

(b) Show that there is a positive integer N such that

If n, m ≥ N =⇒ |an − am| < δ.

(c) Show that {f(an)}∞n=1 is a Cauchy sequence in R (and therefore by Ex-ercise 7.7 is convergent).

Solution.(a) This is just the definition of uniform continuity.(b) This is just the definition of Cauchy sequence.(c) For the given ε > 0 we can find a positive integer N such that if n,m ≥ Nwe have |an − am| < δ by (b). Using (a), |an − am| < δ implies that |f(an)−f(am)| < ε. Thus, {f(an)}∞n=1 is Cauchy in R and therefore is convergent

Exercise 13.9Consider the function f(x) = tan x on the interval −π

2< x < π

2.

(a) Show that the sequence {π2− 1

n}∞n=1 is convergent.

(b) Show that the sequence in (a) is also Cauchy.(c) Show that the sequence {f

(π2− 1

n

)}∞n=1 is not Cauchy.

(d) Show that the function f(x) = tan x is not uniformly continuous on theinterval −π

2< x < π

2.

Solution.(a) limn→∞

(π2− 1

n

)= π

2.

(b) This follows from Exercise 5.2.(c) Since limn→∞ f

(π2− 1

n

)= limn→∞ tan

(π2− 1

n

)= limn→∞ cot 1

n= cot

(limn→∞

1n

)=

cot 0 which is undefined, the sequence {f(

π2− 1

n

)}∞n=1 is not convergent and

therefore is not Cauchy by Exercise 7.7.(d) This follows from Exercise 13.8

Exercise 13.10Let f : D → R and g : D′ → R be two uniformly continuous functions withthe range of f contained in D′. Looking closely at Exercise 12.3, show thatthe composite function g(f(x)) is also uniformly continuous.

76

Solution.Let ε > 0 be given. Since g is uniformly continuous, there is a δ′ > 0 suchthat if |u− v| < δ′ then |g(u)− g(v)| < ε, where u and v are in D′.Now, since f is uniformly continuous, there is a δ′′ > 0 such that if |x1−x2| <δ′′ then |f(x1)−f(x2)| < δ′, where x1 and x2 are in D. Let δ = min{δ′, δ′′}. Letx1 and x2 be in D such that |x1− x2| < δ. Then |x1− x2| < δ′′ which impliesthat |f(x1)−f(x2)| < δ′. But f(x1) and f(x2) are in D′ with |f(x1)−f(x2)| <δ′. Then |g(f(x1)) − g(f(x2))| < ε. This establishes the fact that g(f(x)) isuniformly continuous

Exercise 13.11Consider the function f(x) = sin x defined on the interval −π

2< x < π

2.

(a) Use Exercise 9.11(a) to show that | sin x| ≤ |x| on the interval −π2

< x <π2.

(b) Using the trigonometric identity sin a − sin b = 2 sin(

a−b2

)cos(

a+b2

)and

(a) to show that| sin a− sin b| ≤ |a− b|.

(c) Show that f is uniformly continuous on the −π2

< x < π2.

Solution.(a) For 0 ≤ x < π

2we know that sin x ≥ 0 so that −|x| ≤ sin x ≤ x = |x|.

That is, | sin x| ≤ |x|. For −π2

< x < 0 we have | sin (−x)| ≤ | − x| which isthe same as | sin x| ≤ |x|.(b) Using (a) and the fact that | cos x| ≤ 1 we can write

| sin a− sin b| =∣∣∣∣2 sin

(a− b

2

)cos

(a + b

2

)∣∣∣∣=2

∣∣∣∣sin(a− b

2

)∣∣∣∣ ∣∣∣∣cos

(a + b

2

)∣∣∣∣≤2

∣∣∣∣sin(a− b

2

)∣∣∣∣ ≤ 2 · |a− b|2

= |a− b|

(c)Let ε > 0 be given. Choose δ = ε. Then for all a and b on the interval−π

2< x < π

2we have

if |a− b| < δ =⇒ | sin a− sin b| ≤ |a− b| < δ = ε

Thus, f(x) = sin x is uniformly continuous on the interval −π2

< x < π2

77

Exercise 13.12Using Exercise 13.10 and Exercise 13.11, show that the function g(x) = cos xis uniformly continuous in the interval −π

2< x < π

2.

Solution.First, recall that cos x = sin

(π2− x). Let f(x) = π

2− x and g(x) = sin x.

Then cos x = g(f(x)) is uniformly continuous (by Exercise 13.10) since bothf and g are uniformly continuous( by Exercise 13.11 and Exercise 13.1)

Exercise 13.13Give an example of two uniformly continuous functions f and g such thatf(x)g(x)

is not uniformly continuous.

Solution.The functions f(x) = sin x and g(x) = cos x are uniformly continuous on theinterval −π

2< x < π

2(See Exercise 13.11 and Exercise 13.12.) However, the

function f(x)g(x)

= tan x is not uniformly continuous in the interval −π2

< x < π2

(See Exercise 13.9)

Exercise 13.14Let g : D → R be a uniformly continuous function with |g(x)| ≥ M > 0 forall x ∈ D. Hence, the function 1

g(x)is bounded and g(x) 6= 0 for all x in D.

Show that 1g(x)

is uniformly continuous.

Solution.Let ε > 0 be given. By uniform continuity, there is a δ > 0 such that

if |a− b| < δ =⇒ |g(a)− g(b)| < εM2.

Thus, if |a− b| < δ we can also have∣∣∣∣ 1

g(a)− 1

g(b)

∣∣∣∣ =

∣∣∣∣g(b)− g(a)

g(a)g(b)

∣∣∣∣=|g(a)− g(b)||g(a)||g(b)|

≤|g(a)− g(b)|M2

<M2ε

M2= ε

This shows that 1g(x)

is uniformly continuous

78

Exercise 13.15Let f, g : D → R be two uniformly continuous functions such that f(x) is

bounded and |g(x)| ≥ M > 0 for all x ∈ D. Show that the function f(x)g(x)

isuniformly continuous on D.

Solution.Note that f(x)

g(x)= f(x) · 1

g(x)which is the product of two uniformly continuous

functions with f(x) and 1g(x)

bounded. By Exercise 13.7(c), the function f(x)g(x)

is uniformly continuous

Exercise 13.16A function f : D → R is said to be Lipschitz if there is a constant K > 0such that |f(x) − f(y)| ≤ K|x − y| for all x, y ∈ D. Show that a Lipschitzfunction is uniformly continuous.

Solution.Let ε > 0 be given. Let δ = ε

K. Then for all x, y ∈ D such that |x − y| < δ

we have|f(x)− f(y)| ≤ K|x− y| < K

ε

K= ε.

This shows that f is uniformly continuous

79


Exercise 14.1(a) Let c0 = a+b

2. Then either [a, c0] or [c0, b] contains an infinite members of

{vn}∞n=1. Let’s call the interval [a1, b1]. Show that b1 − a1 = b−a2

.

(b) Let c1 = a1+b12

. Then either [a1, c1] or [c1, b1] contains an infinite membersof {vn}∞n=1. Let’s call the interval [a2, b2]. Show that b2 − a2 = b−a

22 . Comparea1 and a2. Compare b1 and b2.(c) Let c2 = a2+b2

2. Then either [a2, c2] or [c2, b2] contains an infinite members

of {vn}∞n=1. Let’s call the interval [a3, b3]. Show that b3 − a3 = b−a23 . Compare

a1, a2 and a3. Compare b1, b2 and b3.

Solution.(a) If a1 = a and b1 = c0 then b1 − a1 = a− a+b

2= b−a

2. If a1 = c0 and b1 = b

then b1 − a1 = b− a+b2

= b−a2

.

(b) If a2 = a1 and b2 = c1 then b2 − a2 = a1+b12

− a1 = b1−a1

2= b−a

22 . If a2 = c1

and b2 = b1 then b2 − a2 = b1 − a1+b12

= b1−a1

2= b−a

22 . Note that a2 ≥ a1 andb2 ≤ b1.(c) If a3 = a2 and b3 = c2 then b3 − a3 = a2+b2

2− a2 = b2−a2

2= b−a

23 . If a3 = c2

and b3 = b2 then b3 − a3 = b2 − a2+b22

= b2−a2

2= b−a

23 . Note that a3 ≥ a2 ≥ a1

and b3 ≤ b2 ≤ b1

Exercise 14.2(a) Show that the sequence {an}∞n=1 is bounded from above. What is anupper bound?(b) Show that there is a constant M such that M = sup{a1, a2, · · · }.(c) Show that a ≤ M ≤ b.

Solution.(a) For all n ≥ 1, we have an ≤ b. That is the sequence {an}∞n=1 is boundedfrom above with an upper bound b.(b) By the Completeness axiom of R there is a finite number M such thatM = sup{a1, a2, · · · }.(c) Since b is an upper bound of {an}∞n=1 and M is the smallest upper bound,we must have M ≤ b. Since M is an upper bound of {an}∞n=1 we musthave an ≤ M for all n ≥ 1. Since a ≤ an for all n ≥ 1 we conclude thata ≤ M ≤ b

80

Moreover, from (??), we have |un − vn| < 1n

< δ2.

(c) We have |un−M | = |(un−vn)+(vn−M)| ≤ |un−vn|+|vn−M | < δ2+ δ

2= δ.

(d) Since |un −M | < δ and |vn −M | < δ2

< δ, by Exericse 14.3(b) we obtain|f(un) − f(vn)| < ε. But this contradicts (1). Hence, f must be uniformlycontinuous

Exercise 14.6Show that the function f : [0, 1] → R defined by f(x) =

√x is uniformly

continuous.

Solution.By Exercise 11.15, f is continuous on [0, 1]. Thus, f is uniformly continuouson [0, 1]

Exercise 14.7(a) A function f : D → R is said to be Lipschitz if there is a constant K > 0such that |f(x)− f(y)| ≤ K|x− y| for all x, y ∈ D. Show that the functionf(x) =

√x is not Lipschitz on [0, 1]. Hint: Assume the contrary and get a

contradiction.(b) Give an example of a uniformly continuous function that is not Lipschitz.Thus, the converse to Exercise 13.16 is false.

Solution.(a) Suppose that f is Lipschitz on [0, 1]. Then there is a positive constant Ksuch that |

√x−√y| ≤ K|x− y| for all x, y ∈ [0, 1]. Letting y = 0 we obtain√

x ≤ Kx for all x ∈ [0, 1]. In particular,√

x ≤ Kx for all 0 < x ≤ 1 andtherefore K ≥ 1√

xfor all x ∈ (0, 1]. Letting x → 0+ we see that K ≥ ∞, a

contradiction.(b) Let f : [0, 1] → R be defined by f(x) =

√x. We know from (a) that f is

not Lipschitz. We also know by Exercise 14.6 that f is uniformly continuous

Exercise 14.8Show, using the definition of uniform continuity (epsilon-delta definition) thefunction f(x) = x

x+1is uniformly continuous on [0, 2].

Solution.Let x, y ∈ [0, 2]. Note that∣∣∣∣ x

x + 1− y

y + 1

∣∣∣∣ =|x− y|

(x + 1)(y + 1).

82

Let ε > 0 be given. Choose δ < ε. Since (x + 1)(y + 1) ≥ 1 for |x− y| < δ wehave ∣∣∣∣ x

x + 1− y

y + 1

∣∣∣∣ =|x− y|

(x + 1)(y + 1)≤ |x− y| < ε.

Hence, the given function is uniformly continuous on [0, 2]

Exercise 14.9Conisder the function f : [0, 1] → R defined by

f(x) =

{sin x

xif 0 < x ≤ 1

1 if x = 0

Show that f is uniformly continuous on [0, 1].

Solution.f(x) is continuous for all x ∈ (0, 1]. Since limx→0+ f(x) = 1 = f(0) thefunction is continuous at x = 0. Thus, the function f is continuous on [0, 1]and therefore uniformly continuous

Exercise 14.10Show that the function f : (−2, 1] → R defined by f(x) = x2 is Lipschitz on(−2, 1].

Solution.We have

|x2 − y2| = |x + y||x− y| ≤ (|x|+ |y|)|x− y| ≤ 4|x− y|

83

Exercise 15.1Give an example of a continuous f : D → R with D a bounded set (i.e.|x| ≤ M for some M > 0 and for all x in D) but f(D) is not bounded.

Solution.Let D be the open interval (0, 1). Then D is bounded. Define f : D → Rby f(x) = 1

x. We know from Exercise 13.3(b) that f is continuous on D.

Moreover, f(D) is the open interval (0,∞) which is not bounded

Exercise 15.2Let D be a bounded subset of R with |x| ≤ M for all x ∈ D. Suppose thatf : D → R is uniformly continuous.(a) Show that there is a δ > 0 such that if u and v belong to D such that|u− v| < δ then |f(u)− f(v)| < 1.(b) Let n be a positive integer such that n > 2M

δ. Divide the interval

[−M, M ] into n equal subintervals:[x0, x1], [x1, x2], · · · , [xn−1, xn]. Show thatxk − xk−1 < δ for all k = 1, 2, · · · , n(c) Let [a1, b1], [a2, b2], · · · , [ak, bk] be those intervals in (b) that intersect D.That is, D ⊆ [a1, b1]∪ [a2, b2]∪· · ·∪ [ak, bk]. For 1 ≤ i ≤ k let ui ∈ [ai, bi]∩D.Show that if v is in D then there is an 1 ≤ i ≤ k such that |v − ui| < δ and|f(v)| < 1 + |f(ui)|.(d) Show that |f(v)| ≤ M for all v in D. That is, f(D) is bounded.

Solution.(a) Let ε = 1. By uniform continuity, there is a δ > 0 such that if u and vbelong to D such that |u− v| < δ then |f(u)− f(v)| < 1(b) We have xk − xk−1 = 2M

n< δ.

(c) Since D ⊆ [a1, b1] ∪ [a2, b2] ∪ · · · ∪ [ak, bk], if v is in D then v belongs to[ai, bi] for some 1 ≤ i ≤ k. But then |v − ui| < |bi − ai| < δ (by (b)). By (a)we must have |f(v) − f(ui)| < 1. Using the triangle inequality, we concludethat |f(v)| < 1 + |f(ui)|.(d) Let M = 1 + |f(u1)|+ |f(u2)|+ · · ·+ |f(uk)|. By part (c), for any v in Dwe have |f(v)| < 1 + |f(ui)| < M. Thus, the range of f is bounded

Exercise 15.3Show that if f : [a, b] → R is continuous then f([a, b]) is bounded. Hint:Exercises 14.5 and 15.2.

84

Solution.Since f is continuous, by Exercise 14.5 it is uniformly continuous. Since [a, b]is bounded, by Exercise 15.2, f([a, b]) is bounded

Exercise 15.4Show that if f : [a, b] → R is continuous then inf{f(x) : a ≤ x ≤ b} andsup{f(x) : a ≤ x ≤ b} exist.

Solution.By Exercise 15.3, f([a, b]) is bounded. Now, the result follows from theCompletenes Axiom of real numbers

Exercise 15.5Let f : [a, b] → R be continuous. Let I = inf{f(x) : a ≤ x ≤ b}. Note that Iexists by Exercise 15.4. Suppose that I < f(x) for all x ∈ [a, b]. That is, theinfimum can not be attained in [a, b]. Define the function g : [a, b] → R by

g(x) =1

f(x)− I.

(a) Show that g is continuous on [a, b].(b) Show that there is a positive number M such that |g(x)| ≤ M for allx ∈ [a, b].(c) Show that I + 1

Mis a lower bound of f([a, b]) and this leads to a contra-

diction.Conclusion: There must be a number x1 ∈ [a, b] such that f(x1) = inf{f(x) :a ≤ x ≤ b}.

Solution.(a) Since the constant function w1(x) = 1 and w2(x) = f(x) − I are con-tinuous, the function g being the ratio of two continuous functions is alsocontinuous.(b) By Exercise 15.3, g([a, b]) is bounded, that is |g(x)| ≤ M for all x ∈ [a, b].(c) From (b), we have 1

f(x)−I≤ M which implies that I + 1

m≤ f(x) for all

x ∈ [a, b]. But this says that I + 1M

is a lower bound of f([a, b]). Since I isthe largest lower bound, we must have I + 1

M≤ I, a contradiction. Hence,

there must be a number x1 ∈ [a, b] such that f(x1) = inf{f(x) : a ≤ x ≤ b}

85

Exercise 15.6Let f : [a, b] → R be continuous. Let S = sup{f(x) : a ≤ x ≤ b}. Notethat S exists by Exercise 15.4. Show that there exists x2 ∈ [a, b] such thatf(x2) = S. Hint: Mimic Exercise 15.5.

Solution.Suppose that f(x) < S for all x ∈ [a, b]. Define the function g(x) = 1

S−f(x).

Since the constant function w1(x) = 1 and w2(x) = S− f(x) are continuous,the function g being the ratio of two continuous functions is also continuous.By Exercise 15.3, g([a, b]) is bounded, that is |g(x)| ≤ M for all x ∈ [a, b].Thus, we have 1

S−f(x)≤ M which implies that f(x) ≤ S− 1

Mfor all x ∈ [a, b].

But this says that S− 1M

is an upper bound of f([a, b]). Since S is the smallestupper bound, we must have S < S − 1

M, a contradiction. Hence, there must

be a number x2 ∈ [a, b] such that f(x2) = sup{f(x) : a ≤ x ≤ b}

Exercise 15.7Let f : [a, b] → R be continuous. Let f(a) ≤ c ≤ f(b).(a) Let D = {x ∈ [a, b] : f(x) ≤ c}. Show that D is non-empty and that Dis bounded from above. By the Completeness Axiom of real numbers thereis a number d such that d = sup{x ∈ D}.(b) Show that d ∈ [a, b].(c) Suppose that f(d) > c. Show that there is a δ > 0 such that if |x−d| < δthen |f(x)− f(d)| < f(d)− c.(d) Show that for x ∈ [a, b] and |x − d| < δ we must have f(x) > c. Hint:Exercise 1.14.(e) Using (d), show that d− δ is an upper bound of D. Thus, f(d) > c leadsto a contradiction.(f) Suppose that f(d) < c. Show that there is a δ > 0 such that if d − δ <x < d + δ and x ∈ [a, b] we must have f(x) < c.(g) Show that f(d + δ

2) < c. Why this leads to a contradiction?

Conclusion: We must have f(d) = c.

Solution.(a) Since a ∈ D, D is non-empty. Moreover, since x ≤ b for all x ∈ D, we seethat D is bounded from above. By the Completeness Axiom of real numbersthere is a number d such that d = sup{x ∈ D}.(b) By the definition of d, we have a ≤ x ≤ d for all x ∈ D. Also, since bis an upper bound of D and d is the smallest upper bound, we must have

86

d ≤ b. Hence, a ≤ d ≤ b.(c) Let ε = f(d) − c > 0. Since f is continuous at d, there is a δ > 0 suchthat if |x− d| < δ then |f(x)− f(d)| < ε = f(d)− c.(d) If |x − d| < δ then |f(x) − f(d)| < f(d) − c which is equivalent toc− f(d) < f(x)− f(d) or f(x) > c.(e) If x ∈ D and x > d − δ then x − d > −δ which implies that f(x) > c acontradiction. Hence, x ≤ d − δ so that d − δ is an upper bound of D. Butthen d < d− δ which is impossible. Hence, f(d) > c cannot happen.(f) If f(d) < c. Letting ε = c−f(d), we can find δ > 0 such that if |x−d| < δwe have |f(x)− f(d)| < c− f(d) or f(x) < c.(g) Let x = d + δ

2. Then d − δ < x < d + δ and therefore f(x) < c, that is

d + δ2∈ D. But the definition of d implies that d + δ

2< d, a contradiction

Exercise 15.8Let f : [a, b] → R be continuous. By Exercise 15.5, there exist x1 ∈ [a, b] andx2 ∈ [a, b] such that m = f(x1) = inf{f(x) : x ∈ [a, b]} and M = sup{f(x) :x ∈ [a, b]}.(a) Show that f([a, b]) ⊆ [m, M ].(b) Use Exercise 15.7 (restricted to [x1, x2]) to show that [m, M ] ⊆ f([a, b]).Conclusion: f([a, b]) = [m, M ].

Solution.(a) Let y ∈ f([a, b]). Then there is x ∈ [a, b] such that y = f(x). Butm ≤ f(x) ≤ M, that is, m ≤ y ≤ M. This shows that y ∈ [m,M ] andtherefore f([a, b]) ⊆ [m, M ].(b) Let y ∈ [m, M ]. By Exercise 15.7 restricted to the interval [x1, x2], thereexists a ≤ x1 ≤ x ≤ x2 ≤ b such that f(x) = y. That is, y ∈ f([a, b]). Thisshows that [m, M ] ⊆ f([a, b])

Exercise 15.9Prove that there exists a number c ∈

(0, π

2

)such that 2c− 1 = sin

(c2 + π

4

).

Solution.Let f(x) = 2x − 1 − sin

(x2 + π

4

). Then f(0) = −1 − 1√

2< 0 and f

(π2

)=

π − 1 − sin(

π2

2+ π

4

)> 0. By the Intermediate value theorem, there is a

c ∈(0, π

2

)such that f(c) = 0 or 2c− 1 = sin

(c2 + π

4

)

87

Exercise 15.10Let f : [a, b] → [a, b] be a continuous function. Prove that there is c ∈ [a, b]such that f(c) = c. We call c a fixed point of f. Hint: Intermediate ValueTheorem applied to a specific function F (to be found) defined on [a, b].

Solution.Define F : [a, b] → R by F (x) = x − f(x). Then F is continuous on [a, b].Since a ≤ f(a) ≤ b and a ≤ f(b) ≤ b we find F (a) = a − f(a) ≤ 0 andF (b) = b−f(b) ≥ 0. By the Intermediate Value Theorem, there is a c ∈ [a, b]such that F (c) = 0 or c− f(c) = 0. Thus, f(c) = c

Exercise 15.11Using the Intermediate Value Theorem, show that(a) the equation 3 tan x = 2 + sin x has a solution in the interval (0, π

4).

(b) the polynomial p(x) = −x4 + 2x3 + 2 has at least two real roots.

Solution.(a) Let f(x) = 3 tan x− sin x− 2. Then f is continuous on [0, π

4] and f(0) =

−2 < 0, f(

π4

)= 1− 1√

2> 0. By IVT, there is a c ∈ (0, π

4) such that f(c) = 0.

This means, the given equation has at least one solution in the interval (0, π4).

(b) Since p(−1) = −1 < 0, p(0) = 2 > 0, and p(3) = −25 < 0 there exist atleast two numbers −1 < c1 < 0 < 3 such that f(c1) = f(c2) = 0

Exercise 15.12Let f, g : [a, b] → R be continuous functions such that f(a) ≤ g(a) andf(b) ≥ g(b). Show that there is a c ∈ [a, b] such that f(c) = g(c).

Solution.Let h(x) = f(x)− g(x). Then h is continuous on [a, b] with h(a) ≤ 0 ≤ h(b).By the IVT, there is a c ∈ [a, b] such that h(c) = 0 or f(c) = g(c)

Exercise 15.13Let f : [a, b] → R be continuous such that f(a) ≤ a and f(b) ≥ b. Prove thatthere is a c ∈ [a, b] such that f(c) = c. We call c a fixed point of f.

Solution.Let g(x) = f(x)−x. Then g is continuous on [a, b] with g(a) ≤ 0 and g(b) ≥ 0.By IVT, there is a c ∈ [a, b] such that g(c) = 0. That is, f(c) = c

88

Exercise 15.14Let f : [a, b] → R\Q be continuous. Prove that f must be a constantfunction. Hint: Exercise 2.6(c).

Solution.Suppose for a contradiction that f is not constant. Then, we can find x, y ∈[a, b] with x < y and such that f(x) 6= f(y). Choose an rational number mlying between f(x) and f(y). Then, by the intermediate value theorem, thereexists z ∈ [x, y] with f(z) = m. Hence, f takes a rational value, contradictingthe hypotheses

Exercise 15.15Prove that a polynomial of odd degree considered as a function from the realsto the reals has at least one real root.

Solution.Let f(x) be a polynomial of odd degree. Then limx→−∞ f(x) = −∞ andlimx→∞ f(x) = ∞ (or limx→−∞ f(x) = ∞ and limx→∞ f(x) = −∞ depend-ing on whether the leading coefficient is positive or negative, respectively).Hence, there exist a 0: Now theIntermediate Value Theorem applies to give an x ∈ (a, b) such that f(x) = 0

Exercise 15.16Suppose f(x) is continuous on the interval [0, 2] and f(0) = f(2) : Provethere must be a number c between 0 and 1 so that f(c + 1) = f(c). Hint:Consider the function g(x) = f(x + 1)− f(x) on [0, 1].

Solution.We let g(x) = f(x + 1)− f(x). f(x) is continuous on [0, 1]. Furthermore,

g(0) = f(1)− f(0)

andg(1) = f(2)− f(1) = f(0)− f(1) = −(f(1)− f(0)).

If f(1) = f(0) we have obtained the desired conclusion upon taking c = 0.We therefore assume f(0) 6= f(1). But then g(0) and g(1) have oppositesigns. The Intermediate Value Theorem therefore guarantees the existenceof a number c in the open interval (0, 1) satisfying g(c) = 0. But by definitionof g(x), this means f(c + 1) = f(c)

89



f(x) =

{x sin

(1x

)if x 6= 0

0 if x = 0

Show that f is not differentiable at a = 0.

Solution.We have

f ′(0) = limh→0

f(0 + h)− f(0)

h= lim

h→0sin

(1

h

).

By Exercise 8.3, the limit does not exist. Hence, f is not differentiable at 0


f(x) =

{x2 sin

(1x

)if x 6= 0

0 if x = 0

Show that f is differentiable at a = 0. What is f ′(0)?

Solution.We have

f ′(0) = limh→0

f(0 + h)− f(0)

h= lim

h→0h sin

(1

h

).

By Exercise 9.4(b), the limit does exist and is equal to 0

Exercise 16.3Show that f(x) = |x| is not differentiable at 0.

Solution.We have

f ′(0) = limh→0

f(h)− f(0)

h= lim

h→0

|h|h

.

By Exercise 8.2, the limit does not exist. Hence f is not differentiable at 0

90

Exercise 16.4Find the derivative of f(x) = sin x. Hint: Recall the trigonometric identitysin a− sin b = 2 cos

(a+b2

)sin(

a−b2

)and use Exercise 9.11.

Solution.Using Exercise 9.11 along with the continuity of the cosine function we have

f ′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

sin (x + h)− sin x

h

= limh→0

2 cos(

2x+h2

)sin(

h2

)h

=

[limh→0

cos

(2x + h

2

)][limh→0

sin(

h2

)h2

]= cos x · 1 = cos x

Exercise 16.5Let f : D → R be differentiable at a.(a) Show that

limx→a

[f(x)− f(a)] = limh→0

[f(h + a)− f(a)].

(b) Show that f is continuous at a. That is,

limx→a

[f(x)− f(a)] = 0.

Solution.(a) Letting h = x − a and noting that limx→a f(x) = limh→0 f(h + a), Wehave

limx→a

[f(x)− f(a)] = limh→0

[f(h + a)− f(a)].

(b) We have

limx→a

[f(x)− f(a)] = limh→0

[f(h + a)− f(a)]

= limh→0

f(a + h)− f(a)

h· h

=

[limh→0

f(a + h)− f(a)

h

][limh→0

h] = f ′(a) · 0 = 0

Hence, f is continuous at a

91

Exercise 16.6Give an example of a function f : D → R that is continuous at a but notdifferentiable there.

Solution.The function f(x) = |x| is continuous at 0 but not differentiable there (Ex-ercise 16.3)

Exercise 16.7Suppose that f, g : D → R are differentiable at a. Show that the functionsf ± g are also differentiable at a.

Solution.We have

(f ± g)(a) = limh→0

(f ± g)(a + h)− (f ± g)(a)

h

= limh→0

[f(a + h)− f(a)

h± g(a + h)− g(a)

h

]= lim

h→0

[f(a + h)− f(a)

h

]± lim

h→0

[g(a + h)− g(a)

h

]=f ′(a)± g′(a)

Exercise 16.8 (Product Rule)Suppose that f, g : D → R are differentiable at a.(a) Show that (fg)(a+h)− (fg)(a) = [f(a+h)− f(a)]g(a+h)+ f(a)[g(a+h)− g(a)].(b) Show that the function f · g is also differentiable at a.


[f(a + h)− f(a)]g(a + h) + f(a)[g(a + h)− g(a)] =f(a + h)g(a + h)− f(a)g(a + h)

+f(a)g(a + h)− f(a)g(a)

=(fg)(a + h)− (fg)(a).

92

(b) Using (a) and the continuity of g We have

(f · g)′(a) = limh→0

(f · g)(a + h)− (f · g)(a)

h

= limh→0

[f(a + h)− f(a)]g(a + h) + f(a)[g(a + h)− g(a)]

h

=

[limh→0

f(a + h)− f(a)

h

][limh→0

g(a + h)] + f(a)

[limh→0

g(a + h)− g(a)

h

]=f ′(a)g(a) + f(a)g′(a)

Exercise 16.9 (Quotient Rule)Suppose that f, g : D → R are differentiable at a with g(a) 6= 0.(a) Show that(

fg

)(a + h)−

(fg

)(a)

h=

f(a + h)− f(a)

h· 1

g(a + h)−g(a + h)− g(a)

h·f(a)

g(a)· 1

g(a + h).

(b) Show that (f

g

)′(a) =

f ′(a)g(a)− f(a)g′(a)

g(a)2.

Solution.(a) We have(

fg

)(a + h)−

(fg

)(a)

h=

f(a + h)g(a)− f(a)g(a + h)

hg(a + h)g(a)

=[f(a + h)− f(a)]g(a)− f(a)[g(a + h)− g(a)]

hg(a + h)g(a)

=f(a + h)− f(a)

h· 1

g(a + h)

−g(a + h)− g(a)

h· f(a)

g(a)· 1

g(a + h)

(b) Since g is differentiable at a, it is continuous there. That is, limh→0 g(a+

93

h) = g(a). Now,

(f

g

)′(a) = lim

h→0

(fg

)(a + h)−

(fg

)(a)

h

= limh→0

f(a + h)− f(a)

h· 1

g(a + h)

− limh→0

g(a + h)− g(a)

h· f(a)

g(a)· 1

g(a + h)

= limh→0

f(a + h)− f(a)

h· 1

limh→0 g(a + h)

−f(a)

g(a)limh→0

g(a + h)− g(a)

h· 1

limh→0 g(a + h)

=f ′(a)g(a)− f(a)g′(a)

g(a)2

Exercise 16.10 (Chain Rule)Let f : D → R and g : D′ → R be two functions with f(D) ⊆ D′. Supposethat f is differentiable at a and g is differentiable at f(a).(a) Define w : D′ → R by

w(y) =

{g(y)−g(f(a))

y−f(a)if y 6= f(a)

g′(f(a)) if y = f(a).

Show that w is continuous at f(a). That is,

limh→0

w(f(a) + h) = w(f(a)).

(b) Show that (w ◦ f)(x) is continuous at a.(c) Show that

(g ◦ f)(a + h)− (g ◦ f)(a)

h= (w ◦ f)(a + h) · f(a + h)− f(a)

h.

(d) Show that(g ◦ f)′(a) = g′(f(a)) · f ′(a).

94

Solution.(a) This follows from

limh→0

w(h + f(a)) = limh→0

g(h + f(a))− g(f(a))

h= g′(f(a)) = w(f(a))

(b) The composition of two conitnuous functions is a continuous functionaccording to Exercise 12.3(a).(c) We have

(g ◦ f)(a + h)− (g ◦ f)(a)

h=

g(f(a + h))− g(f(a))

f(a + h)− f(a)· f(a + h)− f(a)

h

=(w ◦ f)(a + h) · f(a + h)− f(a)

h.

(d) We have

(g ◦ f)′(a) = limh→0

g(f(a + h))− g(f(a))

f(a + h)− f(a)· f(a + h)− f(a)

h

= limh→0

(w ◦ f)(a + h) ·[limh→0

f(a + h)− f(a)

h

]=w(f(a)) · f ′(a) = g′(f(a)) · f ′(a)

Exercise 16.11 (Power Rule)Let f(x) = xn where n is a non-negative integer.(a) By letting h = ax− x, show that

f ′(x) = lima→1

f(ax)− f(x)

ax− x.

(b) What is the quotient of the division of an−1 by a−1? Hint: use syntheticdivision.(c) Use (a) and (b) to show that f ′(x) = nxn−1.

Solution.(a) From the definition of f ′(x) we have

f ′(x) = limh→0

f(x + h)− f(x)

h.

95

Letting h = ax−x we see that a → 1 as h → 0. Thus, with this substitutionwe can write

f ′(x) = lima→1

f(ax)− f(x)

ax− x.

(b) Using synthetic division we find

an − 1 = (a− 1)(1 + a + a2 + · · ·+ an−1).

(c) Using (a) and (b) we find

f ′(x) = lima→1

(ax)n − xn

ax− x= xn−1 lim

a→1

an − 1

a− 1

=xn−1 lima→1

(1 + a + a2 + · · ·+ an−1) = nxn−1

Exercise 16.12(a) Show that the derivative of a constant function is zero and that thederivative of f(x) = x is f ′(x) = 1.(b) Show that the function h(x) = x sin

(1x

)is differentiable for all x 6= 0.

Solution.(a) Suppose that f(x) = C for all x. Then

f ′(x) = limh→0

f(x + h)− f(x)

h

= limh→0

C − C

h= 0

Now, suppose that f(x) = x. Then

f ′(x) = limh→0

f(x + h)− f(x)

h

= limh→0

x + h− x

h= 1

(b) Let f(x) = 1x. Then f is differentiable for all x 6= 0. Let g(x) = sin x.

Then g(x) is differentiable for all x. But h(x) = g(f(x)) so that by the chainrule h is differentiable for all x 6= 0

Exercise 16.13Let f(x) =

√2x− 1. Find f ′(2) by using only the definition of derivative.

96

Solution.We have

f ′(2) = limh→0

f(2 + h)− f(2)

h

= limh→0

√2h + 3−

√3

h

= limh→0

(√

2h + 3−√

3)(√

2h + 3 +√

3)

h(√

2h + 3 +√

3)

= limh→0

2h

h(√

2h + 3 +√

3)

= limh→0

2√2h + 3 +

√3

=

√3

3

Exercise 16.14Let

f(x) =

{2x + 5 if x ≤ 19x2 − 2 if x > 1.

Show that f(x) is continuous but not differentiable at x = 1.

Solution.Since

limx→1−

f(x) = limx→1−

(2x + 5) = 7

andlim

x→1+f(x) = lim

x→1+(9x2 − 2) = 7

we conclude that f is continuous at x = 1.Now, since

limh→0−

f(1 + h)− f(1)

h= lim

h→0−

2(1 + h) + 5− 7

h= 2

and

limh→0+

f(1 + h)− f(1)

h= lim

h→0+

9(1 + h)2 − 2− 7

h= lim

h→0+(18 + 9h) = 18

the function is NOT differentiable at x = 1

97

Exercise 16.15Find constants a and b such that the piecewise defined function

f(x) =

{ax2 − 4 if x ≤ 1bx + a if x > 1

is differentiable at x = 1.

Solution.Since f is differentiable at x = 1, it is continuous there. Thus, limx→1− f(x) =limx→1+ f(x). That is, a−4 = b+a and this implies that b = −4. Now, sincef is differentiable at 1 we must have

limh→0−

f(1 + h)− f(1)

h= lim

h→0+

f(1 + h)− f(1)

h.

That is, −4 = 2a or a = −2

Exercise 16.16Let f(x) = x2 cos

(1x

)if x 6= 0 and f(0) = 0. Show that f is differentiable at

x = 0 and find f ′(0).

Solution.We have

f ′(0) = limh→0

f(h)− f(0)

h= lim

h→0h cos

(1

h

)= 0.

Note that the cosine function is bounded and limh→0 h = 0

Exercise 16.17(a) Let f(x) = xn with n a negative integer. Prove that f ′(x) = nxn−1.

(b) Let f(x) = xpq where p and q are integers with q 6= 0. Prove that f ′(x) =

pqx

pq−1. Hint: Let y = x

pq so that yq = xp and use Exercise 16.10.

Solution.(a) We have f(x) = 1

x−n so that by the quotient rule we obtain f ′(x) =

− (−n)x−n−1

(x−n)2= nxn−1.

(b) Let y = xpq . The yq = xp. Differentiate both sides and use (a) and Exercise

16.10 we find qyq−1y′ = pxp−1. Thus,

f ′(x) = y′ =p

qxp−1y1−q =

p

qxp−1x

pq−p =

p

qx

pq−1

98

Exercise 16.18We define the number e to be the unique number satisfying

limh→0

eh − 1

h= 1.

It is an irrational number whose value is approximately 2.718281828459045.Define the function f(x) = ex. Find f ′(x) using the definition of the deriva-tive.

Solution.We have

f ′(x) = limh→0

ex+h − ex

h= ex lim

h→0

eh − 1

h= ex

Exercise 16.19The natural logarithmic function is the function f(x) = ln x definedas follows: y = ln x if and only if x = ey. Find the derivative of f. Hint:Differentiate x = ey with respect to x.

Solution.Differentiating x = ey with respect to x we obtain 1 = y′ey and thereforey′ = 1

ey = 1x

Exercise 16.20Consider the function f(x) = xn where n is a real number.(a) Suppose that x > 0 and x in the domain of f. Using the fact thatxn = en ln x, show that f ′(x) = nxn−1.(b) Suppose that x < 0 and x in the domain of f. Show that f ′(x) = nxn−1.Hint: xn = (−1)n(−x)n.

Solution.(a) We have f ′(x) = eln x(n ln x)′ = n

xxn = nxn−1.

(b) We have f ′(x) = (−1)nn(−x)n−1(−1) = (−1)n+1n(−x)n−1 = (−1)n−1n(−x)n−1 =nxn−1

99

Exercise 17.1(a) Find the local extrema (if they exist) of the function f(x) = |x|.(b) Find the local extrema (if they exist) of the function f(x) = x3.(c) Find the local extrema (if they exist) of the function f(x) = x on theinterval [0, 1].

Solution.(a) Since |x| ≥ 0 for all x, we find that f has a local minimum at x = 0.However, f has no local maximum since f is always increasing for x ≥ 0 andalways decreasing for x < 0.(b) The graph of f(x) = x3 is always increasing so that f has no localextrema.(c) The graph is a straight line that is rising to the right. Thus, f(x) has alocal minimum at x = 0 and a local maximum at x = 1

Exercise 17.2Let f : [a, b] → R. Suppose that c ∈ (a, b) is a local maximum (or localminimum) of f such that f ′(c) exists. Let ε > 0 such that f(x) ≤ f(c) forall x ∈ (c− ε, c + ε) ⊆ [a, b].(a) Let h > 0 be small enough so that c + h ∈ (c− ε, c + ε). Using Exercise9.8, show that f ′(c) ≤ 0.(b) Leth < 0 be large enough so that c + h ∈ (c − ε, c + ε). Using Exercise9.8, show that 0 ≤ f ′(c) and therefore f ′(c) = 0.

Solution.(a) Since c+h ∈ (c− ε, c+ ε), we have f(c+h) ≤ f(c). Thus, f(c+h)−f(c)

h≤ 0.

By Exercise 9.8, we have

f ′(c) = limh→0+

f(c + h)− f(c)

h≤ lim

h→0+0 = 0.

(b) Since c+h ∈ (c− ε, c+ ε), we have f(c+h) ≤ f(c). Thus, f(c+h)−f(c)h

≥ 0.By Exercise 9.8, we have

f ′(c) = limh→0−

f(c + h)− f(c)

h≥ lim

h→0−0 = 0.

From (a) and (b) we conclude that f ′(c) = 0

100

Exercise 17.3The condition a < c < b is critical in the previous exercise. Give an exampleof a function f : [a, b] → R such that either a or b is a local extremum butwith non-zero derivative there.

Solution.See Exercise 17.1(c). f has a local minimum at 0 with f ′(0) = 1 6= 0 and alocal maximum at 1 with f ′(1) = 1 6= 0

Exercise 17.4Suppose f : [a, b] → R is continuous. Then there exists x1, x2 ∈ [a, b] suchthat

f(x1) ≤ f(x) ≤ f(x2)

for all x ∈ [a, b]. Show that x1 and x2 are either the endpoints of [a, b] orcritical points of f in a < x < b.

Solution.Since x1 ∈ [a, b], either x1 = a, x1 = b, or a < x1 < b. If a < x1 < b then byExercise 17.2, f ′(x1) = 0. That is, x1 is a critical point of f. Similar argumentholds for x2

Exercise 17.5 (Rolle’s Theorem)Suppose f : [a, b] → R is continuous for a ≤ x ≤ b and differentiable fora < x < b. By Exercise 15.8 there exist x1, x2 ∈ [a, b] such that f(x1) ≤f(x) ≤ f(x2) for all x ∈ [a, b]. Suppose that f(a) = f(b).(a) Show that if f(x) = C for all a ≤ x ≤ b then there is at least a numbera < c < b such that f ′(c) = 0.(b) Suppose that f is a non-constant function. Let d ∈ [a, b] such thatf(d) 6= f(a). Show that if f(d) < f(a) then a < x1 < b. What can you sayabout the value of f ′(x1).(c) Show that if f(a) < f(d) then a < x2 < b. What can you say about thevalue of f ′(x2).

Solution.(a) If f(x) = C for all a ≤ x ≤ b then f ′(x) = 0 for all a ≤ x ≤ b. Inparticular, f ′

(a+b2

)= 0 with a < a+b

2< b. So c = a+b

2.

(b) Suppose f(d) < f(a). If x1 = a then f(d) < f(a) = f(x1) which isimpossible since f(x1) ≤ f(x) for all x ∈ [a, b]. If x1 = b then f(d) < f(a) =

101

f(b) = f(x1) which is again impossible. So we must have a < x1 < b. Now,by Exercise 17.2 we must have f ′(x1) = 0.(c) Suppose f(a) < f(d). If x2 = a then f(x2) < f(a) which is impossiblesince f(x) ≤ f(x2) for all x ∈ [a, b]. If x2 = b then f(x2) = f(b) < f(d) whichis again impossible. So we must have a < x2 < b. Now, by Exercise 17.2 wemust have f ′(x2) = 0

Exercise 17.6Find the number c of Rolle’s theorem for the function f : [0, 1] → R definedby f(x) =

√x− x.

Solution.The function f satisfies the conditions of Rolle’s theorem. We have 0 =f ′(c) = 1

2√

c− 1. Solving this equation for c we find c = 1

4

Exercise 17.7Assume a0, a1, · · · , an are real numbers such that

an

n + 1+

an−1

n+ · · ·+ a1

2+ a0 = 0

Show that the polynomial function

f(x) = anxn + an−1x

n−1 + · · ·+ a1x + a0

has at least one root in (0, 1).

Solution.Let

F (x) =an

n + 1xn+1 +

an−1

nxn + · · ·+ a1

2x2 + a0x.

Note that F is continuous in [0, 1] and differentiable in (0, 1) with derivativeF ′(x) = f(x). Moreover, F (0) = F (1) = 0. By Rolle’s theorem, there is ac ∈ (0, 1) such that F ′(c) = 0. Hence, f(c) = 0

Exercise 17.8(a) Show that the function f(x) = x3 − 4x2 − 3x + 1 has a root in [0, 2].(b) Use Rolle’s theorem to show that there is exactly one root in [0, 2].

102

Solution.(a) We have f(0) = 1 > 0 and f(2) = −13 < 0 so that by IVT there is aroot in [0, 2].(b) Suppose that x1 and x2 are two roots of f in [0, 2]. Then by Rolle’stheorem we must have c ∈ (0, 2) such that f ′(c) = 0. But the solutions tof ′(x) = 0 = 3x2 − 8x − 3 are x = 3 and x = −1

3where neither is in [0, 2].

Hence, f has exactly one solution in [0, 2]

Exercise 17.9Let f, g : R → R be differentiable, and let a, b ∈ R be such that a < b. Showthat there is a c ∈ (a, b) such that

f ′(c)[g(b)− g(a)] = g′(c)[f(b)− f(a)].

Hint: Apply Rolle’s theorem to the function h(x) = f(x)[g(b) − g(a)] −g(x)[f(b)− f(a)].

Solution.Let h; [a, b] → R be the function defined by

h(x) = [f(b)− f(a)]g(x)− [g(b)− g(a)]f(x).

Then h is continuous on [a, b] and differentiable in a < x < b with derivative

h′(x) = [f(b)− f(a)]g′(x)− [g(b)− g(a)]f ′(x).

Moreoverh(a) = h(b) = f(b)g(a)− g(b)f(a).

By Rolle’s theorem there is a a < c < b such that h′(c) = 0. That is

[f(b)− f(a)]g′(c)− [g(b)− g(a)]f ′(c) = 0

or[f(b)− f(a)]g′(c) = [g(b)− g(a)]f ′(c)

Exercise 17.10Suppose f : [a, b] → R is continuous for a ≤ x ≤ b and differentiable fora < x < b. Show that there is a < c < b such that

f ′(c) =f(b)− f(a)

b− a.

103

Hint: Apply Rolle’s theorem to the function g : [a, b] → R defined by

g(x) = f(x)− f(a)−(

f(b)− f(a)

b− a

)(x− a).

Solution.The function g(x) is continuous on [a, b] being a combination of continuousfunctions on [a, b]. Furthermore, g(x) is differentiable for a < x < b withderivative

g′(x) = f ′(x)− f(b)− f(a)

b− a.

Also, g(a) = g(b) = 0. By Exercise 17.5, there is a < c < b such that g′(c) = 0which is equivalent to

f ′(c) =f(b)− f(a)

b− a

104

Exercise 18.1 (Mean Value Theorem)Suppose f : [a, b] → R is continuous for a ≤ x ≤ b and differentiable fora < x < b. Show that there is a < c < b such that

f ′(c) =f(b)− f(a)

b− a.

Hint: Use Exercise 17.5 with the function g : [a, b] → R defined by

g(x) = f(x)− f(a)−(

f(b)− f(a)

b− a

)(x− a).

Solution.The function g(x) is continuous on [a, b] being a combination of continuousfunctions on [a, b]. Furthermore, g(x) is differentiable for a < x < b withderivative

g′(x) = f ′(x)− f(b)− f(a)

b− a.

Also, g(a) = g(b) = 0. By Exercise 17.5, there is a < c < b such that g′(c) = 0which is equivalent to

f ′(c) =f(b)− f(a)

b− a

Exercise 18.2 (Cauchy Mean Value Theorem)Suppose f, g : [a, b] → R are continuous for a ≤ x ≤ b and differentiable fora < x < b. Show that there is a < c < b such that

[g(b)− g(a)]f ′(c) = [f(b)− f(a)]g′(c).

Hint: Use Exercise 17.5 with the function h : [a, b] → R defined by

h(x) = [f(b)− f(a)]g(x)− [g(b)− g(a)]f(x).

Solution.Let h; [a, b] → R be the function defined by

h(x) = [f(b)− f(a)]g(x)− [g(b)− g(a)]f(x).

105

Then h is continuous on [a, b] and differentiable in a < x < b with derivative

h′(x) = [f(b)− f(a)]g′(x)− [g(b)− g(a)]f ′(x).

Moreoverh(a) = h(b) = f(b)g(a)− g(b)f(a).

By Rolle’s theorem there is a a < c < b such that h′(c) = 0. That is

[f(b)− f(a)]g′(c)− [g(b)− g(a)]f ′(c) = 0

or[f(b)− f(a)]g′(c) = [g(b)− g(a)]f ′(c)

Exercise 18.3Let f : [a, b] → R be continuous for a ≤ x ≤ b and differentiable for a <x < b. We say that f is one-to-one if and only if for any a ≤ x1 ≤ b anda ≤ x2 ≤ b such that f(x1) = f(x2) we must have x1 = x2. Suppose thatf ′(x) 6= 0 for all a < x < b.(a) Let a ≤ x1 ≤ b and a ≤ x2 ≤ b such that f(x1) = f(x2). Show thatif x1 < x2 then there is a < x1 < c < x2 < b such that f ′(c) = 0 whichcontradicts the assumption that f ′(x) 6= 0 for all a < x < b. Hint: Use theMean Value Theorem on the interval [x1, x2]..(b) Answer the same question for x2 < x1.Conclusion: We must have x1 = x2. This shows that f is 1-1.

Solution.(a) Applying the MVT on the interval x1 ≤ x ≤ x2, we can find a numberx1 < c < x2 such that f(x2)−f(x1) = f ′(c)(x2−x1). Since f(x1) = f(x2) = 0we obtain (x2 − x1)f

′(c) = 0. Since x1 6= x2 we find that f ′(c) = 0 whichcontradicts the assumption on f ′

(b) Argument similar to (a)

Exercise 18.4Let f : [a, b] → R be continuous for a ≤ x ≤ b and differentiable for a < x <b. We say that f is increasing in [a, b] if and only if for every x1 and x2 in[a, b], if x1 ≤ x2 then f(x1) ≤ f(x2). Show that if f ′(x) ≥ 0 for all a < x < bthen f(x) is increasing in [a, b]. Hint: Use the MVT restricted to the interval[x1, x2].

106

Solution.Let x1, x2 ∈ [a, b]. Clearly, if x1 = x2 then f(x1) = f(x2). So assume thatx1 < x2. By the MVT there is a x1 < c < x2 such that f(x2) − f(x1) =f ′(c)(x2 − x1) ≥ 0 which implies that f(x1) ≤ f(x2). Thus, we have shownthat x1 ≤ x2 implies f(x1) ≤ f(x2). That is, f is increasing in [a, b]

Exercise 18.5Consider Case (i). We have either f(y) < f(x) < f(z) or f(z) < f(x) < f(y).(a) Suppose that f(z) < f(x) < f(y). Use the Intermediate Value theoremrestricted to [y, z] to show that such a double inequality can not occur.(b) Suppose that f(x) < f(z) < f(y). Use the Intermediate Value theoremrestricted to [x, y] to show that such a double inequality can not occur.We conclude that Case (i) does not hold.

Solution.(a) If f(z) < f(x) < f(y), we can apply the Intermediate value theorem to[y, z] to find y < w < z such that f(w) = f(x). Since f is one-to-one wemust have w = x < y which contradicts the inequality y < w.(b) If f(x) < f(z) < f(y), we can apply the Intermediate value theorem to[x, y] to find x < w < y such that f(w) = f(z). Since f is one-to-one wemust have w = z < y which contradicts the inequality y < z

Exercise 18.6Consider Case (ii). We have either f(y) < f(x) < f(z) or f(y) < f(z) <f(x).(a) Suppose that f(y) < f(x) < f(z). Use the Intermediate Value theoremrestricted to [y, z] to show that such a double inequality can not occur.(b) Suppose that f(y) < f(z) < f(x). Use the Intermediate Value theoremrestricted to [x, y] to show that such a double inequality can not occur.We conclude that Case (ii) does not hold.

Solution.(a) If f(y) < f(x) < f(z), we can apply the Intermediate value theorem to[y, z] to find y < w < z such that f(w) = f(x). Since f is one-to-one wemust have w = x < y which contradicts the inequality y < w.(a) If f(y) < f(z) < f(x), we can apply the Intermediate value theorem to[x, y] to find x < w < y such that f(w) = f(z). Since f is one-to-one wemust have w = z < y which contradicts the inequality y < z

107

Exercise 18.7Suppose that f : [a, b] → R is differentiable such that f ′(x) 6= 0 for alla < x < b. We know from the above discussion that f is monotone.(a) Show that if f is increasing in [a, b] then f ′(x) ≥ 0 for all a ≤ x ≤ b. Hint:Let x ∈ [a, b) and choose h > 0 small enough so that x + h ∈ [a, b). If x = b,choose h < 0 so that b + h < b. Now use the definition of the derivative.(b) Show that if f is decreasing in [a, b] then f ′(x) ≤ 0 for all a ≤ x ≤ b.

Solution.(a) Let x ∈ [a, b). Choose h > 0 so that x + h ∈ [a, b). Since x < x + h and

f is increasing, we find that f(x+h)−f(x)h

≥ 0. Thus,

f ′(x) = limh→0

f(x + h)− f(x)

h≥ 0.

If x = b choose h < 0 so that b + h < b. In this case, f(b + h) ≤ f(b) and

f ′(b) = limh→0−

f(b + h)− f(b)

h≥ 0.

(b) Let x ∈ [a, b). Choose h > 0 so that x + h ∈ [a, b). Since x < x + h and

f is decreasing, we find that f(x+h)−f(x)h

≤ 0. Thus,

f ′(x) = limh→0

f(x + h)− f(x)

h≤ 0

If x = b choose h < 0 so that b + h < b. In this case, f(b + h) ≥ f(b) and

f ′(b) = limh→0−

f(b + h)− f(b)

h≤ 0

Exercise 18.8Let f : [a, b] → R be continuous for a ≤ x ≤ b and differentiable for a <x < b. We say that f is a constant function on [a, b] if and only if there is aconstant C such that f(x) = C for all a ≤ x ≤ b. Suppose that f ′(x) = 0 forall a < x < b.Let x1 and x2 be any two numbers in the interval [a, b] with x1 < x2. Supposethat f(x1) 6= f(x2). Show that by applying the Mean Value Theorem on theinterval [x1, x2] we obtain the contradiction f(x1) = f(x2). Thus, we musthave f(x1) = f(x2) = C for any x1 and x2 in [a, b]. That is, f(x) = C for alla ≤ x ≤ b.

108

Solution.Applying the MVT on the interval x1 ≤ x ≤ x2, we can find a numberx1 < c < x2 such that f(x2) − f(x1) = f ′(c)(x2 − x1). Since f ′(c) = 0 weobtain f(x1) = f(x2), a contradiction. Since x1 and x2 were arbitrary, wehave f(x) = C for all x ∈ [a, b]

Exercise 18.9Let f : [a, b] → R be continuous for a ≤ x ≤ b and differentiable for a < x <b. Suppose that f ′(x) = g′(x) for all a < x < b. Show that f(x) = g(x) + Cfor all a ≤ x ≤ b, where C is a constant. Hint: Exercise 18.8

Solution.Let h(x) = f(x)− g(x). Then h(x) is continuous in [a, b] being the differenceof two continuous functions and h′(x) = 0 for all a < x < b. By Exercise18.8, there is C such that h(x) = C for all a ≤ x ≤ b or equivalentlyf(x) = g(x) + C for all a ≤ x ≤ b

Exercise 18.10Let f : [a, b] → R be continuous for a ≤ x ≤ b and differentiable for a < x <b. We say that f is decreasing in [a, b] if and only if for every x1 and x2 in[a, b], if x1 ≤ x2 then f(x1) ≥ f(x2). Show that if f ′(x) ≤ 0 for all a < x < bthen f(x) is decreasing in [a, b]. Hint: Use the MVT restricted to the interval[x1, x2].

Solution.Let x1, x2 ∈ [a, b]. Clearly, if x1 = x2 then f(x1) = f(x2). So assume thatx1 < x2. By the MVT there is a x1 < c < x2 such that f(x2) − f(x1) =f ′(c)(x2 − x1) ≤ 0 which implies that f(x1) ≥ f(x2). Thus, we have shownthat x1 ≤ x2 implies f(x1) ≥ f(x2). That is, f is decreasing in [a, b]

Exercise 18.11Consider the function f(x) = (1 + x)p where 0 0.(a) Apply the MVT to the interval [0, h] to show that f(h) = p(1+ t)p−1h+1for some 0 < t < h.(b) Use (a) to show that (1 + h)p < 1 + ph.In annuity theory, (1 + h)p may represent compound interest and 1 + phrepresent simple interest. A result in annuity theory says that for time pless than a year compound interest formula can be estimated by the simpleinterest formula.

109

Solution.(a) Applying the mean value theorem to the interval [0, h], we can find a0 < t < h such that f(h)− f(0) = f ′(t)h or f(h)− 1 = p(1 + t)p−1h. Hence,f(h) = (1 + h)p = p(1 + t)p−1h + 1.(b) Since t > 0, we have 1 + t > 1 → (1 + t)1−p > 1 → (1 + t)p−1 < 1 →p(1 + t)p−1h < ph → 1 + p(1 + t)p−1h < 1 + ph. Hence, (1 + h)p < 1 + ph

Exercise 18.12Suppose that f : [a, b] → R is differentiable in [a, b]. Let λ be a real numbersuch that either f ′(a) < λ < f ′(b) or f ′(b) < λ < f ′(a).(a) Define g(x) = f(x) − λx. Show that if f ′(a) < λ < f ′(b) then g′(x)changes sign between a and b.(b) Establish the same result for f ′(b) < λ < f ′(a).(c) Show that the condition g′(c) 6= 0 for all c ∈ [a, b] leads to a contradiction.Hint: Exercise 18.7. Conclude that there must be a a < c < b such thatf ′(c) = λ.

Solution.(a) Note that g is continuous in [a, b] and differentiable there with derivativeg′(x) = f ′(x) − λ. Since f ′(a) < λ < f ′(b), we find g′(a) = f ′(a) − λ < 0 <g′(b) = f ′(b)− λ. So g′ changes sign from negative to positive.(b) Since f ′(b) < λ < f ′(a), we find g′(b) = f ′(b)−λ < 0 < g′(a) = f ′(a)−λ.So g′ changes sign from positive to negative.(c) If g′(c) 6= 0 for all c ∈ [a, b] then by Exercise 18.7 either g′ is alwaysnonnegative in [a, b] or always nonpositive which contradict (a) and (b). Weconclude that there must be a a < c < b such that g′(c) = 0 which is thesame as f ′(c) = λ

Exercise 18.13Let f, g : [a, b] → R be two differentiable functions on [a, b] such that f(a) =g(a). Show that if f ′(x) = g′(x) for all x ∈ (a, b) then f(x) = f(x) for allx ∈ [a, b]. Hint: Exercise 18.8.

Solution.Let F : [a, b] → R be given by F (x) = f(x)− g(x). Then F ′ is differentiableon [a, b] and F ′(x) = 0 for all x ∈ (a, b). By Exercise 18.8, there is a constantC such that F (x) = C for all x ∈ [a, b]. But F (a) = 0 so that C = 0. Thus,F (x) = 0 for all x ∈ [a, b]. This is equivalent to f(x) = g(x) for all x ∈ [a, b]

110

Exercise 18.14Let f : R → R be differentiable such that |f ′(x)| < 1 for all x ∈ R. Showthat f can have at most one fixed point. That is, There is at most one c ∈ Rsuch that f(c) = c. Hint: Mean Value Theorem.

Solution.Suppose the contrary. Let a, b ∈ R such that a < b, f(a) = a, and f(b) = b.We have that f is continuous in [a, b] and differentiable in (a, b). By theMVT, there is a c ∈ (a, b) such that

f ′(c) =f(b)− f(a)

b− a=

b− a

b− a= 1.

This is impossible since |f ′(x)| < 1 for all x ∈ R. We conclude that f has atmost one fixed point

Exercise 18.15Let f : R → R be differentiable everywhere and that f ′(a) < 0 and f ′(b) > 0for some a < b. Prove that there is a c ∈ (a, b) such that f ′(c) = 0.

Solution.This is just Exercise 18.12 with λ = 0

Exercise 18.16Let f : R → R be differentiable and |f ′(x)| ≤ K < 1 for all x ∈ R. Leta0 ∈ R. Define the numbers an = f(an−1).(a) Show that |an+1 − an| ≤ Kn|a1 − a0| for all n ∈ N.(b) Show that for all m,n ∈ N such that m > n we have

|am − an| ≤Kn

1−K.

Solution.(a) By the MVT there is a c1 ∈ (a1, a0) such that f(a1)−f(a0) = f ′(c1)(a1−a0). Thus, |a2 − a1| ≤ K|a1 − a0| since |f ′(c1)| ≤ K. Likewise, we can write|a3 − a2| ≤ K|a2 − a1| ≤ K2|a1 − a0|. Now, suppose that |an − an−1| ≤Kn|a1 − a0|. Then |an+1 − an| ≤ K|an − an−1| ≤ Kn+1|a1 − a0|.(b) Let m, n ∈ N such that m > n. Then we have |am − an| ≤ |am − am−1|+· · ·+ |am − am−1| ≤ |a1 − a0|

∑mi=n Ki = Kn

1−K|a1 − a0|

111

Exercise 18.17Show that if 0 < a < b then 1− a

b< ln

(ba

)< b

a− 1. Hint: Apply the MVT

for the function f(x) = ln x.

Solution.The function f(x) = ln x is continuous on [a, b] and differentiable in (a, b).By the Mean value theorem there is a a < c < b such that f ′(c) = ln b−ln a

b−a.

Thus,1

b<

1

c=

ln b− ln a

b− a<

1

aor

1− a

b< ln

(b

a

)<

b

a− 1

112

Exercise 19.1Let f, g : [a, b] → R be continuous on [a, b] and differentiable in a < x < bwith g′(x) 6= 0 for all a < x < b. Suppose that f(c) = g(c) = 0 for somea ≤ c ≤ b. Also, suppose that

limx→c

f ′(x)

g′(x)= A.

(a) Let {cn}∞n=1 ⊂ [a, b] be an arbitrary sequence with the properties cn 6= cfor all n ≥ 1 and limn→∞ cn = c. Show that there is a dn between cn and csuch that

[f(cn)− f(c)]g′(dn) = [g(cn)− g(c)]f ′(dn).

(b) Show that dn 6= c for all n ≥ 1 and limn→∞ dn = c.(c) Show that g(dn) 6= g(c) for all n ≥ 1. Hint: Exercise 18.3.(d) Show that

f ′(dn)

g′(dn)=

f(cn)

g(cn).

(e) Show that limn→∞f ′(dn)g′(dn)

= A. Hint: See Exercise 10.1.

(f) Show that limn→∞f(cn)g(cn)

= A.

(g) Show that limx→cf(x)g(x)

= A.

Solution.(a) Since both f and g are continuous in the closed interval with endpointscn and c and differentiable in the open interval with the same endpoints, wecan use Exercise 13.5 to find a point dn in that interval such that

[f(cn)− f(c)]g′(dn) = [g(cn)− g(c)]f ′(dn).

(b) Since dn is in the open interval with endpoints cn and c we have dn 6= cfor all n ≥ 1. Also, by the Squeeze rule we have limn→∞ dn = c.(c) Since g′(x) 6= 0 for all a < x < b, by Exercise 18.3 we find that g isone-to-one. Since dn 6= c for all n ≥ 1, we must have g(dn) 6= g(c) for alln ≥ 1.(d) This follows from (a) and the assumption that f(c) = g(c) = 0.

113

(e) Since limx→cf ′(x)g′(x)

= A, dn 6= c, dn → c, we can apply Exercise 10.1 towrite

limn→∞

f ′(dn)

g′(dn)= A.

(f) Using (d) and (e) the result follows.(g) The result follows from Exercise 10.1

Exercise 19.2Find

limx→2

√x−

√2 +

√x− 2√

x2 − 4.

Solution.Let f(x) =

√x −

√2 +

√x− 2 and g(x) =

√x2 − 4. Both functions are

continuous in [2, 3] and differentiable in 2 < x < 3. Moreover, f(2) = g(2) =0. Applying L’Hopital’s rule we find

limx→2

√x−

√2 +

√x− 2√

x2 − 4= lim

x→2

12√

x+ 1

2√

x−2x√

x2−4

= limx→2

(√

x− 2 +√

x)(√

x2 − 4)

2x√

x√

x− 2

= limx→2

(√

x− 2 +√

x)(√

x + 2)

2x√

x

=2√

2

4√

2=

1

2

Exercise 19.3Let f : [a, b] → R be a one-to-one function.(a) Define g : f([a, b]) → [a, b] by g(y) = x if and only if f(x) = y. Show thatg is indeed a function. That is, if y1, y2 ∈ f([a, b]) are such that y1 = y2 theng(y1) = g(y2).(b) Show that f(g(y)) = y for all y ∈ f([a, b]) and g(f(x)) = x for allx ∈ [a, b]. Thus, conclude that f is invertible.

Solution.(a) Define g : f([a, b]) → [a, b] by g(y) = x if and only if f(x) = y. Then gis a well-defined function: Suppose that y1, y2 ∈ f([a, b]) such that y1 = y2.

114

Let x1, x2 ∈ [a, b] such that f(x1) = y1 and f(x2) = y2. Then x1 = g(y1)and x2 = g(y2). Moreover, we have f(x1) = f(x2). Since f is one-to-one, weconclude that x1 = x2 which implies that g(y1) = g(y2).(b) Let y ∈ f([a, b]). Then f(x) = y for some a ≤ x ≤ b. By the definitionof g we have g(y) = x. Thus, f(g(y)) = f(x) = y. Likewise, let x ∈ [a, b].Then y = f(x) ∈ f([a, b]). Hence, g(f(x)) = g(y) = x. We conclude that f−1

exists

Exercise 19.4Let f : [a, b] → R be continuous in [a, b] and differentiable in [a, b] withf ′(x) 6= 0 for all a < x < b. Let the range of f be denoted by [m,M ].(a) Show that f is one-to-one, monotone, and invertible with inverse f−1 :[m, M ] → [a, b].(b) Assume that f is strictly increasing. That is, if x1 < x2 then f(x1) <f(x2). In this case, [m, M ] = [f(a), f(b)]. Let f(a) < y0 < f(b). Show thatthere is a a < x0 0 be given. Let ε1 = min{ε, x0−a, b−x0}. Show that if x satisfies|x− x0| < ε1 then a < x 0 so that (y0− δ, y0 + δ) ⊂ (y1, y2). Show that if |y−y0| < δthen |f−1(y)−f−1(y0)| < ε. This shows that f−1 is continuous in (f(a), f(b)).(f) Show that f−1 is right continuous at f(a) and left continuous at f(b).We conclude from this problem that f−1 is continuous on the closed interval[f(a), f(b)].

Solution.(a) By Exercise 18.3, f is one-to-one in [a, b]. By Exercises 18.5 and 18.6, f ismonotone. Now, by Exercise 19.3, f is invertible with inverse f−1 : [m, M ] →[a, b].(b) Since f(a) < y0 < f(b), by the IVT, there is a x0 ∈ [a, b] such thatf(x0) = y0. Since f(a) < f(x0) < f(b), we must have a < x0 < b.(c) Suppose that |x − x0| < ε1. Since ε1 ≤ ε we can write that |x − x0| < ε.Since |x − x0| is less than or equal to both x0 − a and b − x0, we concludethat a < x < b.(d) First we show that x0 − ε1 ∈ [a, b]. Indeed, we have a ≤ a + ε1 ≤ x0 <b ≤ b + ε1. This show that x0 − ε1 ∈ [a, b]. Now, let y ∈ f [(x0 − ε1, x0 + ε1)].Then y = f(x) for some x ∈ (x0 − ε1, x0 + ε1). Since f is strictly increasing

115

we find y1 < y < y2. That is, y ∈ (y1, y2). The converse is similar.(e) Suppose that |y − y0| < δ. Then y ∈ (y1, y2) which implies that f−1(y) ∈(x0 − ε1, x0 + ε1). Hence, |f−1(y) − f−1(y0)| < ε1 ≤ ε. Thus, we have shownthat for any positive number ε, we can find a number δ > 0 such that if|y − y0| < δ then |f−1(y)− f−1(y0)| < ε. This says that f−1 is continuous in(f(a), f(b)).(f) Let ε > 0 be given. There is a δ1 > 0 such that if x − a < δ1 thenf(x) − f(a) < ε. Let δ2 = min δ1, ε. Define δ = f(a + δ2) − f(a). Supposethat y − f(a) < δ. Then there is a < x < a + δ2 such that f(x) = y. Thus,|f−1(y) − a| = |x − a| < δ2 ≤ ε. This shows that f−1 is continuous at f(a).A similar argument holds for the left continuity of f−1 at f(b)

Exercise 19.5 (Inverse Function Theorem)Let f : [a, b] → R be continuous in [a, b] and differentiable in [a, b] withf ′(x) 6= 0 for all a < x < b. Let c ∈ f([a, b]). Then there is a d ∈ [a, b] suchthat f(d) = c.(a) Let {cn}∞n=1 ⊆ f([a, b]) such that cn 6= c for all n ≥ 1 and limn→∞ cn = c.Show that there is a sequence {dn}∞n=1 ⊆ [a, b] such that

limn→∞

dn = d.

Hint: Exercise 14.9(b).(b) Show that dn 6= d for all n ≥ 1.(c) Show that

limn→∞

f(dn)− f(d)

dn − d= f ′(d).

Hint: Exercise 10.1.(d) Show that f(dn)−f(d)

dn−d6= 0 for all n ≥ 1. Hint: Exercise 18.3.

(e) Show that

limn→∞

f−1(cn)− f−1(c)

cn − c=

1

f ′(d).

Thus, conclude that

(f−1)′(f(d)) =1

f ′(d)

for all d ∈ [a, b]. That is f−1 is differentiable in f([a, b]). Hint: Exercise 10.2.

116

Solution.(a) Since {cn}∞n=1 ⊆ f([a, b]), there is a sequence {dn}∞n=1 ⊆ [a, b] such thatf(dn) = cn for all n ≥ 1. By the continuity of f−1 we have

limn→∞

dn = limn→∞

f−1(cn) = f−1( limn→∞

cn)) = f−1(c) = d.

(b) If dn = d then f−1(dn) = f−1(d) which implies cn = c. This contradictsthe fact that cn 6= c for all n ≥ 1.(c) We have

limn→∞

f(dn)− f(d)

dn − d== f ′(d)

since f is differentiable at d. (Exercise 10.1).(d) Since f is one-to-one (Exercise 18.3) and dn 6= d, we have f(dn) 6= f(d).Thus, the ratio

f(dn)−f(d)dn−d

6= 0 for all n ≥ 1.

(e) We have

limn→∞

f−1(cn)− f−1(c)

cn − c= lim

n→∞

dn − d

f(dn)− f(d)

= limn→∞

1f(dn)−f(d)

dn−d

=1

f ′(d)

Applying Exercise 10.2 we find

(f−1)′(c) = (f−1)′(f(d)) =1

f ′(d)

for all d ∈ [a, b]. That is f−1 is differentiable in f([a, b])

Exercise 19.6Find limx→∞

(ln xx· sin

(xπ+2

2x

)Solution.By L’Hopital’s rule we have

limx→∞

ln x

x= lim

x→∞

1

x= 0.

117

By the continuity of the sine function we have

limx→∞

sin

(xπ + 2

2x

)= sin

π

2= 1.

Thus,

limx→∞

(ln x

x· sin

(xπ + 2

2x

))= 0

Exercise 19.7Let f, g : [a, b] → R be continuous on [a, b] and differentiable in a < x < bwith g′(x) 6= 0 for all a < x < b. Suppose that limx→c f(x) = limx→c g(x) =∞ for some a ≤ c ≤ b. Also, suppose that

limx→c

f ′(x)

g′(x)= A.

Prove that

limx→c

f(x)

gx(x)= A.

Solution.We have

limx→c

f(x)

g(x)= lim

x→c

1g(x)

1f(x)

= limx→c

g′(x)

f ′(x)·(

limx→c

f(x)

g(x)

)2

Thus, we have

limx→c

f(x)

g(x)= lim

x→c

f ′(x)

g′(x)= A

Exercise 19.8Use L’Hopital’s rule to evaluate limx→0+ xx. Note that 00 is an undeterminateform.

Solution.We have

limx→0+

xx = limx→0+

ex ln x.

118

But

limx→0+

x ln x = limx→0+

ln x1x

= limx→0+

−x = 0.

Thus,lim

x→0+xx = elimx→0+ x ln x = e0 = 1

Exercise 19.9Let f and g be invertible differentiable functions such that

f(1) = 2; g(2) = 1; f ′(1) = g′(2) = 3.

Find the derivative (f−1 ◦ g−1)′(1).

Solution.We have

(f−1 ◦ g−1)′(1) =[f−1]′(g−1(1)) · [g−1]′(1) =1

f ′(f−1(g−1(1)))· 1

g′(g−1(1))

=1

f ′(f−1(2))

1

g′(2)=

1

f ′(1)

1

g′(2)=

1

9

Exercise 19.10Let f(x) = x tan2 x for x ∈ (0, π

2). Calculate (f−1)′(π). Note that f(π

3) = π.

Solution.By the inverse function theorem we have

(f−1)′(π) =1

f ′(f−1(π))=

1

f ′(π3)

= (3 +8√3π)−1

119

Exercise 20.1(a) Show that m ≤ mi(f) ≤ Mi(f) ≤ M.(b) Show that m(b− a) ≤ L(f, P ) ≤ U(f, P ) ≤ M(b− a).

Solution.(a) Since m is a lower bound of f in [a, b] we must have m ≤ mi(f). Sincemi(f) is a lower bound of f and Mi(f) is an upper bound of f we must havemi(f) ≤ Mi(f). Finally, since M is an upper bound of f in [a, b], we haveMi(f) ≤ M.(b) For all 1 ≤ i ≤ n, we have m(xi−xi−1) ≤ mi(f)(xi−xi−1) ≤ Mi(f)(xi−xi−1) ≤ M(xi − xi−1). Adding these inequalities, we obtain the desired in-equality

Exercise 20.2Let Q be a refinement of P. Suppose that P = {a = x0 < x1 < · · · < xn−1 <xn = b} and Q = {a = x0 < x1 < · · · < xi−1 < z < xi < · · · < xn = b}.(a) Show that U(f, Q) ≤ U(f, P ).(b) Show that L(f, P ) ≤ L(f, Q).

Solution.Suppose first that P is a partition of [a, b] and that Q is the partition obtainedfrom P by adding an additional point xi−1 < z < xi. The general case followsby induction, adding one point at at time. In particular, we let

P = {a = x0 < x1 < · · · < xn−1 < xn = b}

andQ = {a = x0 < x1 < · · · < xi−1 < z < xi < · · · < xn = b}.

Observe that

U(f, P ) =n∑

i=1

Mi(f)(xi − xi−1)

and

U(f, Q) =i−1∑j=1

Mj(f)(xj−xj−1)+M(z−xi−1)+M ′(xi−z)+n∑

j=i+1

Mj(f)(xj−xj−1)

where

120

M = sup{f(x) : x ∈ [xi−1, z]} and M ′ = sup{f(x) : x ∈ [z, xi]}.

Since M(z− xi−1) + M ′(xi− z) ≤ Mi(xi− xi−1) we conclude that U(f, Q) ≤U(f, P ).(b) Similar argument as in (a)

Exercise 20.3Suppose that f is bounded on [a, b]. Show that

∫ b

af(x)dx ≤

∫ b

af(x)dx. Hint:

Exercise 20.2.

Solution.Let P be a partition of [a, b]. For any partition Q of [a, b] we let R = P ∪Q.Since P ⊂ R and Q ⊂ R we can use Exercise 20.2 to write

L(f, P ) ≤ L(f, R) ≤ U(f, R) ≤ U(f, Q).

Since Q was arbitrary, L(f, P ) is a lower bound of SU . But the Riemannupper integral is the largest lower bound of SU . We conclude that

L(f, P ) ≤∫ b

a

f(x)dx.

Since P was arbitrary, the above inequality says∫ b

af(x)dx is an upper bound

of SL. But the lower Riemann integral is the smallest upper bound of SL.That is, ∫ b

a

f(x)dx ≤∫ b

a

f(x)dx

Exercise 20.4Consider the function f : [a, b] → R defined by

f(x) =

{2 if a ≤ x < b3 if x = 3

(a) Find numbers m and M such that m ≤ f(x) ≤ M for all x ∈ [a, b]?(b) Show that for any partition P of [a, b] we have L(f, P ) = 2(b − a).Conclude that ∫ b

a

f(x)dx = 2(b− a).

121

(c) Show that∫ b

af(x)dx ≥ 2(b − a). (d) Suppose

∫ b

af(x)dx > 2(b − a). Let

ε =∫ b

af(x)dx− 2(b− a) > 0. Let Q be the partition

Q = {a = x0 < x1 < x2 < · · · < xn = b}

such that b − xn−1 < ε. Show that U(f, Q) <∫ b

af(x)dx. Why this is impos-

sible?(e) Is f(x) Riemann integrable? If so, what is the value of the integral∫ b

af(x)dx?

Solution.(a) Since 2 ≤ f(x) ≤ 3 for all a ≤ x ≤ b, we have m = 2 and M = 3.(b) Let P be a parition of [a, b] given by

P = {a = x0 < x1 < x2 < · · · < xn = b}.

Note that Mi(f) = 2 for all 1 ≤ i ≤ n − 1, Mn(f) = 3, mi(f) = 2 for all1 ≤ i ≤ n. Thus,

L(f, P ) =n∑

i=1

mi(f)(xi − xi−1) = 2(b− a).

Since P was arbitrary, it follows that∫ b

a

f(x)dx = 2(b− a).

(c) This follows from Exercise 20.3.(d) We have U(f, Q) =

∑n−1i=1 Mi(f)(xi−xi−1)+Mn(f)(b−xn−1) = 2(xn−1−

a)+3(b−xn−1) = 2(b−a)+(b−xn−1) < 2(b−a)+∫ b

af(x)dx−2(b−a) = infSU

,

which contradicts the definition of infimum. We conclude that∫ b

af(x)dx =

2(b− a).

(e) Since∫ b

af(x)dx =

∫ b

af(x)dx = 2(b − a), the function f is Riemnann

integrable with ∫ b

a

f(x)dx = 2(b− a)

122

Exercise 20.5Consider the function f : [0, 1] → R defined by f(x) = 1 if x is rational andf(x) = 0 if x is irrational.(a) Compute the upper Riemann integral and the lower Riemann integral.Hint: Exercise 2.6(c).(b) Is f Riemann integrable on [0, 1]?

Solution.(a) Let P be a partition of [0, 1] :

P = {0 = x0 < x1 < x2 < · · · < xn = 1}.

Clearly, Mi(f) = 1 and mi(f) = 0 for all 1 ≤ i ≤ n. Hence, U(f, P ) = 1 and

L(f, P ) = 0. It follows that∫ 1

0f(x)dx = 1 and

∫ 1

0f(x)dx = 0.

(b) Since the lower Riemann sum is different from the upper Riemann sum,the function is not Riemann integrable

Exercise 20.6Let f : [a, b] → R be a bounded function. Suppose that f is Riemannintegrable. We want to show that f satisfis the following property:

(P)∀ε > 0, there is a partition P of [a, b] such that U(f, P )− L(f, P ) < ε.

(a) Let ε > 0 be given. Show that there is a partition P of [a, b] such that∫ b

a

f(x)dx− ε

2< L(f, P ).

Hint: Assume the contrary and get a contradiction.(b) Show that there is a partition Q of [a, b] such that

U(f, Q) <

∫ b

a

f(x)dx +ε

2.

(c) Let R = P ∪Q. Use Exercise 20.2 to show that∫ b

a

f(x)dx− ε

2< L(f, R) ≤ U(f, R) <

∫ b

a

f(x)dx +ε

2.

(d) Show that

123

∣∣∣L(f, R)−∫ b

af(x)dx

∣∣∣ < ε2

and∣∣∣U(f, R)−

∫ b

af(x)dx

∣∣∣ < ε2

(e) Use the triangle inequality to show that U(f, R)− L(f, R) < ε.

Solution.(a) Suppose the contrary. That is,∫ b

a

f(x)dx− ε

2≥ L(f, P )

for all partition of [a, b]. This means that∫ b

af(x)dx − ε

2is an upper bound

of SL. But∫ b

af(x)dx is the smallest upper bound of SL. Hence,

∫ b

af(x)dx <∫ b

af(x)dx− ε

2, which is impossible.

(b) Suppose the contrary. That is,

U(f, Q) ≥∫ b

a

f(x)dx +ε

2

for all partition of [a, b]. This means that∫ b

af(x)dx + ε

2is a lower bound of

SU . But∫ b

af(x)dx is the largest lower bound of SU . Hence,

∫ b

af(x)dx + ε

2<∫ b

af(x)dx, which is impossible.

(c) We have∫ b

a

f(x)dx− ε

2=

∫ b

a

f(x)dx− ε

2

<L(f, P ) < L(f, R) ≤ U(f, R) ≤ U(f, Q)

<

∫ b

a

f(x)dx +ε

2

=

∫ b

a

f(x)dx− ε

2

(d) We conclude from (c) that∣∣∣L(f, R)−∫ b

af(x)dx

∣∣∣ < ε2

and∣∣∣U(f, R)−

∫ b

af(x)dx

∣∣∣ < ε2

124

(e) By the triangle inequality we have U(f, R) − L(f, R) = |U(f, R) −L(f, R)| =

∣∣∣(U(f, R)−∫ b

af(x)dx) + (

∫ b

af(x)dx− L(f, R))

∣∣∣ ≤ ∣∣∣U(f, R)−∫ b

af(x)dx

∣∣∣+∣∣∣∫ b

af(x)dx− L(f, R)

∣∣∣ < ε2

+ ε2

= ε

Exercise 20.7Let f : [a, b] → R be a bounded function. Suppose that f satisfies property(P) above.(a) Show that for each positive integer n, there is a partition Pn such that

U(f, Pn)− L(f, Pn) <1

n.

(b) Using (a), show that

L(f, Pn) ≤∫ b

a

f(x)dx ≤∫ b

a

f(x)dx < L(f, Pn) +1

n.

(c) Show that

0 ≤∫ b

a

f(x)dx−∫ b

a

f(x)dx <1

n.

(d) Show that ∫ b

a

f(x)dx =

∫ b

a

f(x)dx.

Hint: Squeeze rule. We conclude that any bounded function that satisfiesproperty (P) is Riemann integrable.

Solution.(a) This follows from property (P).(b) We have

L(f, Pn) ≤∫ b

a

f(x)dx ≤∫ b

a

f(x)dx

≤U(f, Pn) < L(f, Pn) +1

n

(c) Using (b) we have

0 ≤∫ b

a

f(x)dx−∫ b

a

f(x)dx < L(f, Pn) +1

n− L(f, Pn) =

1

n.

(d) This follows from the squeeze rule

125

Exercise 20.8Let f : [0, 1] → R be the function f(x) = x2. For any ε > 0, choose apartition P = {0 = x0 < x1 < · · · < xn = 1} such that

xi − xi−1 < ε2

for all 1 ≤ i ≤ n

Show that U(f, P )− L(f, P ) < ε. Hence, f is Riemann integrable.

Solution.For 1 ≤ i ≤ n we have Mi(f) = x2

i and mi(f) = x2i−1. Then U(f, P ) =∑n

i=1 x2i (xi − xi−1) and L(f, P ) =

∑ni=1 x2

i−1(xi − xi−1). Hence,

U(f, P )− L(f, P ) =n∑

i=1

(x2i − x2

i−1)(xi − xi−1)

=n∑

i=1

(xi + xi−1)(xi − xi−1)(xi − xi−1)

<n∑

i=1

2( ε

2

)(xi − xi−1)

≤n∑

i=1

ε(xi − xi−1) = ε

Exercise 20.9Suppose that f(x) = x for x ∈ [1, 2].(a) Find U(f, P ) and L(f, P ). Hint: Consider a partition with equal subin-tervals.(b) Show that f is Riemann integrable. Hint: Exercise 20.7.(c) Show that U(f, P ) ≥ 3

2and L(f, P ) ≤ 3

2.

(d) Find∫ 2

1xdx.

Solution.(a) Let P = {1 = x0 < x1 < · · · < xn = 2} be a partition of [1, 2] withxi = 1 + i

n. Then for 1 ≤ i ≤ n we have mi(f) = 1 + i−1

nand Mi(f) = 1 + i

n.

126

Hence,

L(f, P ) =n∑

i=1

mi(f)(xi − xi−1) =n∑

i=1

(1 +

i− 1

n

)· 1

n

=1

n

(n +

1

n

n∑i=1

(i− 1)

)

=1

n

(n +

1

n

(n(n + 1)

2− n

))=1 +

n− 1

2n

and

U(f, P ) =n∑

i=1

Mi(f)(xi − xi−1) =n∑

i=1

(1 +

i

n

)· 1

n

=1

n

(n +

1

n

n∑i=1

i

)

=1

n

(n +

1

n· n(n + 1)

2

)=1 +

n + 1

2n

(b) Let ε > 0 be given. Choose n large enough so that n > 1ε. Let P = {1 =

x0 < x1 < · · · < xn = 2} be a partition of [1, 2] with xi = 1 + in. Then

U(f, P )− L(f, P ) =1

n< ε.

By Exercise 20.7, f is Riemann integrable.(c) Since n+1

2nis a decreasing function of n and converges to 1

2, we must have

U(f, P ) ≥ 32. Likewise, since n−1

2nis increasing and converges to 1

2we must

have L(f, P ) ≤ 32. Hence ∫ b

a

xdx ≤ 3

2≤∫ b

a

xdx.

(c) From (c) we have∫ 2

1

xdx =

∫ 2

1

xdx =

∫ 2

1

xdx =3

2

127

Exercise 20.10Let f : [a, b] → R be bounded. Let P and Q be any two partitions of [a, b].Prove that L(f, P ) ≤ U(f, Q).

Solution.Let R = P ∪Q. Then L(f, R) ≤ L(f, P ) ≤ U(f, P ) ≤ U(f, R) ≤ U(f, Q)

Exercise 20.11Let f : [a, b] → R be such that m ≤ f(x) ≤ M for all x ∈ [a, b]. Prove that∫ b

a

f(x)dx−∫ b

a

f(x)dx ≤ (M −m)(b− a).

Solution.Let P be a paritition of [a, b]. Then L(f, p) ≤ M(b−a) and U(f, P ) ≥ m(b−a). Thus, M(b − a) is an upper bound of SL so that

∫ b

af(x)dx ≤ M(b − a).

Likewise, m(b − a) is a lower bound of SU so that m(b − a) ≤∫ b

af(x)dx.

Hence,∫ b

af(x)dx−

∫ b

af(x)dx ≤ (M −m)(b− a)

Exercise 20.12Let f : [a, b] → R be bounded functions such that f(x) ≤ g(x) for allx ∈ [a, b]. Prove the following:

(a)∫ b

af(x)dx ≤

∫ b

ag(x)dx

(b)∫ b

af(x)dx ≤

∫ b

ag(x)dx

Solution.(a) We prove (a) since (b) is similar. Let P be any partition of [a, b]. Then

L(f, P ) ≤ L(g, P ) ≤∫ b

ag(x)dx. That is,

∫ b

ag(x)dx is an upper bound of

SL(f). This implies that∫ b

af(x)dx ≤

∫ b

ag(x)dx

Exercise 20.13Let f : [a, b] → R be bounded functions. Let P be any partition of [a, b].Prove

U(f + g, P ) ≤ U(f, P ) + U(g, P ).

128

Solution.Let P = {a = x0 < x1 < · · · < xn = b}. Then for all x ∈ [xi−1, xi] we have

f(x) + g(x) ≤ Mi(f) + Mi(g).

This implies thatMi(f + g) ≤ Mi(f) + Mi(g).

Hence,U(f + g, P ) ≤ U(f, P ) + U(g, P )

Exercise 20.14Let f : [a, b] → R be Riemann integrable. Prove that there is a sequence of

partitions {Pn}∞n=1 such that limn→∞ U(f, Pn) = limn→∞ L(f, Pn) =∫ b

af(x)dx.


Exercise 20.15Consider the function f : [0, 1] → R defined by f(x) = ax+b where a > 0 andb > 0. Assume that this function is Riemann integrable. For each positiveinteger n consider the partition Pn = {0 = x0 < x1 < · · · < xn = 1} withequal subintervals.(a) Compute L(f, Pn) and U(f, Pn).

(b) Show that∫ 1

0f(x)dx = a

2+ b.


L(f, Pn) =n∑

i=1

(axi−1 + b)1

n=

a

n

n∑i=1

i− 1

n+ b =

a

2

n− 1

n+ b

and

U(f, Pn) =n∑

i=1

(axi + b)1

n=

a

n

n∑i=1

i

n+ b =

a

2

n + 1

n+ b

(b) We have∫ 1

0

f(x)dx = limn→∞

L(f, Pn) = limn→∞

U(f, Pn) =a

2+ b

129

Exercise 21.1Let f : [a, b] → R be an increasing function on [a, b].(a) Show that f is bounded on [a, b].

(b) Let ε > 0 be given. Choose a positive integer N such that f(b)−f(a)N

< ε.Let P = {a = x0 < x1 < · · · < xn = b} be a partition of [a, b] such thatxi−xi−1 < 1

Nfor all 1 ≤ i ≤ n. For each 1 ≤ i ≤ n, express Mi(f) and mi(f)

in terms of f(x).(c) Show that U(f, P ) − L(f, P ) < ε. Thus, conclude that f is Riemannintegrable.

Solution.(a) Since f is increasing, we must have f(a) ≤ x ≤ f(b) for all x ∈ [a, b]. Letm = f(a) and M = f(b). Then f is bounded.(b) Since f is increasing, we must have Mi(f) = f(xi) and mi(f) = f(xi−1)for 1 ≤ i ≤ n.(c) We have

U(f, P )− L(f, P ) =n∑

i=1

[f(xi)− f(xi−1)](xi − xi−1)

<n∑

i=1

[f(xi)− f(xi−1)]

N

=f(b)− f(a)

N< ε

It follows from Exercise 20.7 that f is Riemann integrable

Exercise 21.2Let f : [a, b] → R be a continuous function on [a, b].(a) Show that there exist numbers m and M such that m ≤ f(x) ≤ M forall a ≤ x ≤ b. That is, f is bounded on [a, b].(b) Show that f is uniformly continuous on [a, b].(c) Let ε > 0. Show that there is a positive number δ > 0 such that if|u− v| < δ then |f(u)− f(v)| < ε

b−a.

(d) Choose a partition P = {a = x0 < x1 < · · · < xn = b} such thatxi − xi−1 < δ for all 1 ≤ i ≤ n. Show that for each interval [xi, xi−1] thereexist si, ti ∈ [xi, xi−1] such that Mi(f) = f(ti) and mi(f) = f(si). Hint:

130

Exercise 17.4.(e) Show that Mi(f)−mi(f) < ε

b−afor each 1 ≤ i ≤ n.

(f) Using (e), show that U(f, P ) − L(f, P ) < ε. Hence, conclude that f isRiemann integrable.

Solution.(a) By Exercise 17.4 there exist x1, x2 ∈ [a, b] such that f(x1) ≤ f(x) ≤ f(x2)for all a ≤ x ≤ b. Let m = f(x1) and M = f(x2).(b) This follows from Exercise 14.5.(c) This follows from the fact that f is uniformly continuous.(d) This follows from Exercise 17.4.(e) Since |ti − si| < δ, we have |f(ti) − f(si)| < ε

b−awhich implies − ε

b−a<

f(ti)− f(si) < εb−a

. Thus, Mi(f)−mi(f) = f(ti)− f(si) < εb−a

.(f) We have U(f, P )−L(f, P ) =

∑ni=1[f(ti)−f(si)](xi−xi−1) <

∑ni=1

εb−a

(xi−xi−1) = ε. We conclude that f is Riemann integrable

Exercise 21.3Suppose f is continuous except at a point c in [a, b]. Let ε > 0 be given andconsider a partition Q = {a = x0 < x1 < · · · < xk−1 < c < xk+1 < · · · <xn = b} such that µ(Q) < ε

12M.

(a) Prove that |xk−1 − xk + 1| < ε6M

.(b) Show that there exist δ′ > 0 and δ′′ > 0 such that for all x, y ∈ [a, xk−1]with |x − y| < δ′ we have |f(x) − f(y)| < ε

3(b−a)and for all x, y ∈ [xk+1, b]

with |x− y| < δ′′ we have |f(x)− f(y)| < ε3(b−a)

.

(c) Let P1 be a refinement of Q on [a, xk−1] such that µ(P1) < δ′ and P2 bea refinement of P on [xk+1, b] such that µ(P2) < δ′′. Let P = P1 ∪ P2. Thenwe have

U(f, P )− L(f, P ) =k−1∑i=1

(Mi −mi)(xi − xi−1) + (Mk −mk)(c− xk−1)

+(Mk+1 −mk+1(xk+1 − c) +n∑

i=k+2

(Mi −mi)(xi − xi−1)

Show thatk−1∑i=1

(Mi −mi)(xi − xi−1) <ε

3

(Mk −mk)(c− xk−1) + (Mk+1 −mk+1(xk+1 − c) <ε

3

131

andn∑

i=k+2


3

(d) Conclude that U(f, P ) − L(f, P ) < ε and therefore f is Riemann inte-grable.

Solution.(a) We have |xk−1 − xk+| ≤ |c− xk−1|+ |xk+1 − c| < ε

12M+ ε

12M= ε

6M.

(b) This follows from the fact that f is uniformly continuous on [a, xk−1] and[xk+1, b].(c) We have

k−1∑i=1


3(b− a)(xk−1 − a) <

ε

3

(Mk−mk)(c−xk−1)+(Mk+1−mk+1(xk+1−c) < 2M(xk+1−xk) < 2Mε

6M=

ε

3n∑

i=k+2


3(b− a)(b− xk+1) <

ε

3

(d) Taking everything together we have:

U(f, P )− L(f, P ) <ε

3+

ε

3+

ε

3= ε.

by Exercise 20.7, we conclude that f is Riemann integrable

Exercise 21.4Suppose f is continuous except at points c1, c2, · · · , cn in [a, b]. We wantto show that f is Riemann integrable on [a, b]. The proof is by inductionon n. For n = 1 the result holds by the previous exercise. Suppose thatthe result holds for c1, c2, · · · , cn. Suppose that f is continuous except atc1 < c2 < · · · < cn < cn+1. Let ε > 0. Choose δ > 0 small enough so thatδ < ε

8Mand (cn+1 − δ, cn+1 + δ) ⊂ [cn, b].

(a) Show that there is a partition P1 of [a, cn+1 − δ] such that U(f, P1) −L(f, P1) < ε

4and a partition P2 of [cn+1, b] such that U(f, P2)−L(f, P2) < ε

4.

(b) Let P = P1 ∪P2. Show that U(f, P )−L(f, P ) < ε. Hence, f is Riemannintegrable on [a, b].

132

Solution.(a) By the induction hypothesis f is Riemann integrable on [a, cn+1−δ]. Sincef is continuous on [cn+1 + δ, b], it is integrable there. Then by Exercise 20.6we can find a partition P1 of [a, cn+1 − δ] such that U(f, P1)− L(f, P1) < ε

4

and a partition P2 of [cn+1, b] such that U(f, P2)− L(f, P2) < ε4.

(b) Let M ′ be the supremum of f and m′ the infimum of f on [cn+1−δ, cn+1+δ]. We have

U(f, P )− L(f, P ) =U(f, P1)− L(f, P1) + (M ′ −m′)(2δ) + U(f, P2)− L(f, P2)

<ε

4+ 2M(2)

ε

8M+

ε

4= ε

Thus, f is Riemann integrable on [a, b]

Exercise 21.5Let f : [a, b] → R be a increasing function on [a, b].(a) Show that f is bounded on [a, b].

(b) Let ε > 0 be given. Choose a positive integer N such that f(a)−f(b)N

< ε.Let P = {a = x0 < x1 < · · · < xn = b} be a partition of [a, b] such thatxi−xi−1 < 1

Nfor all 1 ≤ i ≤ n. For each 1 ≤ i ≤ n, express Mi(f) and mi(f)

in terms of f(x).(c) Show that U(f, P ) − L(f, P ) < ε. Thus, conclude that f is Riemannintegrable.

Solution.(a) Since f is decreasing, we must have f(b) ≤ x ≤ f(a) for all x ∈ [a, b]. Letm = f(b) and M = f(a). Then f is bounded.(b) Since f is decreasing, we must have Mi(f) = f(xi−1) and mi(f) = f(xi)for 1 ≤ i ≤ n.(c) We have

U(f, P )− L(f, P ) =n∑

i=1

[f(xi−1)− f(xi)](xi − xi−1)

<

n∑i=1

[f(xi−1)− f(xi)]

N

=f(a)− f(b)

N< ε

It follows from Exercise 20.7 that f is Riemann integrable

133

Exercise 21.6Suppose f : [a, b] → R is continuous and f ≥ 0 on [a, b]. Let [c, d] ⊂ [a, b].

Prove that∫ b

af(x)dx ≥

∫ d

cf(x)dx.

Solution.Let P = {c = x0 < x1 < · · · < xn = d} be a partition of [c, d]. Since

f ≥ 0 We have L(f, P ) ≤∫ b

af(x)dx =

∫ b

af(x)dx. Since P was arbitrary, we

can say that∫ b

af(x)dx is an upper bound of SL([c, d]). Hence,

∫ d

cf(x)dx =∫ d

cf(x)dx ≤

∫ b

af(x)dx

Exercise 21.7(a) Suppose f : [0, 1] → R is continuous and f ≥ 0 on [0, 1]. Let a ∈ [0, 1] be

such that f(a) > 0. Show that∫ 1

0f(x)dx > 0.

(b) Construct a nonnegative function f on [0, 1] such that f(0.5) > 0 but∫ 1

0f(x)dx = 0.

Solution.(a) Let ε = f(a)

2> 0. Since f is continuous at a we can find a δ > 0 such that

if |x− a| < δ then |f(x)− f(a)| < ε = f(a)2

. Using the triangle inequality we

end up with |f(x) > f(a)4

> 0 for all x ∈ (a − δ, a + δ). Thus,∫ 1

0f(x)dx >∫ a+δ

a−δf(x)dx >

∫ a+δ

a−δf(a)

4dx = f(a)δ

4> 0.

(b) Let

f(x) =

{0 if x 6= 1

2

1 if x = 12.

Then f(0.5) = 1 > 0 and∫ 1

0f(x)dx = 0

Exercise 21.8Suppose that f : [a, b] → R is differentiable on [a, b]. Prove that f is Riemannintegrable on [a, b].

Solution.Since f is differentiable, f is continuous; this implies f is integrable

Exercise 21.9Let f : [a, b] → R be defined by

f(x) =

{1 if x is rational−1 if x is irrational

134

(a) Prove that f is not Riemann integrable on [a, b]. Hint Show that the lowerRiemann integral is different from the upper Riemann integral.(b) Prove that |f | is Riemann integrable.

Solution.(a) Since between any two real numbers we can find a rational number andan irrational number, we can write U(f, P ) = b− a and L(f, P ) = a− b forall P. Thus ∫ b

af(x)dx = b− a and

∫ b

af(x)dx = a− b

It follows that f is not integrable.(b) Since |f(x)| = 1 for all x ∈ [a, b], |f | is continuous on [a, b] and thereforeintegrable

Exercise 21.10Suppose f is a continuous function on [a, b] and that f(x) ≥ 0 for all x ∈ [a, b].

Show that if∫ b

af(x)dx = 0, then f(x) = 0 for all x ∈ [a, b]. Hint: Assume

the contrary and get a contradiction.

Solution.Suppose that there is an c ∈ [a, b] such that f(c) > 0. Let ε = f(c)

2> 0.

Since f is continuous at c we can find a δ > 0 such that if |x − c| < δ

then |f(x) − f(c)| < ε = f(c)2

. Using the triangle inequality we end up with

|f(x) > f(c)4

> 0 for all x ∈ (c − δ, c + δ). Thus,∫ 1

0f(x)dx >

∫ c+δ

c−δf(x)dx >∫ c+δ

c−δf(c)4

dx = f(c)δ4

> 0 which is a contradiction. Hence, f(x) = 0 for allx ∈ [a, b]

135

Exercise 22.1Let f : [a, b] → R be a bounded function. Suppose that limµ(P )→0 S(f, P ) =A.(a) Let ε > 0. Show that there is a δ > 0 such that for any partition P of[a, b] such that µ(P ) < δ we must have |S(f, P )− A| < ε

4.

(b) Let Q = {a = x0 < x1 < · · · < xn = b} be a partition of [a, b] such thatµ(Q) < δ, that is, xi − xi−1 < δ for all 1 ≤ i ≤ n. Fix 1 ≤ i ≤ n. Showthat if f(ui) ≥ mi(f) + ε

4(b−a)for all xi−1 ≤ ui ≤ xi then this contradicts the

definition of mi(f).(c) With Q as above, show that if f(vi) ≤ Mi(f)− ε

4(b−a)for all xi−1 ≤ vi ≤ xi

then this contradicts the definition of Mi(f).(d) Show that for every 1 ≤ i ≤ n, there exists ui, vi ∈ [xi−1, xi] such thatf(ui) < mi(f) + ε

4(b−a)and f(vi) > Mi(f)− ε

4(b−a)

(e) Show that∑n

i=1 f(ui)(xi−xi−1) < L(f, Q)+ ε4

and∑n

i=1 f(vi)(xi−xi−1) >U(f, Q)− ε

4

(f) Show that

A− ε4

<∑n

i=1 f(ui)(xi − xi−1) < A + ε4

andA− ε

4<∑n

i=1 f(vi)(xi − xi−1) < A + ε4

(g) Use (f) to show that

A− ε

2< L(f, Q) ≤ U(f, Q) < A +

ε

2.

(h) Show that U(f, Q)− L(f, Q) < ε. That is, f is Riemann integrable.(i) Show that ∣∣∣∣∫ b

a

f(x)dx− A

∣∣∣∣ < ε.

(k) Use the Squeeze rule to show that∫ b

af(x)dx = A.

Conclusion: Suppose that there is a number A such that limµ(P )→0 S(f, P ) =

A. Then f is Riemann integrable with∫ b

af(x)dx = A.

Solution.(a) This follows from the hypothesis that limµ(P )→0 S(f, P ) = A.(b) Let Q = {a = x0 < x1 < · · · < xn = b} be a partition of [a, b] such thatµ(Q) < δ, that is, xi − xi−1 < δ for all 1 ≤ i ≤ n. If f(ui) ≥ mi(f) + ε

4(b−a)

136

for all xi−1 ≤ ui ≤ xi then mi(f) + ε4(b−a)

is a lower bound of f in [xi−1, xi]

so by the definition of mi(f) we must have mi(f) + ε4(b−a)

< mi(f), which isimpossible.(c) If f(vi) ≤ Mi(f) − ε

4(b−a)for all xi−1 ≤ vi ≤ xi then Mi(f) − ε

4(b−a)is

an upper bound of f on [xi−1, xi]. By the definition of Mi(f), we must haveMi(f) ≤ Mi(f)− ε

4(b−a), which is impossible.

(d) This follows from (b) and (c).(e) We have

n∑i=1

f(ui)(xi − xi−1) <

n∑i=1

mi(f)(xi − xi−1) +ε

4

n∑i=1

xi − xi−1

b− a= L(f, Q) +

ε

4.

Similar argument for the second part of the question.(f) Since

∑ni=1 f(ui)(xi − xi−1) and

∑ni=1 f(vi)(xi − xi−1) are Riemann sums

with µ(Q) < δ they are within ε4

of A.(g) we have A − ε

2<∑n

i=1 f(ui)(xi − xi−1) − ε4≤ L(f, Q) ≤ U(f, Q) ≤∑n

i=1 f(vi)(xi − xi−1) + ε4

< A + ε2.

(h) Since both U(f, Q) and L(f, Q) are inside the interval centered at A andof length at most ε, we must have U(f, Q)− L(f, Q) < ε. By Exercise ??, fis Riemann integrable.

(i) Since L(f, Q) ≤∫ b

af(x)dx ≤ U(f, Q), we must have

∣∣∣∫ b

af(x)dx− A

∣∣∣ < ε.

(k) Let ε → 0 and apply the Squeeze rule

Exercise 22.2Prove that A− L(f, R) < ε

2and U(f, R)− A < ε

2.

Solution.by Exercise ?? we have L(f, R) ≥ L(f, P ) and U(f, R) ≤ U(f, P ). Thus,A− L(f, R) ≤ A− L(f, P ) < ε

2and U(f, R)− A ≤ U(f, P )− A < ε

2

Exercise 22.3(a) For 1 ≤ i ≤ m such that L(f, Ri)−mi(zi− zi−1) 6= 0 and Mi(zi− zi−1)−U(f, Ri) prove that

L(f, Ri)−mi(zi − zi−1) < 2Mδ and Mi(zi − zi−1)− U(f, Ri) < 2Mδ.

(b) Use (a) and the sums above to show that

L(f, R)− L(f, Q) < ε2

and U(f, Q)− U(f, R) < ε2

>

137

L(f, Ri)−mi(zi − zi−1) ≤ 2M(zi − zi−1) < 2Mδ

andMi(zi − zi−1)− U(f, Ri) ≤ 2M(zi − zi−1) < 2Mδ

(b) Because there are at most n− 1 such terms, we obtain the bounds

L(f, R)− L(f, Q) < 2nMδ =ε

2

andU(f, Q)− U(f, R) < 2nMδ =

ε

2

Exercise 22.4Use Exercise 22.2 and 22.3 to prove that

U(f, Q) < A + ε and L(f, Q) > A− ε.

Solution.We have

U(f, Q)− A = [U(f, Q)− U(f, R)] + [U(f, R)− A] <ε

2+

ε

2

and

A− L(f, Q) = [A− L(f, R)] + [L(f, R)− L(f, Q)] <ε

2+

ε

2= ε

Exercise 22.5Using the previous problem, show that

A− ε < S(f, Q) < A + ε.

That is,|S(f, Q)− A| < ε.

Solution.We have A− ε ≤ L(f, Q) ≤ S(f, Q) ≤ U(f, Q) < A + ε

138

Exercise 22.6Suppose that f : [a, b] → R is bounded and Riemann integrable. The goal ofthis problem is to show that for any sequence {Pn}∞n=1 of partitions of [a, b]

such that limn→∞ µ(Pn) = 0 we have limn→∞ S(f, Pn) =∫ b

af(x)dx.

(a) Let ε > 0. Show that there is a δ > 0 such that if P is a partition of [a, b]with µ(P ) < δ we have ∣∣∣∣S(f, P )−

∫ b

a

f(x)dx

∣∣∣∣ < ε.

(b) Show that there is a positive integer N such that if n ≥ N then µ(Pn) < δ.(c) Use (a) and (b) to conclude that for n ≥ N we have∣∣∣∣S(f, Pn)−

∫ b

a

f(x)dx

∣∣∣∣ < ε.

Hence,

limn→∞

S(f, Pn) =

∫ b

a

f(x)dx.

Solution.(a) This follows from Exercise 16.6.(b) This follows from the definition of convergence of a sequence.(c) If n ≥ N then µ(Pn) < δ and this implies by (a) that∣∣∣∣S(f, Pn)−

∫ b

a

f(x)dx

∣∣∣∣ < ε

Exercise 22.7Let f : [a, b] → R be bounded and Riemann integrable. Let ε > 0 be given.Show that there is a δ > 0 such that for any partition P of [a, b] with µ(P ) < δwe have

U(f, P )− L(f, P ) < ε.


Exercise 22.8Suppose that f : [a, b] → R is differentiable in [a, b] and that f ′ : [a, b] → R isRiemann integrable. Let Pn = {a = x0 < x1 < · · · < xn = b} be a partition

139

of [a, b] such that xi − xi−1 = b−an

.(a) For each 1 ≤ i ≤ n, show that there exists xi−1 < ti < xi such thatf(xi)− f(xi−1) = f ′(ti)(xi − xi−1).(b) Show that S(f ′, Pn) =

∑ni=1 f ′(ti)(xi − xi−1) = f(b)− f(a).

(c) Show that limn→∞ µ(Pn) = 0.

(d) Show that limn→∞ S(f ′, Pn) =∫ b

af ′(x)dx.

(e) Show that ∫ b

a

f ′(x)dx = f(b)− f(a).

Solution.(a) This follows from the Mean Value Theorem.(b) We

∑ni=1 f ′(ti)(xi − xi−1) = (f(x1) − f(a)) + (f(x2) − f(x1)) + · · · +

(f(xn−1)− f(xn−2) + (f(b)− f(xn−1)) = f(b)− f(a).(c) We have limn→∞ µ(Pn) = limn→∞

b−an

= 0.(d) This follows from the previous exercise.(e) We have∫ b

a

f ′(x)dx = limn→∞

S(f ′, Pn) = limn→∞

(f(b)− f(a)) = f(b)− f(a)

Exercise 22.9 (Fundamental Theorem of Calculus)Suppose that f : [a, b] → R is continuous and let F : [a, b] → R be adifferentiable function such that F ′(x) = f(x) for all a ≤ x ≤ b. Show that∫ b

a

f(x)dx = F (b)− F (a).

The function F (x) is called an antiderivative of f.

Solution.Using the previous problem we see that∫ b

a

f(x)dx =

∫ b

a

F ′(x)dx = F (b)− F (a)

140

Exercise 23.1Let f, g : [a, b] → R be Riemann integrable functions and α, β be real num-bers. Let ε > 0.(a) Show that there is a δ1 > 0 such that if µ(P ) < δ1 then

∣∣∣S(f, P )−∫ b

af(x)dx

∣∣∣ <ε

|α|+|β| .

(b) Show that there is a δ2 > 0 such that if µ(P ) < δ2 then∣∣∣S(g, P )−

∫ b

ag(x)dx

∣∣∣ <ε

|α|+|β| .

(c) Show that there is a δ > 0 such that if µ(P ) < δ then∣∣∣∣S(αf + βg, P )−[α

∫ b

a

f(x)dx + β

∫ b

a

g(x)dx

]∣∣∣∣ < ε.

We conclude that αf + βg is Riemann integrable and∫ b

a

(αf(x) + βg(x))dx = α

∫ b

a

f(x)dx + β

∫ b

a

g(x)dx.

Solution.(a) Follows from Exercises 22.2 - 22.5.(b) Follows from Exercises 22.2 - 22.5.(c) Let δ = min{δ1, δ2}. Suppose that µ(P ) < δ. Then µ(P ) < δ1 andµ(P ) < δ2. Moreover, we have∣∣∣∣S(αf + βg, P )−

[α

∫ b

af(x)dx + β

∫ b

ag(x)dx

]∣∣∣∣=∣∣∣∣αS(f, P ) + βS(g, P )−

[α

∫ b

af(x)dx + β

∫ b

ag(x)dx

]∣∣∣∣≤|α|

∣∣∣∣S(f, P )−∫ b

af(x)dx

∣∣∣∣+ |β|∣∣∣∣S(g, P )−

∫ b

ag(x)dx

∣∣∣∣<|α| ε

|α|+ |β|+ |β| ε

|α|+ |β|= ε.

By Exercise 22.1, αf + βg is Riemann integrable and∫ b

a

(αf(x) + βg(x))dx = α

∫ b

a

f(x)dx + β

∫ b

a

g(x)dx

141

Exercise 23.2Let f, g : [a, b] → R be Riemann integrable functions such that f(x) ≤ g(x)for all x ∈ [a, b].(a) Show that for any partition P of [a, b] we have L(f, P ) ≤ L(g, P ).

(b) Show that∫ b

af(x)dx ≤

∫ b

ag(x)dx.

(c) Show that∫ b

af(x)dx ≤

∫ b

ag(x)dx.

Solution.(a) Since f(x) ≤ g(x) for all x ∈ [a, b], we have mi(f) ≤ mi(g) for all1 ≤ i ≤ n. Thus,

∑ni=1 mi(f)(xi − xi−1) ≤

∑ni=1 mi(g)(xi − xi−1), that is,

L(f, P ) ≤ L(f, g).(b) From (b), we have

L(f, P ) ≤ L(g, P ) ≤∫ b

a

g(x)dx

for all partitions P of [a, b]. Thus,∫ b

ag(x)dx is an upper bound for SL =

{L(f, P ) : P a partition of [a, b}. But∫ b

af(x)dx is the smallest upper bound

so that ∫ b

a

f(x)dx ≤∫ b

a

g(x)dx.

(c) This follows from the fact that∫ b

af(x)dx =

∫ b

af(x)dx and

∫ b

ag(x)dx =

∫ b

ag(x)dx

Exercise 23.3Let f : [a, b] → R be a Riemann integrable function such that m ≤ f(x) ≤ Mfor all x ∈ [a, b].(a) Show that m(b − a) ≤ L(f, P ) ≤ U(f, P ) ≤ M(b − a) for any partitionP of [a, b].

(b) Show that∫ b

af(x)dx =

∫ b

af(x)dx ≤ M(b− a).

(c) Show that m(b− a) ≤∫ b

af(x)dx =

∫ b

af(x)dx.

Conclusion: m(b− a) ≤∫ b

af(x)dx ≤ M(b− a).

Solution.(a) Let P = {a = x0 < x1 < · · · < xn = b}. Then

m(xi − xi−1) ≤ mi(f)(xi − xi−1) ≤ Mi(f)(xi − xi−1) ≤ M(xi − xi−1).

142

Summing from i = 1 to i = n to obtain

m(b− a) ≤ L(f, P ) ≤ U(f, P ) ≤ M(b− a)

(b) M(b − a) is an upper bound of SL. But∫ b

af(x)dx is the smallest upper

bound so that ∫ b

a

f(x)dx ≤ M(b− a).

Since f is Riemann integrable, we have∫ b

a

f(x)dx =

∫ b

a

f(x)dx ≤ M(b− a).

(c) m(b−a) is a lower bound of SU . But∫ b

af(x)dx is the largest lower bound

so that

m(b− a) ≤∫ b

a

f(x)dx.

Since f is Riemann integrable, we have

m(b− a) ≤∫ b

a

f(x)dx =

∫ b

a

f(x)dx

Exercise 23.4Let f : [a, b] → R be a Riemann integrable function and a < c < b.(a) Let ε > 0. Show that there is a partition P of [a, b] such that U(f, P )−L(f, P ) < ε.(b) Let Q = P ∪ {c}, Q1 = Q ∩ [a, c], and Q2 = Q ∩ [c, b]. That is, Q ispartition of [a, b], Q1 is a partition of [a, c], and Q2 is a partition of [c, b].Show that

[U(f, Q1)− L(f, Q1)] + [U(f, Q2)− L(f, Q2)] < ε.

(c) Show that U(f, Q1) − L(f, Q1) < ε. Thus, by Exercise 21.1,∫ c

af(x)dx

exists and is finite.(d) Show that U(f, Q2) − L(f, Q2) < ε. Thus, by Exercise 21.1,

∫ b

cf(x)dx

exists and is finite.

143

Solution.(a) This follows from Exercise 21.1.(b) We have

[U(f, Q1)− L(f, Q1)] + [U(f, Q2)− L(f, Q2)] =[U(f, Q1) + U(f, Q2]

−[L(f, Q1)] + L(f, Q2)]

=U(f, Q)− L(f, Q)

≤U(f, P )− L(f, P ) < ε

(c) Since U(f, Q1)−L(f, Q1) > 0 we must have U(f, Q1)−L(f, Q1) < ε. ByExercise 21.1, f is Riemann integrable in [a, c].(d) Since U(f, Q2)−L(f, Q2) > 0 we must have U(f, Q2)−L(f, Q2) < ε. ByExercise 21.1, f is Riemann integrable in [c, b]

Exercise 23.5Let f : [a, b] → R be a Riemann integrable function and a < c < b. Let ε > 0.(a) Show that there is a δ1 > 0 such that if P1 is a partition of [a, c] suchthat µ(P1) < δ1 then

∣∣S(f, P1)−∫ c

af(x)dx

∣∣ < ε2.

(b) Show that there is a δ2 > 0 such that if P2 is a partition of [c, b] such

that µ(P2) < δ2 then∣∣∣S(f, P2)−

∫ b

cf(x)dx

∣∣∣ < ε2.

(c) Let P = P1 ∪ P2. Then P is a partition of [a, b]. Show that there is δ > 0such that µ(P ) < δ and∣∣∣∣S(f, P )−

[∫ c

a

f(x)dx +

∫ b

c

f(x)dx

]∣∣∣∣ < ε.

That is, ∫ b

a

f(x)dx =

∫ c

a

f(x)dx +

∫ b

c

f(x)dx.

Solution.(a) By the previous exercise, f is Riemann integrable in [a, c]. Now, the resultfollows from Exercises 22.2 - 22.5.(b) Similar to (a).

144

(c) Let δ = min{δ1, δ2}. Then µ(P ) < δ. Also, we have∣∣∣∣S(f, P )−[∫ c

a

f(x)dx +

∫ b

c

f(x)dx

]∣∣∣∣≤∣∣∣∣S(f, P1)−

∫ c

a

f(x)dx

∣∣∣∣+ ∣∣∣∣S(f, P2)−∫ b

c

f(x)dx

∣∣∣∣<

ε

2+

ε

2= ε

Now the result follows from Exercise 22.1

Exercise 23.6Let f : [a, b] → R be continuous. Use the Intermediate Value Theorem toprove the existence of a number c ∈ [a, b] such that∫ b

a

f(x)dx = (b− a)f(c).

The number f(c) is called the average value of f on [a, b].

Solution.We know that m(b − a) ≤

∫ b

af(x)dx ≤ M(b − a) (Exercise 23.3). Let

d = 1b−a

∫ b

af(x)dx. Then d ∈ [m, M ]. By the IVT there is a number c ∈ [a, b]

such that f(c) = d. This implies that∫ b

a

f(x)dx = (b− a)f(c)

Exercise 23.7Suppose that f and g are continuous function on [a, b] such that

∫ b

af(x)dx =∫ b

ag(x)dx. Prove there is a c ∈ [a, b] such that f(c) = g(c).

Solution.Let h(x) = f(x) − g(x). Then h is continuous on [a, b]. By the previous

exercise, there is a c ∈ [a, b] such that 1b−a

∫ b

ah(x)dx = h(c). But

∫ b

ah(x)dx =

0. Thus, h(c) = 0 or f(c) = g(c)

Exercise 23.8(a) For any set S, one can see that M(f, S)−m(f, S) = sups,t∈S |f(s)−f(t)|.

145

Let f be a function defined on a set S. Show that M(|f |, S) − m(|f |, S) ≤M(f, S)−m(f, S).(b) Suppose that f : [a, b] → R is Riemann integrable. Show that |f is alsoRiemann integrable.

Solution.(a) For any s, t ∈ S we have

||f(s)| − |f(t)|| ≤ |f(s)− f(t)|.

It follows that

M(|f |, S)−m(|f |, S) = sups,t∈S

||f(s)|−|f(t)|| ≤ sups,t∈S

|f(s)−f(t)| = M(f, S)−m(f, S).

(b) Let ε > 0 be given. Then there is a partition P = {a = x0 < x1 < · · · <xn = b} such that U(f, P ) − L(f, P ) < ε. From part (a) we have that for1 ≤ i ≤ n

Mi(|f |)−mi(|f |) ≤ Mi(f)−mi(f).

Hence,U(|f |, P )− L(|f |, P ) < U(f, P )− L(f, P ) < ε.

That is, |f | is integrable

Exercise 23.9Let f : [a, b] → R be defined by

f(x) =

{1 if x ∈ Q−1 if x 6∈ Q

(a) Compute∫ b

af(x)dx and

∫ b

af(x)dx.

(b) Is f Riemann integrable?(c) Show that |f | is Riemann integrable.

Solution.Let P = {a = x0 < x1 < · · · < xn = b} be any partition of [a, b]. Thenmi = −1 and Mi = 1 for all 1 ≤ i ≤ n. Hence, L(f, P ) = a − b andU(f, P ) = b− a. It follows that∫ b

af(x)dx = a− b and

∫ b

af(x)dx = b− a.

146

(b) It follows from (a) that f is not Riemann integrable.(c) |f |(x) = 1 for all x ∈ [a, b]. Since |f | is a continuous function, |f | isintegrable on [a,b]

Exercise 23.10Let f : [a, b] → R be Riemann integrable with |f(x)| ≤ M for all x ∈ [a, b].(a) Prove that |f 2(x) − f 2(y)| ≤ 2M |f(x) − f(y)| for all x, y ∈ [a, b] wheref 2(x) = (f(x))2.(b) Let ε > 0. Show that there is a partition P of [a, b] such that

U(f, P )− L(f, P ) <ε

2M.

(c) Prove that U(f 2, P )− L(f 2, P ) < ε. That is, f 2 is Riemann integrable.

Solution.(a) We have |f 2(x)−f 2(y)| = |f(x)+f(y)||f(x)−f(y)| ≤ (|f(x)|+|f(y)|)|f(x)−f(y)| ≤ 2M |f(x)− f(y)|.(b) This follows from Exercise 20.6.(c) Let P = {a = x0 < x1 < · · · < xn = b}. We have

U(f 2, P )− L(f 2, P ) =n∑

i=1

sups,t∈[xi−1,xi]

|f 2(s)− f 2(t)|(xi − xi−1)

≤2Mn∑

i=1

sups,t∈[xi−1,xi]

|f(s)− f(t)|(xi − xi−1)

=2M(U(f, P )− L(f, P )) < 2M · ε

2M= ε

Exercise 23.11Let f, g : [a, b] → R be two Riemann integrable functions.(a) Show that

f · g =1

2[(f + g)2 − f 2 − g2].

(b) Prove that f · g is Riemann integrable.

Solution.(a) Trivial algebra.(b) Since f and g are integrable so are the functions f 2, g2, and (f + g)2

according to the previous problem. Now the result follows from Exercise23.1

147

Exercise 24.1Suppose that f : [a, b] → [c, d] is a Riemann integrable function on [a, b] andthat g : [c, d] → R is continuous (and hence integrable by Exercise ??).(a) Show that the set {|g(x)| : x ∈ [c, d]} is bounded. Hence, by the Com-pleteness Axiom of R there exists K > 0 such that K = sup{|g(x)| : x ∈[c, d]}.(b) Let ε > 0. Chosse ε′ so that ε′ < ε

b−a+2K. Show that there is a δ < ε′ such

that if |s− t| < δ, where s, t ∈ [c, d], then |g(s)− g(t)| < ε′.(c) Show that there is a partition P = {a = x0 < x1 < · · · < xn = b} of [a, b]such that U(f, P )− L(f, P ) < δ2.(d) Let A = {1 ≤ i ≤ n : Mi(f) − mi(f) < δ}. Show that if i ∈ A thenMi(g ◦ f)−mi(g ◦ f)| < ε′.(e) Let B = {1 ≤ i ≤ n : Mi(f)−mi(f) ≥ δ}. Show that δ

∑i∈B(xi−xi−1) <

δ2 and hence∑

i∈B(xi − xi−1) < δ.(f) Show that for all 1 ≤ i ≤ n we have Mi(g ◦ f) −mi(g ◦ f) < 2K. Hint:Use Exercise 15.8 and the triangle inequality.(g) Use (d) (e) and (f) to show that U(g ◦ f, P )−L(g ◦ f, P ) < ε. Hence, byExercise 20.7, g ◦ f : [a, b] → R is Riemann integrable.

Solution.(a) This follows from Exercise 15.3.(b) This is just the uniform continuity of g. See Exercise 14.5(d).(c) This follows from Exercise 20.6 since f is Riemann integrable on [a, b].(d) Let i ∈ A and s, t ∈ [xi−1, xi]. Since Mi(f) − mi(f) < δ, we must have|f(s)−f(t)| < δ. Now by (b) we have |g(f(s))−g(f(t))| < ε′ and this impliesthat Mi(g ◦ f)−mi(g ◦ f)| < ε′.(e) We have δ

∑i∈B(xi−xi−1) <

∑i∈B[Mi(f)−mi(f)](xi−xi−1) ≤ U(f, P )−

L(f, P ) < δ2. Divide both sides by δ to obtain the required result.(f) Note that by Exercise 15.8 we have Mi(g ◦ f) ∈ g([c, d]) and mi(g ◦ f) ∈g([c, d]). Hence, Mi(g ◦ f) − mi(g ◦ f) ≤ |Mi(g ◦ f) − mi(g ◦ f)| ≤ |Mi(g ◦f)|+ |mi(g ◦ f)| < K + K = 2K.

148

(g) We have

U(g ◦ f, P )− L(g ◦ f, P ) =n∑

i=1

[Mi(g ◦ f)−mi(g ◦ f)](xi − xi−1)

=∑i∈A

[Mi(g ◦ f)−mi(g ◦ f)](xi − xi−1)

+∑i∈B

[Mi(g ◦ f)−mi(g ◦ f)](xi − xi−1)

<ε′∑i∈A

(xi − xi−1) + 2K∑i∈B

(xi − xi−1)

<ε′n∑

i=1

(xi − xi−1) + 2Kδ

=ε′(b− a) + 2Kδ

<ε′(b− a) + 2Kε′

=ε′(b− a + 2K) < ε

Thus, by Exercise 20.7, g ◦ f is Riemann integrable

Exercise 24.2Let f, g : [a, b] → R be Riemann integrable and bounded such that |f(x)| ≤M1 and |g(x)| ≤ M2 for all x ∈ [a, b].(a) Find a positive constant M such that |f(x)| ≤ M and |g(x)| ≤ M. Thus,f([a, b]) ⊆ [−M, M ] and g([a, b]) ⊆ [−M, M ](b) Consider the continuous function h : [−2M, 2M ] → R given by h(x) = x2.Show that (f + g)2 and (f − g)2 are Riemann integrable on [a, b]. Hint: Notethat h ◦ (f + g) = (f + g)2 and h ◦ (f − g) = (f − g)2.(c) Show that f · g is Riemann integrable on [a, b].

Solution.(a) Let M = M1 + M2.(b) This follows from Exercise 24.4.(c) Since f · g = 1

4[(f + g)2 − (f − g)2], by Exercise 23.1 we conclude that

f · g is Riemann integrable on [a, b]

Exercise 24.3Let f : [a, b] → R be Riemann integrable and bounded such that |f(x)| ≤ M

149

for all x ∈ [a, b].(a) consider the continuous function g : [−M, M ] → R defined by g(x) = |x|.Show that |f | is Riemann integrable on [a, b].(b) Using the fact that −|f(x)| ≤ f(x) ≤ |f(x)| for all x ∈ [a, b], show that

−∫ b

a

|f(x)|dx ≤∫ b

a

f(x)dx ≤∫ b

a

|f(x)|dx.

Hence, show that ∣∣∣∣∫ b

a

f(x)dx

∣∣∣∣ ≤ ∫ b

a

|f(x)|dx.

Solution.(a) This follows from Exercise 24.4.(b) This follows from Exercise 23.2. The last equality follows from Exercise1.14

Exercise 24.4 (Integration by Parts)Let f, g : [a, b] → R be continuous and f ′, g′ : [a, b] → R be Riemannintegrable.(a) Show that f and g are Riemann integrable on [a, b].(b) Show that f ′ · g and f · g′ are Riemann integrable on [a, b].

(c) Show that∫ b

af ′gdx +

∫ b

afg′dx = (fg)(b) − (fg)(a). Hint: Use product

rule and Exercise 22.8.

Solution.(a) This follows from Exercise 21.2.(b) This follows from Exercise 24.5.(c) By the product rule we have (fg)′ = f ′g+fg′. Hence, using Exercise 22.8we can write∫ b

a

f ′gdx +

∫ b

a

fg′dx =

∫ b

a

(fg)′dx = (fg)(b)− (fg)(a)

or ∫ b

a

f ′gdx = (fg)(b)− (fg)(a)−∫ b

a

fg′dx

150

Exercise 24.5Consider the function f : [0, 1] → R defined by

f(x) =

{1 if 0 < x ≤ 10 if x = 0.

Show that f is Riemann integrable on [0, 1]. What is the value of∫ 1

0f(x)dx?

Solution.Let P be a parition of [0, 1] given by

P = {0 = x0 < x1 < x2 < · · · < xn = 1}.

Note that Mi(f) = 1 for all 1 ≤ i ≤ n, m1(f) = and mi(f) = 1 for all1 ≤ i ≤ n− 1. Thus,

U(f, P ) =n∑

i=1

Mi(f)(xi − xi−1) = 1.

Since P was arbitrary, it follows that∫ 1

0

f(x)dx = 1.

We know from Exercise 20.3 that∫ 1

0f(x)dx ≤ 1. Suppose that

∫ 1

0f(x)dx < 1.

Let ε = 1 −∫ 1

0f(x)dx > 0. Let Q = {0 = x0 < x1 < · · · < xn = 1} be

a partition of [0, 1] such that x1 > ε. Then L(f, Q) = m1(f)(x1 − x0) +∑n−1i=2 mi(f)(xi−xi−1) = 1−x1 <

∫ 1

0f(x)dx which contradicts the definition

of supremum. We conclude that∫ 1

0f(x)dx = 1.

Since∫ 1

0f(x)dx =

∫ 1

0f(x)dx = 1, the function f is Riemnann integrable with∫ 1

0

f(x)dx = 1

Exercise 24.6Consider the function g : [0, 1] → R defined by

g(x) =

1 if x = 0 or x = 11q

if x = pq

is rational with p and q > 0 in lowest terms

0 if x is irrational.

151

(a) Let ε > 0 and ε′ = min{0.5, ε}. Thus, 0 < ε′ ≤ 0.5 and 0 < ε′ ≤ ε. Showthat there is a finite number of rationals in [0, 1] such that g(x) ≥ ε′

2. Denote

the rationals by {r0, r1, · · · , rn} where r0 = 0 and rn = 1.(b) Define the partition Q = {0 = x0 < x1 < x2 < · · · < x2n < x2n+1 = 1}where x0 = 0; x1 < r1 with x1 < ε′

2(n+1); x1 < x2 < r1 < x3 with x3 − x2 <

ε′

2(n+1); · · · ; x2n−2 < rn−1 < x2n−1 with x2n−1 − x2n−2 < ε′

2(n+1); x2n−1 < x2n <

1 with 1 − x2n < ε′

2(n+1)and x2n+1 = 1. Show that U(g,Q) < ε′. Hint: Note

that the sum involves intervals containing r′is and those that do not.(c) Show that L(g,Q) = 0. Hint: Exercise 2.6.(d) Using (b) and (c) show that U(g,Q)− L(g,Q) < ε. Thus, g is Riemannintegrable.(e) What is the value of the integral

∫ 1

0g(x)dx?

Solution.(a) Let x = p

q∈ (0, 1) such that g(x) ≥ ε′

2. Then 1

q≥ ε′

2which implies that

0 < q ≤ 2ε′. This shows that there are only a finite number of positive integers

q that satisfy this inequality. Moreover, g(0) = g(1) > ε′

2. Let’s denote these

rationals by {r0, r1, · · · , rn} where r0 = 0 and rn = 1.(b) The terms of U(g,Q) consists of two types of intervals: The intervals[xi−1, xi] not containing ris and in this case we have Mi(g)(xi − xi−1) ≤ε′

2(xi − xi−1) and there are n such intervals. The second type consists of

those intervals containing ris and in this case Mi(g)(xi − xi−1) < ε′

2(n+1)and

there are n + 1 such intervals. Thus,

U(g,Q) =n∑

i=0

Mi(g)(x2i+1 − x2i) +n∑

i=1

Mi(g)(x2i − x2i−1)

<n∑

i=0

(x2i+1 − x2i) +ε′

2

n∑i=1

(x2i − x2i−1)

<(n + 1)ε′

2(n + 1)+

ε′

2= ε′ ≤ ε.

(c) By Exercise 2.6, we have mi(g) = 0 for all i. Hence, L(g,Q) = 0. There-fore, U(g,Q)− L(g,Q) < ε.(d) By Exercise 20.7 and (c) , we conslude that g is Riemann integrable.

(e) Since L(g, P ) = 0 for all partitions P of [0, 1] we find∫ 1

0g(x)dx = 0 =∫ 1

0g(x)dx

152

Exercise 24.7Consider the functions f and g introduced in the previous two exercises. Leth(x) = (f ◦ g)(x).(a) Write explicitly the formula of h(x) as a piecewise defined function.(b) Show that h is not Riemann integrable on [0, 1].


h(x) =

{1 if 0 < g(x) ≤ 10 if g(x) = 0

=

{1 if x is rational0 if x is irrational.

(b) This follows from Exercise 20.5

Exercise 24.8Let f, g : [a, b] → R be Riemann integrable.

(a) Show that max{f(x), g(x)} = |f(x)−g(x)|+f(x)+g(x)2

.(b) Show that the function max{f(x), g(x)} is Riemann integrable.

Solution.(a) If f(x) ≥ g(x) then max{f(x), g(x)} = f(x) and |f(x)−g(x)|+f(x)+g(x)

2=

f(x)−g(x)+f(x)+g(x)2

= f(x). Similar argument when f(x) < g(x).(b) All functions |f − g|, f, and g are Riemann integrable so the combinationmax{f(x), g(x)} is also Riemann integrable

Exercise 24.9Let f, g : [a, b] → R be Riemann integrable.

(a) Show that min{f(x), g(x)} = f(x)+g(x)−|f(x)−g(x)|2

.(b) Show that the function min{f(x), g(x)} is Riemann integrable.

Solution.Similar to the previous exercise

153

Solutions to Section 25Exercise 25.1Let f and F as defined in Definition 24. Let M be such that |f(x)| ≤ M forall x ∈ [a, b]. Fix c in [a, b].(a) Show that for any x ∈ [a, b] we have

−M(x− c) ≤∫ x

c

f(t)dt ≤ M(x− c).

Hence, we can write ∣∣∣∣∫ x

c

f(t)dt

∣∣∣∣ ≤ M |x− c|.

Hint: Exercise 23.2.(b) Let ε > 0 and δ = ε

M. Show that for any x ∈ [a, b] such that |x− c| < δ

we must have |F (x) − F (c)| < ε. Hence, F is continuous at c. Since c wasarbitrary in [a, b], we conclude that F is continuous on [a, b].

Solution.(a) Since −M ≤ f(t) ≤ M for all t ∈ [a, b] we can use Exercise 23.2 to obtain

−M

∫ x

c

dt ≤∫ x

c

f(t)dt ≤ M

∫ x

c

dt

which implies

−M(x− c) ≤∫ x

c

f(t)dt ≤ M(x− c)

or ∣∣∣∣∫ x

c

f(t)dt

∣∣∣∣ ≤ M |x− c|.

(b) We have

|F (x)− F (c)| =∣∣∣∣∫ x

a

f(t)dt−∫ c

a

f(t)dt

∣∣∣∣=

∣∣∣∣∫ a

c

f(t)dt +

∫ x

a

f(t)dt

∣∣∣∣=

∣∣∣∣∫ x

c

f(t)dt

∣∣∣∣ ≤ M |x− c|

<Mδ = ε

Hence, F is continuous at c. Since c was arbitrary in [a, b], we conclude thatF is continuous on [a, b]

154

Exercise 25.2Let f and F as above. Suppose furthermore that f is continuous at c ∈ [a, b].(a) Show that

F (c + h)− F (c)

h− f(c) =

1

h

∫ c+h

c

[f(t)− f(c)]dt.

(b) Show that F ′(c) exists and is equal to f(c).


F (c + h)− F (c)

h− f(c) =

1

h

∫ c+h

c

f(t)dt− 1

h

∫ c+h

c

dt

=1

h

∫ c+h

c

[f(t)− f(c)]dt.

(b) Let ε > 0. By the continuity of f at c we can find a δ > 0 such that if|x− c| < δ then |f(x)− f(c)| < ε. Choose h such that |h| < δ. Then for any tbetween c and c+h we have |t− c| < δ and therefore |f(t)− f(c)| < ε. Thus,∣∣∣∣F (c + h)− F (c)

h− f(c)

∣∣∣∣ =1

|h|

∣∣∣∣∫ c+h

c

[f(t)− f(c)]dt

∣∣∣∣<

1

|h|ε|h| = ε.

It follows that F ′(c) exists and is equal to f(c)

Exercise 25.3Suppose that f : [a, b] → R is differentiable on [a, b] and f ′ continuous on[a, b].(a) Show that f ′ is Riemann integrable on [a, b].(b) Define F : [a, b] → R by F (x) =

∫ x

af ′(t)dt. Show that F ′(x) = f ′(x) for

all x ∈ [a, b].(c) Show that F (x) = f(x)− f(a) for all x ∈ [a, b].(d) Use (c) to show that ∫ b

a

f ′(x)dx = f(b)− f(a).

155

Solution.(a) This follows from Exercise 21.2.(b) This follows from Exercise 25.2.(c) From Exercise 18.9 we have F (x) = f(x) + C for all x ∈ [a, b], where Cis a constant. Letting x = a we find 0 = F (a) = f(a) + C or C = −f(a).Thus, F (x) = f(x)− f(a).(d) The result follows by letting x = b in (c)

Exercise 25.4Suppose that f : [a, b] → R is continuous on [a, b] and g : [c, d] → [a, b] isdifferentiable on [a, b]. Define F : [c, d] → R by

F (x) =

∫ g(x)

a

f(t)dt.

(a) Show that f is Riemann integrable on [a, b].(b) Define G : [a, b] → R by G(x) =

∫ x

af(t)dt. Show that G is differentiable

and G′(x) = f(x) for all x ∈ [a, b].(c) Write F in terms of G and g. Show that F is differentiable on [c, d] with

F ′(x) = f(g(x)) · g′(x).

Solution.(a) This follows from Exercise 21.2.(b) This follows from Exercise 25.2.(c) We have F (x) = (G ◦ g)(x). Since both G and g are differentiable, byExercise 16.10, we find that F is differentiable with derivative

F ′(x) = G′(g(x)) · g′(x) = f(g(x)) · g′(x)

Exercise 25.5 (Mean Value Theorem for Integrals)Let f : [a, b] → R be continuous.(a) Show that f is Riemann integrable on [a, b].(b) Define F : [a, b] → R by

F (x) =

∫ x

a

f(t)dt.

Show that F is differentiable with F ′(x) = f(x).(c) Show that there is a < c < b such that F (b)− F (a) = F ′(c)(b− a).(d) Use (c) to show that ∫ b

a

f(x)dx = f(c)(b− a).

156

Solution.(a) This follows from Exercise 21.2.(b) This follows from Exercise 25.2.(c) This follows from the Mean Value Theorem (Exercise 18.1).

(d) Since F (a) = 0, we have F (b) = F (b)− F (a) =∫ b

af(x)dx. Also, F ′(c) =

f(c). Hence, ∫ b

a

f(x)dx = f(c)(b− a)

Exercise 25.6 (Change of Variables Formula)Let φ : [a, b] → [c, d] be differentiable with continuous derivative and suchthat φ(a) = c, φ(b) = d. Let f : [c, d] → R be continuous.(a) Show that the functions f and (f ◦ φ) · φ′ are Riemann integrable.(b) Define F (x) =

∫ x

cf(t)dt. Show that F is differentiable with F ′(x) = f(x)

for all x ∈ [c, d].(c) Define G(x) =

∫ x

af(φ(t))φ′(t)dt. Show that G is differentiable with

G′(x) = f(φ(x))φ′(x) for all x ∈ [a, b].(d) Show that F ◦ φ is differentiable on [a, b] with (F ◦ φ)′(x) = G′(x) for allx ∈ [a, b]. Hint:Exercise ??.(e) Use (d) and Exercise 18.9 to show that (F ◦φ)(x) = G(x) for all x ∈ [a, b].(f) Use (e) to show that∫ b

a

f(φ(x))φ′(x)dx =

∫ d

c

f(x)dx.

Solution.(a) Since f, φ, and φ′ are continuous, they are Riemann integrable by Exercise21.2. By Exercise 24.1, f ◦φ is Riemann integrable. Also, f ◦φ is continuousbeing the composition of two continuous functions. Since f ◦ φ : [a, b] → R,by Exercise 15.3 , f ◦ φ is bounded. Since φ′ : [a, b] → R, φ′ is also bounded.Thus, by Exercise 24.2, the product (f ◦ φ) · φ′ is integrable.(b) Since f is continuous, from Exercise 25.2 we conclude that F is differen-tiable with F ′(x) = f(x) for all x ∈ [c, d].(c) Since (f ◦ φ) · φ′ is continuous, by Exercise 25.2, G is differentiable withG′(x) = f(φ(x))φ′(x) for all x ∈ [a, b].(d) Since both F and φ are differentiable, the composition F ◦ φ is alsodifferentiable (Exercise 18.1) with derivative

(F ◦ φ)′(x) = F ′(φ(x)) · φ′(x) = f(φ(x)) · φ′(x) = G′(x).

157

(e) Using (d) and Exercise 10.1 we can write (F ◦ φ)(x) = G(x) + C forall x ∈ [a, b], where C is a constant. In particular, letting x = c we find0 = F (c) = (F ◦ φ)(a) = G(a) + C = C. Thus, (F ◦ φ)(x) = G(x) for allx ∈ [a, b].(f) This follows by letting x = b in (e)

Exercise 25.7Find the derivative of

F (x) =

∫ √x

1

cos (t2)dt.

Solution.From Exercise 25.4 with g(x) =

√x and f(x) = cos x2 we find

F ′(x) = f(g(x)) · g′(x) =cos x

2√

x

Exercise 25.8 (Mean Value Theorem for Monotone Functions)Let f : [a, b] → R be increasing on [a, b].(a) Show that f is Riemann integrable on [a, b].(b) Define g : [a, b] → R by g(x) = f(a)(x− a) + f(b)(b− x). Show that g iscontinuous on [a, b].

(c) Show that g(b) ≤∫ b

af(x)dx ≤ g(a).

(d) Show that there is c ∈ [a, b] such that∫ b

a

f(x)dx = f(a)(c− a) + f(b)(c− b).

Solution.(a) This follows from Exercise 21.1.(b) This follows from the fact that g is a combination of continuous functions.(c) Since a ≤ x ≤ b and f is increasing, we find f(a) ≤ f(x) ≤ f(b). Integrateeach side from a to b we find

f(a)(b− a) ≤∫ b

a

f(x)dx ≤ f(b)(b− a)

or

g(b) ≤∫ b

a

f(x)dx ≤ g(a).

158

(d) By the Intermediate value Theorem we can find a c ∈ [a, b] such that

g(c) =

∫ b

a

f(x)dx

or ∫ b

a

f(x)dx = f(a)(c− a) + f(b)(c− b)

Exercise 25.9Use change of variables to evaluate

∫ 3

1(3x + 1)100dx.

Solution.Let f(x) = x100 and φ(x) = 3x + 1. Then∫ 3

1

(3x + 1)100dx = 3

∫ 10

4

φ100dφ =3

101φ101

∣∣∣∣103

Exercise 25.10Find the smallest positive critical point of

F (x) =

∫ x

0

cos (t32 )dt.

Solution.We have F ′(x) = cos (x

32 ), so the smallest positive critical point is x =

(π2

) 23

Exercise 25.11Suppose f : R → R is continuous at a ∈ R. Find

limx→a

1

x− a

∫ x

a

f(t)dt.

Solution.Let F (x) =

∫ x

af(t)dt. Then F is differentiable at a with F ′(a) = f(a). Thus,

limx→a

1

x− a

∫ x

a

f(t)dt = limx→a

F (x)− F (a)

x− a= F ′(a) = f(a)

159

Exercise 25.12Let f : R → R be continuous and A, B : R → R be differentiable functions.Define g : R → R by

g(x) =

∫ B(x)

A(x)

f(t)dt.

Prove that g is differentiable and find a formula for g′(x).

Solution.Define F (x) =

∫ x

0f(t)dt. Since f is continuous, the First Fundamental Theo-

rem of Calculus shows that F is differentiable everywhere and F ′(x) = f(x).Note that

g(x) =

∫ B(x)

0

f(t)dt−∫ A(x)

0

f(t)dt = F (B(x))− F (A(x)).

Since A, B, F are differentiable, so is g. By the chain rule,

g′(x) = F ′(B(x))B′(x)− F ′(A(x))A′(x) = f(B(x))B′(x)′f(A(x))A′(x)

Exercise 25.13Suppose f : R → R is continuous at 2 and f(2) = 4. Find

limx→2

1

x− 2

∫ x

2

xf(t)dt.

Solution.We have

limx→2

1

x− 2

∫ x

2

xf(t)dt = limx→2

x · 1

x− 2

∫ x

2

f(t)dt = 2f(2) = 8

Exercise 25.14Use a definite integral to define a function F (x) having derivative cos 2x3

√1+x4 for

all x and satisfying F ( 3√

2) = 0.

Solution.The answer is

F (x) =

∫ x

3√2

cos 2t3√1 + t4

dt

160


Exercise 26.1Show that the series

∑∞n=1

1n(n+1)

converges to 1. Hint: Show that for each

n ≥ 1 we have Sn = 1− 1n+1

.

Solution.Using partial fractions we can write

1

n(n + 1)=

1

n− 1

n + 1.

Thus,S1 = 1− 1

2

S2 = (1− 12) + (1

2− 1

3) = 1− 1

3

S3 = S2 + (13− 1

4) = (1− 1

3) + (1

3− 1

4) = 1− 1

4...

Sn = 1− 1n+1

.

It follows that limn→∞ Sn = 1.

Exercise 26.2Is the series Σ∞

n=1(−1)n convergent or divergent?

Solution.The series Σ∞

n=1(−1)n diverges since the sequence of partial sums alternatesbetween the values −1 and 0.

Exercise 26.3Suppose that

∑∞i=1 an = L. Show that limn→∞ an = 0. Hint: Note that

Sn+1 − Sn = an.

Solution.We know that Sn = a1 + a2 + · · ·+ an and Sn+1 = a1 + a2 + · · ·+ an + an+1 =Sn + an so it follows that Sn+1 − Sn = an. Suppose that the series convergesto a number L. Then limn→∞ Sn = limn→∞ Sn+1 = L. Thus, limn→∞ an =limn→∞(Sn+1 − Sn) = L− L = 0.

Exercise 26.4Consider the series

∑ni=1 log

(n+1

n

).

(a) Show that limn→∞ an = 0.(b) Show that limn→∞ Sn = ∞. Hence, the series is divergent.

161

Solution.(a) We have limn→∞ an = limn→∞ log

(1 + 1

n

)= log 1 = 0.

(b) We have that Sn = log (n + 1). Hence, limn→∞ Sn = limn→∞ log (n + 1) =∞ so that the given series is divergent

Exercise 26.5Consider the sequence {rn}∞n=1.(a) Show that if r = −1 the sequence is divergent.(b) Show that if |r| > 1, i.e. r < −1 or r > 1, the sequence is divergent.(c) Show that if |r| < 1, the sequence is convergent.

Solution.(a) This follows from Exercise .(b) |r| > 1 implies r > 1 or r < −1. Suppose first that r > 1. Let ε > 0. LetN be a positive integer greater than ε

r−1. Then for n ≥ N we have

rn = (1 + (r − 1))n

≥ 1 + n(r − 1)(by the binomial formula)> 1 + N(r − 1)> 1 + ε> ε.

This shows that for any given positive number we can find a term in thesequence {rn}∞n=1 which is greater than the number. This means that rn →∞as n →∞.If r < −1 then rn = (−1)n(−r)n with −r > 1. Thus, as n becomes large, rn

alternates between large positive numbers and negative numbers with largeabsolute value so that again the limit rn as n →∞ does not exist.(c) If 0 < r < 1 then rn = 1

(r−1)n with (r−1)n → ∞ as n → ∞. (See (b)).Hence, rn → 0 as n → ∞. If −1 < r < 0 then 0 < −r < 1. In this case,rn = (−1)n(−r)n → 0 as n →∞. If r = 0 then rn = 0 and limn→∞ rn = 0

Exercise 26.6The series

∑∞n=1 arn−1 is called a geometric series with ratio r.

(a) Show that

Sn = a1−rn

1−rfor r 6= 1.

Hint: Calculate Sn − rSn.(b) Show that the series converges to a

1−rfor |r| < 1 and diverges for |r| ≥ 1.

162

Solution.(a) We have Sn − rSn = a− arn

so that Sn = a1−rn

1−rfor r 6= 1.

(b) If |r| < 1, using Exercise 26.5, we find limn→∞ Sn = a1−r

. If r = 1 thenSn = na and this series diverges to either ∞ (if a > 0) or −∞ (if a < 0.)If r = −1 the sequence {(−1)n}∞n=1 is divergent and therefore the sequence{Sn}∞n=1 is divergent.The same applies if |r| > 1

Exercise 26.7 (Harmonic Series)Consider the Harmonic series

∑∞n=1

1n.

(a) Let n = 2m where m is a positive integer. Then

Sn =1 +1

2+

1

3+ · · ·+ 1

2m

=1 +1

2+

(1

3+

1

4

)+

(1

5+

1

6+

1

7+

1

8

)+ · · ·+

(1

2m−1 + 1+ · · ·+ 1

2m

)Show that Sn ≥ 1 + m

2.

(b) Use (a) to show that limn→∞ Sn = ∞. Thus, the Harmonic series isdivergent.

Solution.(a) We have Sn ≥ 1+ 1

2+2· 1

4+4· 1

8+· · ·+2m−1 · 1

2m = 1+ 12+ 1

2+· · ·+ 1

2= 1+ m

2.

(b) If n goes to ∞ then m goes to ∞ and therefore limm→∞(1 + m

2

)= ∞.

By (a), we conclude that limn→∞ Sn = ∞ and therefore the Harmonic seriesis divergent

Exercise 26.8Show that if

∑∞n=1 an = L1 and

∑∞n=1 bn = L2 then

∑∞n=1(αan + βbn) =

αL1 + βL2 for all α, β ∈ R.

Solution.Let S2

n = a1+a2+ · · ·+an and S2n = b1+b2+ · · ·+bn. Then lim n →∞(αS1

n +βS2

n) = αL1 + βL2

Exercise 26.9Find the value of the infinite sum

∑∞n=1

(34n + 5

n(n+1)

).

163

Solution.We have

∑∞n=1

34n = 3

4

∑∞n=1

(14

)n−1= 3

4· 1

1− 14

= 1 and∑∞

n=15

n(n+1)= 5.

Thus,∑∞

n=1

(34n + 5

n(n+1)

)= 6

Exercise 26.10Show that the sequence {

√n2 − 1− n}∞n=1 is convergent and find its limit.

Solution.We have

limn→∞

(√

n2 − 1− n) = limn→∞

(√

n2 − 1− n)(√

n2 − 1 + n)

(√

n2 − 1 + n)

= limn→∞

−1√n2 − 1 + n

= 0

Exercise 26.11Let

∑∞n=1 an be a conditionally convergent series. Define bn = 1

2(an+|an|) and

cn = 12(an − |an|). Prove that the two series

∑∞n=1 and

∑∞n=1 are divergent.

Solution.Note first that |an| = bn − cn for all n ∈ N. Since

∑∞n=1 |an| is divergent,

either∑∞

n=1 bn is divergent or∑∞

n=1 |an|. Suppose that∑∞

n=1 bn is convergentand

∑∞n=1 cn is divergent. Since bn = |an| + cn we must have that

∑∞n=1 bn

is divergent which is a contradiction. A similar contradiction is obtained if∑∞n=1 bn is divergent and

∑∞n=1 cn is convergent. It follows that both

∑∞n=1 bn

and∑∞

n=1 cn must be divergent

Exercise 26.12Let Sn be the n-th partial sum of the series

∑∞n=1

n−2n(n+1)(n+2)

.

(a) Show that Sn = 3n+1

− 2n+1

− 2n+2

. Hint: Partial fractions.

(b) Find the value of the series∑∞

n=1n−2

n(n+1)(n+2).

Solution.(a) Using partial fraction decomposition we find that

n− 2

n(n + 1)(n + 2)=

3

n + 1− 2

n + 2− 1

n.

164

Thus,

Sn =3

2− 3

n∑i=3

1

i− 2

n∑i=3

1

i−

n∑i=3

1

i− 1− 1

2

3

n + 1− 2

n + 1− 2

n + 2

=3

n + 1− 2

n + 1− 2

n + 2

Thus, the series is convergent with sum

∞∑n=1

n− 2

n(n + 1)(n + 2)= lim

n→∞Sn = 0

Exercise 26.13Let {an}∞n=1 be a decreasing sequence such that

∑∞n=1 an is convergent.

(a) Show that an ≥ 0 for all n ∈ N.(b) Let ε > 0. Show that there is a positive integer N such that if n > m ≥ Nwe have

|am+1 + am+2 + · · ·+ an| < ε.

(c) Show that (n−N)an < ε.(d) Let n > 2N. Show that n

2< n−N.

(e) Show that nan

2< ε.

(f) Show that limn→∞ nan = 0.

Solution.(a) If aN < 0 for some N then an ≤ aN < 0 for all n ≥ N. But thenlimn→∞ an 6= 0 which contradicts the fact that


(b) This follows from the fact that the sequence of partial sums is Cauchy.(c) From (b), we have aN+1+aN+2+· · ·+an < ε. But an ≤ · · · ≤ aN+2 ≤ aN+1.Thus, (n−N)an < ε.(d) We have n > 2N → −n

2< −N → n

2< n−N.

(e) If n > 2N > N we have nan

2< (n−N)an < ε.

(f) It follows from (e) that limn→∞nan

2= 0 and therefore limn→∞ nan =

2 limn→∞nan

2= 0

Exercise 26.14Let N be a positive integer. Suppose that an = bn for all n ≥ N. Thenthe series

∑∞n=1 an and

∑∞n=1 bn either both converge or both diverge. Thus,

changing a finite number of terms in a series does not change whether or notit converges, although it may change the value of its sum if it does converge.

165

solution.The proof follows from the equality

∑∞n=1 an =

∑N−1n=1 (an − bn) +

∑∞n=1 bn.

166


Exercise 27.1 (Comparison test))Let {an}∞n=1 and {bn}∞n=1 be two series such that 0 ≤ an ≤ bn for all n ≥ 1.Let {Sn}∞n=1 be the sequence of partial sums of {an}∞n=1 and {Tn}∞n=1 that of{bn}∞n=1.(a) Show that the sequences {Sn}∞n=1 and {Tn}∞n=1 are increasing.(b) Show that Sn ≤ Tn for all n ≥ 1.(c) Show that if {bn}∞n=1 is convergent then {Sn}∞n=1 and {Tn}∞n=1 are bounded.(d) Show that if {bn}∞n=1 is convergent then {an}∞n=1 is also convergent.(e) Show that if {an}∞n=1 is divergent then {bn}∞n=1 is also divergent.

Solution.(a) Sn+1 − Sn = an+1 ≥ 0 so that Sn+1 ≥ Sn. Thus, {Sn}∞n=1 is increasing.Likewise, {Tn}∞n=1 is increasing.(b) Since an ≤ bn for all n ≥ 1, we must have Sn ≤ Tn for all n ≥ 1.(c) If {bn}∞n=1 is convergent then the sequence {Tn}∞n=1 is also convergent. ByExercise 3.9, the sequence {Tn}∞n=1 is bounded say by M. Since Sn ≤ Tn ≤ Mfor all n ≥ 1, the sequence {Sn}∞n=1 is also bounded.(d) Since {Sn}∞n=1 is increasing and bounded from above, it must be conver-gent according to Exercise 5.5.(e) This is just the contrapositive of (d)

Exercise 27.2(a) Show that for n ≥ 1 we have 1

(n+1)2≤ 1

n(n+1).

(b) Show that the series∑∞

n=11

(n+1)2is convergent.

Solution.(a) Since 1 ≥ 0 we have n + 1 ≥ n → 1

n+1≤ 1

n→ 1

(n+1)2≤ 1

n(n+1).

(b) Since the series∑∞

n=11

n(n+1)is convergent so does the given series


∑∞n=1

1√n2−n+1

is divergent.

Solution.Indeed, for n ≥ 1 we have n2 +(1−n) ≤ n2 so that

√n2 − n + 1 ≤

√n2 = n.

This implies that 0 ≤ 1n≤ 1√

n2−n+1. Since the series

∑∞n=1

1n

is divergent

(harmonic series), the comparison test asserts that the series∑∞

n=11√

n2−n+1is divergent.

167

Exercise 27.4 (Limit Comparison Test)Let

∑∞n=1 an and

∑∞n=1 bn be two series with positive terms. Suppose that

limn→∞

an

bn

= L > 0

(a) Let ε = L2. Show that there exists a positive integer N such that∣∣∣an

bn− L

∣∣∣ < L2

for all n ≥ N.

(b) Use (a) to establish

L2bn < an < 3

2Lbn for all n ≥ N.

(c) Show that∑∞

n=1 an is divergent if and only if∑∞

n=1 bn is divergent.

Solution.(a) This follows from the definition of convergence of a sequence.(b) The inequality in (a) is equivalent to

L− L

2<

an

bn

< L +L

2

orL

2<

an

bn

<3

2L.

orL

2bn < an <

3

2Lbn, n ≥ N.

(c) If the series∑∞

n=1 an converges then by the comparison test the series∑∞n=1

L2bn is convergent. By Exercise 26.8, the series

∑∞n=1 bn is also conver-

gent.Conversely, if

∑∞n=1 bn is convergent then

∑∞n=1

32Lbn is convergent and by

the comparison test∑∞

n=1 an is convergent. Similarly,∑∞

n=1 an is divergentif and only if

∑∞n=1 bn is divergent.

Exercise 27.5Determine whether the series

∑∞n=1

3n+14n3+n2−2

converges or diverges.

168

Solution.For large n we have an = 3n+1

4n3+n2−2≈ 3n

4n3 = 34n2 . So let bn = 1

n2 . Then

limn→∞

an

bn

= limn→∞

3n3 + n2

4n3 + n2 − 2=

3

4.

Since the series∑∞

n=11n2 is convergent then so does the series

∑∞n=1

3n+14n3+n2−2

.

Exercise 27.6Let {an}∞n=1 be a bounded sequence of nonnegative terms. Show that if theseries

∑∞n=1 an is divergent so does the series

∑∞n=1

an

1+an. Hint: Comparison

test.

Solution.Since the sequence {an}∞n=1 is bounded there is M > 0 such that an ≤ M forall n ∈ N. Thus,

an

1 + an

≥ 1

1 + Man.

But the series∑∞

n=1 an is divergent so does the series 11+M

∑∞n=1 an. Hence,

by the comparison test, the series∑∞

n=1an

1+anis divergent

Exercise 27.7Use the limit comparison test to show that the series

∑∞i=1

12n+ln n

is divergent.

Solution.We have

limn→∞

12n+ln n

1n

= limn→∞

n

2n + ln n= lim

n→∞

1

2 + 1n

=1

2< 1

Since the Harmonic series is divergent, the given series is also divergent

Exercise 27.8Suppose that an ≥ 0 and that the series

∑∞n=1 an diverges. Suppose that

{an}∞n=1 is unbounded. Show that limn→∞an

1+an6= 0. Hint: assume the con-

trary and get a contradction. Conclude that the series∑∞

n=1an

1+anis diver-

gent.

169

Solution.If limn→∞

an

1+an= 0 then we can find a positive integer N such that for

n ≥ N we have an

1+an< 1

2which implies that an < 1 for all n ≥ N. Let

M = a1 + a2 + · · ·+ aN−1 + 1. Then |an| ≤ M for all n ∈ N. This shows thatthe sequence is bounded contradicting the assumption that it is unbounded.Thus, limn→∞

an

1+an6= 0 and the given series is divergent

Exercise 27.9Suppose that an ≥ 0 for all n ∈ N and that the series

∑∞n=1 an converges.

(a) Show that there is a positive integer N such that an < 1 for all n ≥ N.(b) Show that the series

∑∞n=1 a2

n converges.

Solution.(a) Since the series is convergent, we have limn→∞ an = 0. Thus, we can finda positive integer N such that an < 1 for all n ≥ N.(b) For n ≥ N we have a2

n < an. By the comparison test the series∑∞

n=1 a2n

converges

Exercise 27.10Use comparison test to show that the series

∑∞n=1(

√n2 + 1−n) is divergent.

Solution.We have

√n2 + 1− n = 1√

n2+1+n> 1

1+√

21n. The Harmonic series is divergent

so that by the comparison test the given series is divergent

170


Exercise 28.1 (Alternating Series Test)Let {an}∞n=1 be a sequence of positive numbers such that(i) an ≥ an+1, that is the sequence {an}∞n=1 is decreasing.(ii) limn→∞ an = 0.Let {Sn}∞n=1 be the sequence of partial sums of the series

∑∞n=1(−1)n−1an.

That is, Sn =∑n

k=1(−1)k−1ak.(a) Show that for each n ≥ 1 we have S2n ≤ S2n+2. That is, the sequence{S2n}∞n=1 is increasing. Hint: Show that S2n+2 − S2n ≥ 0.(b) Show that the sequence {S2n+1}∞n=1 is decreasing.(c) Show that for all n ≥ 1, we have S2n ≤ a1. Hence, the sequence {S2n}∞n=1

is bounded from above. Conclude that the sequence {S2n}∞n=1 is convergent,say to L1.(d) Show that for all n ≥ 1, we have S2n+1 ≥ (a1 − a2). Hence, the sequence{S2n+1}∞n=1 is bounded from below. Conclude that the sequence {S2n+1}∞n=1

is convergent, say to L2.(e) Show that L1 = L2. Hint: S2n+1 = S2n + a2n+1.(f) Let L = L1 = L2. Show that limn→∞ Sn = L. We conclude that theseries

∑∞n=1(−1)n−1an is convergent. Hint: Look at the sequence {cn}∞n=1 in

Exercise 10.4.

Solution.(a) We have S2n+2 − S2n = a2n+1 − a2n+2 ≥ 0. Thus, the sequence {S2n}∞n=1

is increasing.(b) We have S2n+1−S2n−1 = a2n+1−a2n ≤ 0. Thus, the sequence {S2n+1}∞n=1

is decreasing.(c) We have

S2n = a1 − (a2 − a3)− (a4 − a5)− · · · − (a2n−2 − a2n−1)− a2n ≤ a1.

By Exercise 5.5, the sequence {S2n}∞n=1 is convergent, say to L1.(d) We have

S2n+1 = (a1 − a2) + (a3 − a4) + · · ·+ (a2n−1 − a2n) + a2n+1 ≥ (a1 − a2).

By Exercise ??, the sequence {S2n+1}∞n=1 is convergent, say to L2.(e) Since S2n+1 = S2n + a2n+1, we have L2 = limn→∞ S2n+1 = limn→∞ S2n +limn→∞ a2n+1 = L1 + 0 = L1.(f) Since {Sn}∞n=1 = {S1, S2, S3, · · · } with limn→∞S2n = limn→∞S2n+1 = L,by Exercise 10.4(c), limn→∞Sn = L

171

∑∞n=1

(−1)n−1

nis convergent.

Solution.To see this, let an = 1

n. Then n < n + 1 implies that 1

n+1< 1

nthat is

an+1 < an. Also, limn→∞ an = limn→∞1n

= 0. Hence, by the previous resultthe given series is convergent.


∑∞n=1

(−1)n−1

n(n+1)is convergent.

Solution.To see this, let an = 1

n(n+1). Now n < n + 2 implies that 1

n+2< 1

nand

this implies that 1(n+1)(n+2)

< 1n(n+1)

that is an+1 < an. Also, limn→∞ an =

limn→∞1

n(n+1)= 0. Hence, by the alternating series test the given series is

convergent.


∑∞n=1(−1)n−1 n

n+1converges or diverges.

Solution.Since

limn→∞

(−1)n−1 n

n + 16= 0

the series is divergent


∑∞n=1(−1)n−1 ln (4n)

nconverges or diverges.

Solution.Let f(x) = ln (4x)

xfor x ≥ 1. Then f ′(x) = 1−ln (4x)

x2 < 0 for x ≥ 1. This

shows that the sequence { ln (4n)n

}∞n=1 is decreasing. Moreover, we have byusing L’Hopital’s rule

limn→∞

ln (4n)

n= lim

n→∞

1

n= 0.

Hence, the given series is convergent

172

Exercise 28.6(a) Show that nn

n!≥ 1 for all n ≥ 1.

(b) Show that ther series∑∞

n=1(−1)n−1 nn

n!is divergent.


nn

n!=

n

n

n

n− 1· · · n

1> 1.

(b) Since limn→∞(−1)n−1 nn

n!6= 0, the given series is divergent


∑∞n=1(−1)n−1 3n+1+2n+1

3n−ndiverges.

Solution.We have

limn→∞

3n+1 + 2n+1

3n − n= lim

n→∞

3n(3 + 2

(23

)n)3n(1− 3n

n

)= lim

n→∞

3 + 2(

23

)n1− 3n

n

=3 + 0

1− 0= 3

Since limn→∞(−1)n−1 3n+1+2n+1

3n−n6= 0 the given series is divergent

173



∑∞n=1

(−1)n−1

n(n+1)is absolutely convergent.


Exercise 29.2Let

∑∞n=1 an be an absolutely convergent series. Define the sequence

∑∞n=1 bn

by bn = |an| and note that an ≤ bn. Show that the sequence∑∞

n=1 an isconvergent. That is, absolute convergence implies convergence.

Solution.This is a direct consequence of the comparison test

Exercise 29.3Give an example of a series that is convergent but not absolutely convergent.

Solution.An example is the series

∑∞n=1

(−1)n−1

n(n+1)

Exercise 29.4Give an example of a series that is conditionally convergent.


∑∞n=1

(−1)n−1

n


∑∞n=1 an is absolutely convergent.

(a) Show that 0 ≤ |an|+an

2≤ |an| and 0 ≤ |an|−an

2≤ |an|


n=1

(|an|+an

2

)and

∑∞n=1

(|an|−an

2

)are convergent.

Solution.(a) Since −|an| ≤ an we have |an|+an

2≥ 0. Since an ≤ |an|, we have an+|an| ≤

2|an| or |an|+an

2≤ |an|. Likewise, one can establish the second inequality.

(b) This follows from the comparison test and the fact that∑∞

n=1 |an| isconvergent

174

Exercise 29.6(a) Show that if

∑∞n=1 an is absolutely convergent then the series

∑∞n=1 a2

n isalso absolutely convergent.(b) Give an example of a convergent series

∑∞n=1 an for which

∑∞n=1 a2

n isdivergent.

Solution.(a) Since limn→∞ |an| = 0 we can find a positive integer N such that forn ≥ N we have |an| < 1. Thus, for n ≥ N we have |an|2 < |an|. Thus,∑∞

n=1 |an|2 =∑N−1

n=1 |an|2+∑∞

n=N |an|2 <∑N−1

n=1 |an|2+∑∞

n=N |an|2 <∑N−1

n=1 |an|2+∑∞n=1 |an|. The series on the right is convergent so that by the comparison

test the series∑∞

n=1 |an|2 is convergent.

(b) By the alternating series test the series∑∞

n=0(−1)n√

n+1is convergent. But

the series∑∞

n=11n

is divergent


∑∞n=1 an is absolutely convergent and {bn}∞n=1 is bounded.

Show that∑∞

n=1 anbn is absolutely convergent (and thus convergent).

Solution.Let M > 0 such that |bn| ≤ M for all n ∈ N. Thus,

∑∞n=1 |anbn| ≤

M∑∞

n=1 |an|. Since∑∞

n=1 |an| is convergent, the series M∑∞

n=1 |an| is con-vergent and so by the comparison test the series

∑∞n=1 |anbn| is convergent.

Thus,∑∞

n=1 anbn is absolutely convergent and thus convergent

Exercise 29.8Test the following series for absolute convergence, conditional convergence,or divergence.(a)∑∞

n=1sin nn2n .

(b)∑∞

n=1(−1)n 5nn2+2n

.

(c)∑∞

n=1(−1)n 2n−2−n

2n+2−n .


∣∣ sin nn2n

∣∣ ≤ 1n2n ≤ 1

2n . The geometric series∑∞

n=112n converges so

that by the comparison test the given series converges absolutely.(b) By the limit comparison test one can show that the series of absolute valueis divergent. By the alternating series test, the given series is convergent.Hence, the given series is conditionally convergent.(c) The limit of the nth term is either 1 or −1 so that by the nth term testthe series is divergent

175


∑∞n=1(−1)n−1 ln 4n

nis absolutely convergent.

Solution.By Exercise ?? we know that the series is convergent. However, since ln 4n

n>

1n

and the Harmonic series is divergent, the series of absolute values is diver-gent. Hence, the given series is conditionally convergent

Exercise 29.10Suppose that the sequence {an}∞n=1 is monotone decreasing with limn→∞ an =0. Let {bn}∞n=1 be a sequence such that |bn| ≤ an − an+1 for all n ∈ N. Showthat

∑∞n=1 bn is absolutely convergent.

Solution.Note first that the conditions on an imply that an ≥ 0 for all n ∈ N. LetSn =

∑nn=1 |bn|. Since |bn| ≤ an − an+1 for each n we have

Sn ≤ (a1 − a2) + (a2 − a3) + · · ·+ (an − an+1) = a1 − an+1 ≤ a1.

It follows that the sequence {Sn}∞n=1 is bounded from above by a1. Moreover,this sequence is increasing. Hence, by Exercise 5.5, the series

∑∞n=1 |bn| is

convergent so that∑∞

n=1 bn converges absolutely

176


Exercise 30.1 (Integral Test)Let

∑∞n=1 an be a series of positive terms and suppose that there is a function

f : [1,∞) → R such that f is decreasing and positive with f(n) = an for alln ≥ 1.(a) Show that {Sn}∞n=1 is increasing.(b) Define F : [1,∞) → R by F (x) =

∫ x

1f(t)dt. Show that F is increasing.

(c) For n ≥ 2 and x ∈ [n − 1, n], show that an ≤ f(x) ≤ an−1 and an ≤∫ n

n−1f(x)dx ≤ an−1.

(d) Show that Sn − a1 ≤ F (n) ≤ Sn−1.(e) Suppose that

∫∞1

f(x)dx = L. Since F is increasing we can write F (n) ≤ Lfor all n ≥ 1. Show that {Sn}∞n=1 is bounded. Hint: Use (d).(f) Show that {Sn}∞n=1 is convergent. Hence,


(g) Conversely, suppose that the series∑∞

n=1 an converges to a number S.Show that for any positive integer n ≥ 2 we have

F (n) ≤ S.

(h) Show that for all R ≥ 1 we have F (R) ≤ S. Thus,∫∞

1f(x)dx =

limR→∞∫ R

1f(x)dx is convergent. Hint: For any R ≥ 1 we have R ≤ [R] + 1

with [R] + 1 ≥ 2.

Solution.(a) We have Sn+1 − Sn = an+1 > 0 so that Sn+1 > Sn.(b) This follows from F ′(x) = f(x) > 0.(c) We have an = f(n) ≤ f(x) ≤ f(n− 1) = an−1. Integrate from n− 1 to nwe obtain the desired result.(d) We have Sn − a1 = a2 + a3 + cdots + an ≤

∫ 2

1f(x)dx +

∫ 3

2f(x)dx + · · ·+∫ n

n−1f(x)dx =

∫ n

1f(x)dx = F (n) ≤ a1 + a2 + · · ·+ an−1 = Sn−1.

(e) Using (d) we have Sn− a1 ≤ F (n) ≤ L so that Sn ≤ L + a1 for all n ≥ 1.That is, {Sn}∞n=1 is bounded.(f) This follows from Exercise 5.5.(g) From (d), we have F (n) ≤ Sn−1 ≤ S.

(h) We have F (R) ≤ F ([R] + 1) =∫ [R]+1

1f(x)dx ≤ S for all R ≥ 1. Hence,

letting R → ∞ we find that∫∞

1f(x)dx ≤ S. That is,

∫∞1

f(x)dx is conver-gent.Note that F is increasing and bounded from above so that limR→∞

∫ R

1f(x)dx

exists

177

Exercise 30.2 (p-series)(a) Show that the series

∑∞n=1

1np is convergent for p > 1.


n=11np is divergent for p ≤ 1.

Solution.If p < 0 then limn→∞

1np = ∞ so that by the nth term test the series is

divergent. If p = 0 then the series is an infinite sum of 1 and so is divergent.So suppose that p > 1. Let f(x) = 1

xp . Then f ′(x) = − pxp+1 < 0 so that f

is decreasing on [1,∞) and positive there. Moreover, f(n) = 1np . So by the

integral test the series∑∞

n=11np converges if and only if

∫∞1

dxxp is convergent.

Now, we have ∫ R

1

dx

xp=

R1−p

1− p− 1

1− p.

The improper integral exists if and only if p > 1. It is divergent if 0 1. Hint:

The integral test.

Solution.It is easy to check that the integrand is nonnegative and decreasing. By theintegral test we have∫ ∞

1

x

(x2 + 1)(ln (x2 + 1))adx = lim

R→∞

∫ R

1

x

(x2 + 1)(ln (x2 + 1))adx

= limR→∞

1

2(1− a)

[(ln (x2 + 1))1−a

]R1

= limR→∞

1

2(1− a)

[1

(ln (R2 + 1)a−1 −1

(ln 2)a−1

]=− 1

2(1− a)(ln 2)a−1

Exercise 30.4Use the integral test to test the convergence of the series

∑∞n=4

1n ln n ln (ln n)

.

178

Solution.We have ∫ ∞

4

dx

x ln x ln ln x=

∫ ∞

ln ln 4

du

u= ∞.

Thus, the given series is divergent

Exercise 30.5Use the Integral Test to show that

∑∞n=1 n2e−n3

is convergent.

Solution.Note that f(x) > 0 and f ′(x) < 0 so that f is decreasing. By the IntegralTest we have ∫ ∞

1

x2e−x3

dx = limR→∞

[−1

3e−R3

+1

3e−1] =

1

3e−1.

Since the improper integral is convergent so does the given series

Exercise 30.6Use the integral test to show that the series

∑∞n=1 e−n2

is convergent.

Solution.Let f(x) = e−x2

> 0. Then f ′(x) = −2xe−x2< 0 for x ≥ 1. Now, for

x ≥ 1 → x2 ≥ x → e−x2 ≤ e−x. Since∫∞

1e−xdx = 1

e, the improper integral∫∞

1e−x2

dx is convergent. Hence, by the integral test, the given series isconvergent

Exercise 30.7Use the integral test to show that the series

∑∞n=1

(ln n)2

nis divergent.

Solution.Let f(x) = (ln x)2

x≥ 0 for x ≥ 1. Moreover, f ′(x) = 2 ln x − ln x)2x2 < 0 for

x ≥ 8. Thus, ∫ ∞

1

(ln x)2

x≥∫ ∞

8

(ln x)2

x.

But ∫ ∞

8

(ln x)2

x= lim

R→∞[(ln R)3

3] = ∞.

Since the improper integral∫∞

1(ln x)2

xis divergent, the given integral is diver-

gent as well

179

Exercise 31.1 (Ratio Test)

Let∑∞

n=1 an be a series of non-zero terms and suppose that limn→∞

∣∣∣an+1

an

∣∣∣ =

L ≥ 0.(a) Suppose 0 ≤ L < 1. Let ε = 1−L

2. Show that there is a positive integer N

such that ∣∣∣an+1

an

∣∣∣ < 1+L2

for all n ≥ N.

Hint: Use definition of convergence and Exercise 1.18.(b) Let r = 1+L

2. Show that 0 < r < 1 and |aN+k| < rk|aN | for all k =

1, 2, · · · .(c) Find the value of the sum

∞∑n=1

rn|aN |.

(d) Let bn =∑n

k=1 |ak|. Show that the sequence {bn}∞n=1 is increasing.

(e) Let M = bN + r|aN |1−r

. Show that |bn| ≤ M for all n ≥ 1.(f) Show that the sequence {bn}∞n=1 is convergent. What can you say aboutthe sequence {an}∞n=1?

Solution.(a) Since limn→∞

∣∣∣an+1

an

∣∣∣ = L, we can find a positive integer N such that∣∣∣an+1

an

∣∣∣− L ≤∣∣∣an+1

an− L

∣∣∣ < 1−L2

for all n ≥ N.

This is equivalent to ∣∣∣an+1

an

∣∣∣ < L+12

for all n ≥ N.

(b) Since L < 1 we have 1 + L < 2 and therefore r = 1+L2

< 1. Clearly,r > 0. By (a) we have aN+1 < r|aN |. Suppose that aN+k < rk|aN . Then|aN+k+1| < r|aN+K | < rk+1|aN . Hence, |aN+k| < rk|aN for all k = 1, 2, · · · .(c) The given sum is a geometric series with ration r and whose sum is givenby

∞∑n=1

rn|aN | = |aN |∞∑

n=1

rn =r|aN

1− r.

180

(d) For any n ≥ 1, we have bn+1 = bn + |an+1| ≥ bn. This shows that thesequence {bn}∞n=1 is increasing.(e) If n ≤ N then |bn| =

∑nk=1 |ak| < M. If n ≥ N then |bn| =

∑Nk=1 |ak| +∑n

k=N |ak| ≤∑N

k=1 |ak| +∑∞

n=1 |an| =∑N

k=1 |ak| + r|aN

1−r= M. It follows that

the sequence {bn}∞n=1 is convergent by Exercise ??. Now, by Exercise 29.2the sequence {an}∞n=1 is convergent

Exercise 31.2 (Ratio Test)

Let∑∞

n=1 an be a series of non-zero terms and suppose that limn→∞

∣∣∣an+1

an

∣∣∣ =

L ≥ 0.(a) Suppose L > 1. Let ε = L − 1. Show that there is a positive integer Nsuch that

L−∣∣∣an+1

an

∣∣∣ < ε for all n ≥ N.

(b) Show that |an+1| > |aN | > 0 for all n ≥ N.(c) Show that the series

∑∞n=1 an is divergent. Hint: The nth term test.

Solution.(a) Since limn→∞

∣∣∣an+1

an

∣∣∣ = L, we can find a positive integer N such that

L−∣∣∣an+1

an

∣∣∣ < L− 1 for all n ≥ N.

(b) From (a) we conclude that∣∣∣an+1

an

∣∣∣ > 1 for all n ≥ N. Hence, |an+1| >

|aN | > 0 for all n ≥ N.(c) From (b) we see that limn→∞ 6= 0 so that by the nth term test thesequence

∑∞n=1 an is divergent

Exercise 31.3Consider the harmoninc series

∑∞n=1

1n

which we know it is divergent. Findlimn→∞

an+1

an.

Solution.

We have limn→∞1

n+11n

= 1

Exercise 31.4Consider the series

∑∞n=1

1n2 .

(a) Show that this series is convergent.(b) Find limn→∞

an+1

an.

181

Solution.(a) This is a p−series with p = 2 > 1 so it is convergent.

(b) We have limn→∞

1(n+1)2

1n2

= 1

Exercise 31.5Use the ratio test to determine the convergence of the series

∑∞n=1(−1)n 100n

n!.

Solution.

We have limn→∞100n+1

(n+1)!100n

n!

= limn→∞100n+1

= 0 < 1. By the ratio test the series

is convergent

Exercise 31.6Use the ratio test to determine the convergence of the series

∑∞n=1

2nn!nn . Hint:

limn→∞(1 + 1

n

)n= e.

Solution.

We have limn→∞

2n+1(n+1)!

(n+1)n+1

2nn!nn

= limn→∞ 2(

nn+1

)n= 2

e< 1. By the ratio test the

series is convergent

Exercise 31.7Find limn→∞

n!n2 .

Solution.Let an = n2

n!. By the ratio test we have

limn→∞

an+1

an

= limn→∞

n + 1

n2= 0 < 1

so that the series∑∞

n=1 an is convergent. Hence,

limn→∞

an = 0.

We conclude that

limn→∞

n!

n2= lim

n→∞

1

an

= ∞

Exercise 31.8 (nth root test)

Consider a series∑∞

n=1 an. Define L = limn→∞ |an|1n .

(a) Suppose first that 0 ≥ L < 1. Let ε = 1−L2

. Show that there is a positiveinteger N such that

182

|an|1n < 1+L

2for all n ≥ N.

(b) Let r = 1+L2

. Show that 0 < r < 1 and |an| < rn for all n ≥ N.(c) Use (b) to conclude that

∑∞n=1 an is absolutely convergent and hence

convergent.

Solution.(a) Since limn→∞ |an|

1n = L, we can find a positive integer N such that

|an|1n − L ≤

∣∣∣|an|1n − L

∣∣∣ < 1−L2

for all n ≥ N.

This is equivalent to

|an|1n < 1+L

2for all n ≥ N.

(b) Since 0 ≤ L < 1 → 1 + L < 2 → r = 1+L2

< 1. Moreover, by (a),we have |an| < rn for all n ≥ N. (c) Since the goemetric series

∑∞n=1 rn

is convergent for 0 < r < 1, by the comparison test the series∑∞

n=1 an isabsolutely convergent and hence convergent

Exercise 31.9 (nth root test)Suppose that L > 1 in the previous exercise. Prove that the series

∑∞n=1 an

is divergent. Hint: nth term test.

Solution.Let ε = L−1. Then there is a positive integer N such that ||an|

1n −L| < L−1.

This implies that L− |an|1n < L− 1 or |an|

1n > 1. Hence, for n ≥ N we have

|an| > 1. But this means that limn→∞ 6= 0. Hence, by the nth term test, thegiven series is divergent

Exercise 31.10The root test is inconclusive if L = 1.(a) We know that the series

∑∞n=1

1n2 is absolutely convergent. Show that

L = 1.(b) We know that the series

∑∞n=1(−1)n−1 1

nis conditionally convergent. Show

that L = 1.(c) We know that the series

∑∞n=1

1n

is divergent. Show that L = 1.

183

Solution.(a) The given series is a p−series with p = 2 so it is convergent by Exercise

??. Let y =(

1n2

) 1n . Then ln y = −2 ln n

n→ 0 as n → ∞. Thus, y → 1 as

n →∞. That is, L = 1.(b) The given series is conditionally convergent by Exercise ??. Let y =(

1n

) 1n . Then ln y = − ln n

n→ 0 as n → ∞. Thus, y → 1 as n → ∞. That is,

L = 1.(c) The given series is the harmonic series so it is divergent. As in (b),L = 1

Exercise 31.11Use the root test to show that the series

∑∞n=1

nn

31+2n is divergent.

Solution.We have

limn→∞

∣∣∣∣ nn

31+2n

∣∣∣∣ 1n

=1

3lim

n→∞

n

9= ∞.

By the root test the given series is divergent

Exercise 31.12Use the root test to show that the series

∑∞n=1

(5n−3n3

7n3+2

)n

is absolutely con-

vergent.

Solution.We have

limn→∞

|an|1n = lim

n→∞

∣∣∣∣5n− 3n3

7n3 + 2

∣∣∣∣ =3

7< 1.

By the root test the given series is convergent

184

Exercise 32.1Define fn : [0,∞) → R by fn(x) = nx

1+n2x2 . Show that the sequence {fn}∞n=1

converges pointwise to the function f(x) = 0 for all x ≥ 0.

Solution.For all x ≥ 0, limn→∞ fn(x) = 0

Exercise 32.2For each positive integer n let fn : (0,∞) → ∞ be given by fn(x) = nx.Show that {fn}∞n=1 does not converge pointwise on D.

Solution.This follows from the fact that limn→∞ nx = ∞ for all x ∈ D

Exercise 32.3For each positive integer n let fn : [0, 1] → ∞ be given by fn(x) = x

n. Show

that {fn}∞n=1 converges uniformly to the zero function. Hint: For a given ε,choose N such that N > 1

ε.

Solution.Let ε > 0 be given. Let N be a positive integer such that N > 1

ε. Then for

n ≥ N we have

|fn(x)− f(x)| = |x|n≤ 1

n≤ 1

N< ε

for all x ∈ [0, 1]

Exercise 32.4Define fn : [0,∞) → R by fn(x) = nx

1+n2x2 . By Exercise 32.1, this sequence

converges pointwise to f(x) = 0. Let ε = 13. Show that there is no positive

integer N with the property n ≥ N implies |fn(x)− f(x)| < ε for all x ≥ 0.Hence, the given sequence does not converge uniformly to f(x).

Solution.For any positive integer N and for n ≥ N we have∣∣∣∣fn

(1

n

)− f

(1

n

)∣∣∣∣ =1

2> ε

185

Exercise 32.5Define fn : [0, 1] → R by fn(x) = xn. Define f : [0, 1] → R by

f(x) =

{0 if 0 ≤ x < 11 if x = 1

(a) Show that the sequence {fn}∞n=1 converges pointwise to f.(b) Show that the sequence {fn}∞n=1 does not converge uniformly to f. Hint:Suppose otherwise. Let ε = 0.5 and get a contradiction by using a point(0.5)

1N < x < 1.

Solution.(a) For all 0 ≤ x < 1 we have limn→∞ fn(x) = limn→∞ xn = 0. Also,limn→∞ fn(1) = 1. Hence, the sequence {fn}∞n=1 converges pointwise to f.(b) Suppose the contrary. Let ε = 1

2. Then there exist a positive integer N

such that for all n ≥ N we have

|fn(x)− f(x)| < 1

2

for all x ∈ [0, 1]. In particular, we have

|fN(x)− f(x)| < 1

2

for all x ∈ [0, 1]. Choose (0.5)1N < x < 1. Then |fN(x)− f(x)| = xN > 0.5 =

ε which is a contradiction. Hence, the given sequence does not convergeuniformly

Exercise 32.6Give an example of a sequence of continuous functions {fn}∞n=1 that convergespointwise to a discontinuous function f.

Solution.See Exercise 32.5(a)

Exercise 32.7Suppose that for each n ≥ 1 the function fn : D → R is continuous in D.Suppose that {fn}∞n=1 converges uniformly to f. Let a ∈ D.(a) Let ε > 0 be given. Show that there is a positive integer N such that ifn ≥ N then |fn(x)− f(x)| < ε

3for all x ∈ D.

186

(b) Show that there is a δ > 0 such that for all |x− a| < δ we have |fN(x)−fN(a)| < ε

3

(c) Using (a) and (b) show that for |x − a| < δ we have |f(x) − f(a)| < ε.Hence, f is continuous in D since a was arbitrary. Symbolically we write

limx→a

limn→∞

fn(x) = limn→∞

limx→a

fn(x)

Solution.(a) This follows from the definition of uniform convergence.(b) This follows from the fact that fN is continuous at a ∈ D.(c) For |x− a| < δ we have |f(x)− f(a)| = |f(a)− fN(a) + fN(a)− fN(x) +fN(x)−f(x)| ≤ |fN(a)−f(a)|+|fN(a)−fN(x)|+|fN(x)−f(x)| < ε

3+ ε

3+ ε

3=

ε

Exercise 32.8Consider the interval [0, 1] and let the rationals in this interval be labeledr1, r2, · · · arranged in increasing order. For each positive integer n we definethe function fn : [0, 1] → R by

fn(x) =

{1 if x ∈ {r1, r2, · · · , rn}0 otherwise.

(a) Show that fn is Riemann integrable on [0, 1]. Hint: Remark 3.(b) Show that {fn}∞n=1 converges pointwise to the function

f(x) =

{1 if x is rational0 if x is irrational

(c) Show that f is not Riemann integrable.

Solution.(a) Since fn is bounded and discontinuous at a finite number of points, byRemark 3, fn is Riemann integrable.(b) If x is irrational then for every n ≥ 1 we have fn(x) = 0. Thus, limn→∞ fn(x) =0 = f(x). Suppose now that x is rational. Then there is a positive in-teger k such that x = rk. In this case, fn(x) = 1 for all n ≥ k. Hence,limn→∞ fn(x) = 1 = f(x). It follows that the given sequence converges point-wise to f.(c) This follows from Exercise 20.5

187

limn→∞

∫D

fn(x)dx =

∫D

limn→∞

fn(x)dx =

∫D

f(x)dx? (3)

That is, can we interchange limit and integration? The answer is no as seenin the next exercise.

Exercise 32.9Consider the functions fn : [0, 1] →∞ defined by fn(x) = n2xe−nx.(a) Show that {fn}∞n=1 converges pointwise to f(x) = 0. Hint: L’Hopital’srule.(b) Find limn→∞

∫ 1

0fn(x)dx. Hint: Integration by parts.

(c) Show that limn→∞∫ 1

0fn(x)dx 6=

∫ 1

0limn→∞ fn(x)dx.


limn→∞

fn(x) = limn→∞

n2xe−nx = limn→∞

n2x

enx= 0

where we apply L’Hopital’s rule twice. Hence, {fn}∞n=1 converges pointwiseto f(x) = 0.(b) Using integration by parts we find∫ 1

0

n2xe−nxdx = n2

[−xe−nx

n− 1

n2e−nx

]1

0

= 1− e−n(n + 1).

Hence,

limn→∞

∫ 1

0

fn(x)dx = limn→∞

[1− e−n(n + 1)] = 1.

(c) From (b) we have

limn→∞

∫ 1

0

fn(x)dx = 1 6= 0 =

∫ 1

0

limn→∞

fn(x)dx

Exercise 32.10Let {fn}∞n=1 be a sequence of Riemann integrable functions on [a, b] thatconverges uniformly to a f defined on [a, b].(a) Let ε > 0 be given. Show that there is a positive integer N such that forall n ≥ N we have

|fn(x)− f(x)| < ε4(b−a)

for all x ∈ [a, b].

188

(b) Let n ≥ N. Show that there is a partition P of [a, b] such that

U(fn, P )− L(fn, P ) <ε

2.

(c) Suppose n ≥ N and P as in (b). Show that

U(f, P ) ≤ U(fn, P ) +ε

4

and thereforeL(f, P ) ≥ L(fn, P )− ε

4.

Hint: |f(x)| ≤ |fn(x)|+ ε4(b−a)

and |fn(x)| ≤ |f(x)|+ ε4(b−a)

(d) Conclude that

U(f, P )− L(f, P ) < ε and therefore f is Riemann integrable on [a, b].

Solution.(a) This is just the definition of uniform convergence.(b) This follows from Exercise 20.6.(c) Since |f(x)| ≤ |fn(x)|+ ε

4(b−a)we have

U(f, P ) ≤ U(fn, P ) + U(ε

4(b− a), P ) = U(fn, P ) +

ε

4.

Likewise, since |fn(x)| ≤ |f(x)|+ ε2(b−a)

L(f, P ) ≥ L(fn, P ) + L(−ε

4(b− a), P ) = L(fn, P )− ε

4.

Hence,

U(f, P )− L(f, P ) ≤ [U(fn, P )− L(fn, P )] +ε

2<

ε

2+

ε

2= ε.

It follows that f is Riemann integrable on [a, b]

Exercise 32.11Let {fn}∞n=1 and f be as in the previous exercise.(a) Let ε > 0 be given. Show that there is a positive integer N such that ifn ≥ N then

|fn(x)− f(x)| ≤ εb−a


189

(b) Show that for every n ≥ N we have∣∣∣∣∫ b

a

fn(x)dx−∫ b

a

f(x)dx

∣∣∣∣ < ε.

Thus, (3) holds. Hint: Exercise 23.1 and Exercise 24.3

Solution.(a) This follows from the definition of uniform convergence.(b) Using (a) and Exercise 24.3 we find∣∣∣∣∫ b

a

fn(x)dx−∫ b

a

f(x)dx

∣∣∣∣ =

∣∣∣∣∫ b

a

[fn(x)− f(x)]dx

∣∣∣∣≤∫ b

a

|fn(x)− f(x)|dx

<ε

b− a

∫ b

a

dx = ε

Exercise 32.12Give an example of a sequence of differentiable functions {fn}∞n=1 that con-verges pointwise to a non-differentiable function f.

Solution.See Exercise 32.5(a)

Exercise 32.13

Consider the family of functions fn : [−1, 1] given by fn(x) =√

x2 + 1n.

(a) Show that fn is differentiable for each n ≥ 1.(b) Show that for all x ∈ [−1, 1] we have

|fn(x)− f(x)| ≤ 1√n

where f(x) = |x|. Hint: Note that√

x2 + 1n

+√

x2 ≥ 1√n.

(c) Let ε > 0 be given. Show that there is a positive integer N such that forn ≥ N we have

|fn(x)− f(x)| < ε for all x ∈ [−1, 1].

190

Thus, {fn}∞n=1 converges uniformly to the non-differentiable function f(x) =|x|.

Solution.(a) fn is the composition of two differentiable functions so it is differentiablewith derivative

f ′n(x) = x

[x2 +

1

n

]− 12

.

(b) We have

|fn(x)− f(x)| =

∣∣∣∣∣√

x2 +1

n−√

x2

∣∣∣∣∣ =

∣∣∣∣∣∣(√

x2 + 1n−√

x2)(√

x2 + 1n

+√

x2)√x2 + 1

n+√

x2

∣∣∣∣∣∣=

1n√

x2 + 1n

+√

x2

notag ≤1n1√n

(4)

(c) Let ε > 0 be given. Since limn→∞1√n

= 0 we can find a positive integer

N such that for all n ≥ N we have 1√n

< ε. Now the answer to the question

follows from this and part (b)

Exercise 32.14Give an example of a sequence of differentiable functions {fn}∞n=1 that con-verges uniformly to a a differentiable function f such that limn→∞ f ′n(x) 6=f ′(x) = [limn→∞ fn(x)]′ . That is, one cannot, in general, interchange limitsand derivatives. Hint: Exercise 32.3

Solution.Exercise ?? with limn→∞ f ′n(x) = g(x) where g(1) = 1 and g(x) = 0 for0 ≤ x < 1

Exercise 32.15Let {fn}∞n=1 defined on a set D be uniformly Cauchy.(a) Show that for each x ∈ D, the sequence of numbers {fn(x)}∞n=1 is con-vergent. Call the limit f(x). Thus, we can define a function f : D → R such

191

that f(x) = limn→∞ fn(x). Hint: Exercise 7.7(b) Show that {fn}∞n=1 converges pointwise to f.(c) Let ε > 0 be given. Show that there is a positive integer N such that forall m, n ≥ N we have

|fm(x)− fn(x)| < ε2

for all x ∈ D.

(d) Fix x ∈ D. Show that there is a positive integer m ≥ N such that|fm(x)− f(x)| < ε

2.

(e) For the fixed x in (d), let n ≥ N. Show that |fn(x)− f(x)| < ε.(f) Conclude that {fn}∞n=1 converges uniformly to f.

Solution.(a) Let x ∈ D. By uniform Cauchy, the sequence {fn(x)}∞n=1 is a Cauchysequence of numbers in R. By Exercise 7.7 it is convergent. Call its limitf(x). Hence, we define a function f : D → R by f(x) = limn→∞ fn(x).(b) This follows from the definition of f.(c) This follows from the definition of uniform Cauchy sequence.(d) Fix x ∈ D. The result follows from the fact that {fn}∞n=1 convergespointwise to f.(e) For n ≥ N we have

|fn(x)− f(x)| ≤ |fn(x)− fm(x)|+ |fm(x)− f(x)| < ε.

(f) Since the previous inequality is true for all x ∈ D we conclude that{fn}∞n=1 converges uniformly to f

Exercise 32.16Let {fn}∞n=1 be a sequence of differentiable functions on [a, b] such that{fn(c)}∞n=1 converges for some c ∈ [a, b]. Assume also that {f ′n}∞n=1 convergesuniformly to g in [a, b].(a) Let ε > 0 be given. Show that there is a positive integer N1 such that forall m, n ≥ N1 we have

|f ′m(x)− f ′n(x)| < ε2(b−a)


(b) Show that there is a positive integer N2 such that for all m, n ≥ N2 wehave

|fm(c)− fn(c)| < ε

2.

192

Hint: Exercise 7.3(c) Show that for all x ∈ [a, b] there is a d between c and x such that

fm(x)− fn(x) = fm(c)− fn(c) + (x− c)[f ′m(d)− fn(d)].

Hint: Apply the Mean Value theorem to the function fm − fn restricted tothe interval [c, x].(d) Let N = N1 + N2. Use (a) - (c) to show that for n ≥ N we have

|fm(x)− fn(x)| < ε for all x ∈ [a, b].

That is, the sequence {fn}∞n=1 is uniformly Cauchy.(e) Show that the sequence {fn}∞n=1 converges uniformly to a a function f.

Solution.(a) Let ε > 0 be given. There is a positive integer N ′ such that for all n ≥ N1

we have

|f ′n(x)− g(x)| < ε4(b−a)


Hence, for all m, n ≥ N1 we have

|f ′m(x)−f ′n(x)| ≤ |f ′m(x)−g(x)|+|f ′n(x)−g(x)| < ε

4(b− a)+

ε

4(b− a)=

ε

2(b− a)

(b) This follows from the fact that {fn(c)}∞n=1 is Cauchy by Exercise 7.3.(c) This is a result from the MVT.(d) For n ≥ N we have |fm(x) − fn(x)| ≤ |fm(c) − fn(c)| + (b − a)|f ′m(d) −f ′n(d)| < ε

2+ (b − a) ε

2(b−a)= ε. Hence, the sequence {fn}∞n=1 is uniformly

Cauchy.(e) This is a consequence of Exercise 32.15

Exercise 32.17In this exercise we want to show that f of the previous exercise is differen-tiable in [a, b] and f ′ = g.(a) Show that there is a positive integer N1 such that for all n ≥ N1 we have

|f ′m(x)− f ′n(x)| < ε3


193

(b) Let x0 ∈ [a, b]. Use the MVT to the function fm−fn to show the existenceof a point d between x0 and x such that

fm(x)− fn(x) = fm(x)− fn(x0) + (x− x0)[f′m(d)− f ′n(d)].

(c) Use (a) and (b) to show that∣∣∣∣fm(x)− fn(x0)

x− x0

− fn(x)− fn(x0)

x− x0

∣∣∣∣ < ε

3.

(d) Show that ∣∣∣∣f(x)− f(x0)

x− x0

− fn(x)− fn(x0)

x− x0

∣∣∣∣ ≤ ε

3.

(e) Show that there is a positive integer N2 such that for all n ≥ N2 we have

|f ′n(x0)− g(x0)| <ε

3.

(f) Let N = N1 + N2. Show that there is a δ > 0 such that

If 0 < |x− x0| < δ then∣∣∣fN (x)−fN (x0)

x−x0− f ′N(x0)

∣∣∣ < ε3.

(g) Use (d) - (f) to conclude that

If 0 < |x− x0| < δ then∣∣∣f(x)−f(x0)

x−x0− g(x0)

∣∣∣ < ε.

That is, f is differentiable at x0 with f ′(x0) = g(x0.

Solution.(a) Similar to (a) in the previous exercise.(b) Easy.(c) We have∣∣∣∣fm(x)− fn(x0)

x− x0

− fn(x)− fn(x0)

x− x0

∣∣∣∣ = |f ′m(d)− f ′n(d)| < ε

3.

(d) This follows by letting m →∞.(e) This follows from the fact that {f ′n(c)}∞n=1 converges to g(x0).

194

(f) This follows from the fact that fN is differentiable at x0.(g) Suppose 0 < |x− x0| < δ. Then∣∣∣∣f(x)− f(x0)

x− x0

− g(x0)

∣∣∣∣ =

∣∣∣∣f(x)− f(x0)

x− x0

− fN(x)− fN(x0)

x− x0

+fN(x)− fN(x0)

x− x0

− f ′N(x0) + f ′N(x0)− g(x0)

∣∣∣∣<

ε

3+

ε

3+

ε

3= ε

Hence, f is differentiable at x0 with f ′(x0) = g(x0)

Exercise 32.18Consider the sequence of functions fn(x) = x− xn

ndefined on [0, 1).

(a) Does {fn}∞n=1 converge to some limit function? If so, find the limit func-tion and show whether the convergence is pointwise or uniform.(b) Does {f ′n}∞n=1 converge to some limit function? If so, find the limit func-tion and show whether the convergence is pointwise or uniform.

Solution.(a) Let ε > 0 be given. Let N be a positive integer such that N > 1

ε. Then

for n ≥ N ∣∣∣∣x− xn

n− x

∣∣∣∣ =|x|n

n<

1

n≤ 1

N< ε.

Thus, the given sequence converges uniformly (and pointwise) to the functionf(x) = x.(b) Since limn→∞ f ′n(x) = 1 for all x ∈ [0, 1), the sequence {f ′n}∞n=1 convergespointwise to f(x) = 1. However, the convergence is not uniform. To seethis, let ε = 1

2and suppose that the convergence is uniform. Then there is a

positive integer N such that for n ≥ N we have

|1− xn−1 − 1| = |x|n−1 <1

2.

In particular, if we let n = N + 1 we must have xN < 12

for all x ∈ [0, 1).

But x =(

12

) 1N ∈ [0, 1) and xN = 1

2which contradicts xN < 1

2. Hence, the

convergence is not uniform

Exercise 32.19Suppose that each fn is uniformly continuous on D and that fn → f uni-formly on D. Prove that f is uniformly continuous on D.

195

Solution.Let ε > 0 be given. By uniform convergence, there is a positive integer Nsuch that |fn(x)− f(x)| < ε

3for all x ∈ D whenever n ≥ N. Now, since fN is

uniformly continuous on D we can find a δ > 0 such that if |x− y| < δ then|fn(x)− fN(y)| < ε

3. Thus, for any x, y ∈ D such that |x− y| < δ we have

|f(x)− f(y)| =|f(x)− fN(x) + fN(x)− fN(y) + fN(y)− f(y)|≤|f(x)− fN(x)|+ |fN(x)− fN(y)|+ |fN(y)− f(y)|

<ε

3+

ε

3+

ε

3= ε

Hence, f is uniformly continuous on D

Exercise 32.20Let fn(x) = xn

1+xn for x ∈ [0, 2].(a) Find the pointwise limit f(x) = limn→∞ fn(x) on [0, 2].(b) Does fn → f uniformly on [0, 2]?

Solution.(a) The pointwise limit is

f(x) =

0 if 0 ≤ x < 112

if x = 11 if 1 < x ≤ 2

(b) The convergence cannot be uniform because if it were f would have tobe continuous

Exercise 32.21Prove that if fn → f and gn → g uniformly on a set D then fn + gn → f + guniformly on D.

Solution.Let ε > 0 be given. Let N1 be a positive integer such that |fn(x)− f(x)| < ε

2

for all n ≥ N1 and all x ∈ D. Likewise, let N2 be a positive integer such that|gn(x)− g(x)| < ε

2for all n ≥ N2 and all x ∈ D. Let N = N1 + N2. Then for

n ≥ N we have

|(fn(x) + gn(x))− (f(x) + g(x))| < ε

2+

ε

2= ε

for all x ∈ D. That is, fn + gn → f + g uniformly on D

196

Exercise 32.22Prove that if fn → f uniformly on a set D then {fn}∞n=1 uniformly Cauchyon D.

Solution.Let ε > 0 be given. Then there is a positive integer N such that if n ≥ Nthen |fn(x)− f(x)| < ε

2for all x ∈ D. Thus, if m, n ≥ N we have

|fn(x)− fm(x)| ≤ |fn(x)− f(x)|+ |fm(x)− f(x)| < ε

2+

ε

2= ε

for all x ∈ D. This shows that {fn}∞n=1 uniformly Cauchy on D

Exercise 32.23Suppose that {fn}∞n=1 is uniformly convergent on a set D where each fn isbounded on D, that is |fn(x)| ≤ Mn for all x ∈ D. Show that there is apositive constant M such that |fn(x)| ≤ M for all n ∈ N and all x ∈ D.

Solution.By the previous exercise {fn}∞n=1 is uniformly Cauchy on D. Let ε = 1. Thenthere is a positive integer N such that

|fn(x)− fN(x)| < 1 for all n ≥ N and all x ∈ D.

Hence,

|fn(x)| < 1 + MN for all n ≥ N and all x ∈ D.

Let M = 1 + M1 + M2 + · · · + MN−1. Then |fn(x)| ≤ M for all x ∈ D andall n ∈ N

Exercise 32.24Suppose that fn → f and gn → g uniformly on D. Moreover, suppose that|fn(x)| ≤ Mn and |gn(x)| ≤ Mn for all n ∈ N and all x ∈ D. Prove thatfngn → fg uniformly on D.

Solution.By the previous exercise there exist M1 > 0 and M2 > 0 such that |fn(x)| ≤M1 and |gn(x)| ≤ M2 for all x ∈ D and all n ∈ N. Let M = M1 + M2.Then |fn(x)| ≤ M and |gn(x)| ≤ M for all x ∈ D and all n ∈ N. Now,|f(x)| − |fn(x) − f(x)| ≤ |f(x) − fn(x) + f(x)| = |fn(x)| ≤ M → |f(x)| ≤

197

M + |fn(x) − f(x)| for all n ∈ N and all x ∈ D. By pointwise convergencewe obtain |f(x)| ≤ M for all x ∈ D. Likewise, |g(x)| ≤ M for all x ∈ D.Let ε > 0 be given. Then there is a positive integer N1 such that if n ≥ N1

then |fn(x)−f(x)| < ε2M

for all x ∈ D. Likewise, there is a positive integer N2

such that if n ≥ N2 then |gn(x)−g(x)| < ε2M

for all x ∈ D. Let N = N1 +N2.Then for n ≥ N we have

|fn(x)gn(x)− f(x)g(x)| ≤|fn(x)||gn(x)− g(x)|+ |g(x)||fn(x)− f(x)|

<Mε

2M+ M

ε

2M= ε

for all x ∈ D. This shows that fngn → fg uniformly on D

Exercise 32.25Let fn(x) = x + 1

nfor all x ∈ R and gn(x) =

(x + 1

n

)2.

(a) Show that fn → f uniformly where f(x) = x.(b) Show that gn does not converge uniformly to the function g(x) = x2.

Solution.(a) Let ε > 0 be given. Choose N such that N > 1

ε. Then for n ≥ N we have

|fn(x)− f(x)| = 1

n≤ 1

N< ε.

This shows that fn converges uniformly to f.(b) Suppose the contrary. Let ε = 1. Then there positive integer N such thatif n ≥ N we have

|gn(x)− g(x)| =∣∣∣∣2xn +

1

n2

∣∣∣∣ < 1

for all n ≥ N and all x ∈ R. But if we choose n = N and x = N we obtain

|gN(N)− g(N)| = 2N2 + 1

N2> 1

a contradiction. Hence, gn does not convergen to g uniformly

Exercise 32.26Give an example of a sequence {fn}∞n=1 and a function f such that fn → funiformly but f 2

n does not converge uniformly to f 2.

Solution.Let fn(x) = x + 1

nand use previous exercise

198

Exercise 32.27Give an example of two sequences {fn}∞n=1 and {gn}∞n=1 such that fn → fand gn → g uniformly but fngn does not converge uniformly to fg. Thus,the condition of boundedness in Exercise 32.24 is crucial.

Solution.Let fn(x) = gn(x) = x + 1

nand f(x) = g(x) = x and use previous exercise

199

∑∞n=0 an(x − a)n is a power series that converges for x = c.

Note that the series converges to a0 if c = a. So we will assume that c 6= a.(a) What is the value of the limit limn→∞ an(c− a)n?(b) Show that there is a positive integer N such that |an(c− a)n| < 1 for alln ≥ N.(c) Let M =

∑N−1n=0 |an(c−a)n|+1. Show that |an(c−a)n| ≤ M for all n ≥ 0.

(d) Let x be such that |x− a| < |c− a|. Show that for any n ≥ 0 we have

|an(x− a)n| ≤ M

∣∣∣∣x− a

c− a

∣∣∣∣n .

(e) Show that the series∑∞

n=0 M∣∣x−a

c

∣∣n is convergent.(f) Show that the series

∑∞n=0 an(x− a)n is absolutely convergent and hence

convergent.We conclude that if a power series

∑∞n=0 an(x− a)n converges for x = c it is

convergent for any x satisfying |x− a| < |c− a|.

Solution.(a) By the nth term test, limn→∞ an(c− a)n = 0.(b) From the definition of convergence of a sequence and part (a), there existsa positive integer N such that |an(c− a)n| < 1 for all n ≥ N.(c) If n ≤ N − 1 then |an(c − a)n| ≤

∑N−1n=0 |an(c − a)n| < M. If n ≥ N we

have |an(c− a)n| < 1 < M.(d) For n ≥ 0 we have

|an(x− a)n| = |an(c− a)n| ·∣∣∣∣x− a

c− a

∣∣∣∣n ≤ M

∣∣∣∣x− a

c− a

∣∣∣∣n .

(e) This series is a geometric series with ratio∣∣x−a

c−a

∣∣ < 1 so it is convergent.(f) This follows from the comparison test and Exercise 29.2


∑∞n=0 an(x − a)n is a power series that diverges for x = d.

Let x be a number satisfying |x − a| > |d − a|. Show that the assumption∑∞n=0 an(x − a)n converges at x leads to a contradiction. Hence, the series∑∞n=0 an(x− a)n must be divergent. Hint: Use Exercise 33.1.

200

Solution.Suppose that |x − a| > |d − a|. If

∑∞n=0 an(x − a)n is convergent then by

Exercise 33.1, the series∑∞

n=0 an(d− a)n is absolutely convergent and henceconvergent. But this contradicts the fact that

∑∞n=0 an(d− a)n is divergent.

Hence,∑∞

n=0 an(x− a)n must be divergent

Exercise 33.3Consider a power series

∑∞n=0 an(x− a)n. Let C be the collection of all real

numbers at which the series∑∞

n=0 an(x− a)n converges. That is,

C = {x ∈ R :∞∑

n=0

an(x− a)n converges}.

(a) Show that C 6= ∅.(b) Explain in words the meaning that C = {a}.(c) Explain in words the meaning that C = (−∞,∞) = R.(d) Suppose that C 6= {a} and C 6= R. That is, there is a real number d 6= asuch that

∑∞n=0 an(d−a)n diverges. Show that if x ∈ C then |x−a| ≤ |d−a|.

Conclude that {|x−a| : x ∈ C} is bounded from above with an upper boundM. What is the value of M?(e) Show that there is a finite number R such that R is the least upper boundof {|x− a| : x ∈ C}. Thus, |x− a| ≤ R for all x ∈ C. Show that R > 0.(f) Show that for any real number x such that |x − a| > R, the series∑∞

n=0 an(x− a)n is divergent.(g) Show that for any real number x such that |x − a| < R, the series∑∞

n=0 an(x − a)n is convergent. Hint: Let ε = R − |x − a| and use the def-inition of supremum to show that there exist an x0 ∈ C such that R − ε <|x0 − a| ≤ R.

Solution.(a) Since the series converges at x = a to a0, we have a ∈ C and so C 6= ∅.(b) If C = {a} the series converges only at x = a and diverges for all x 6= a.(c) If C = (−∞,∞) then the series converges for all values of x.(d) This follows from Exercise 33.2. Thus, for any x ∈ C we have |x− a| ≤|d − a|. This shows that the set {|x − a| : x ∈ C} bounded from above byM = |d− a|.(e) The existence of R follows from the completeness axiom of R. Since|x− a| ≤ R for all x ∈ C and a, d ∈ C with d 6= a we conclude that R > 0.

201

(f) Let x ∈ R such that |x−a| > R. Then x 6∈ C and therefore∑∞

n=0 an(x−a)n

is divergent.(g) Let x ∈ R such that |x− a| < R. Let ε = R−|x− a|. By the definition ofsupremum, there exists an x0 ∈ C such that R − ε < |x0 − a| ≤ R. But thisimplies that |x− a| < |x0 − a|. By Exercise 33.1, the series

∑∞n=0 an(x− a)n

is (absolutely) convergent

Exercise 33.4Find the radius of convergence of each of the following series:(a)∑∞

n=0xn

n!.

(b)∑∞

n=0 n!xn.(c)∑∞

n=0 xn.

Solution.(a) By the ratio test the series converges for any value of x. Thus, R = ∞.(b) By the ratio test the series converges only when x = 0. Thus, R = 0.(c) This is a geomteric series that converges for |x| < 1 so that R = 1


∑∞n=0 an(x − a)n is a power series with an 6= 0 for all n ≥ 0.

Suppose that

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = L ≥ 0.

(a) Find limn→∞

∣∣∣an+1(x−a)n+1

an(x−a)n

∣∣∣ .(b) Suppose that L = 0. Show that R = ∞. That is, a power series convergesfor all x ∈ R.(c) Suppose that L > 0. Show that R = 1

L.

(d) Suppose that L = ∞. Show that R = 0, that is, the series diverges forall x 6= a.

Solution.(a) By the ratio test we have

limn→∞

∣∣∣∣an+1(x− a)n+1

an(x− a)n

∣∣∣∣ = limn→∞

∣∣∣∣an+1

a)n

∣∣∣∣ |x− a| = L|x− a|.

(b) If L = 0 then L|x − a| = 0 < 1 for all x ∈ R. Hence, R = ∞ and thepower series converges for all x ∈ R.

202

(c) A power series converges if and only if L|x − a| < 1 and diverges forL|x− a| > 1. Thus, R = 1

L.

(d) If L = ∞ > 1 then the series diverges for all x 6= a. That is, R = 0

Exercise 33.6Find the interval of convergence of the power series

∑∞n=1(−1)n (x−1)n

n.

Solution.We have 1

R= limn→∞

nn+1

= 1 so that R = 1. Hence, the series converges for|x− 1| < 1 and diverges for |x− 1| > 1. So the series converges for all x suchthat 0 < x < 2. What about the endpoints x = 0 and x = 2? If we replacex by 0 we obtain the series −

∑∞n=1

1n

which is divergent (harmonic series).If we replace x by 2 we obtain the alternating series

∑∞n=1(−1)n−1 1

nwhich

converges by the alternating series test. Thus, the interval of convergence is0 < x ≤ 2


∑∞n=1

nn2+2

xn.

Solution.Let R be the radius of convergence. Then

1

R= lim

n→∞

n + 1

(n + 1)2 + 2· n2 + 2

n= 1.

Thus, R = 1 so the series converges for |x| < 1 and diverges for |x| > 1. Ifx = 1 then the series becomes

∑∞n=1

nn2+2

. Since

limn→∞

nn2+2

1n

= 1 > 0

and the series∑∞

n=11n

is divergent, by the limit comparison test, the series∑∞n=1

nn2+2

is also divergent.

If x = −1 then we get the alternating series∑∞

n=1(−1)nnn2+2

. By the alternatingseries test one can show that this series is convergent. Thus, the interval ofconvergence is −1 ≤ x < 1


∑∞n=1

(−1)n−1xn

n2+1.

203

Solution.Use the absolute ratio test:∣∣∣∣an+1

an

∣∣∣∣ =|x|n+1

(n + 1)2 + 1· n2 + 1

|x|n= |x|

(n2 + 1

n2 + 2n + 2

)→ |x| as n →∞.

Therefore the series converges absolutely if |x| < 1 and diverges if |x| > 1.If |x| = 1, the ratio test gives no information, so we have to look at theendpoints separately:

• At x = 1 we have the series∑∞

n=1(−1)n−1

n2+1an absolutely convergent series

by the alternating series test.• At x = −1 we have the series

∑∞n=1

1n2+1

an absolutely convergent series

by comparison to the series∑∞

n=11n2 .

Conclusion: This power series is absolutely convergent on [−1, 1] and divergeseverywhere else


∑∞n=1(−1)n−1

(e2

)n (x−1)n

n.

Solution.Use the absolute ratio test:∣∣∣∣an+1

an

∣∣∣∣ =en+1|x− 1|n+1

2n+1(n + 1)· 2nn

en|x− 1|n=

e

2· n

n + 1|x−1| → e

2|x−1| as n →∞.

Therefore the series converges absolutely if |x − 1| < 2e

and diverges if|x− 1| > 2

e. If |x− 1| = 2

e, the ratio test gives no information, so we have to

look at the endpoints separately:• At x− 1 = −2

ewe have the series −

∑∞n=1

1n

a divergent series.

• At x − 1 = 2e

we have the series∑∞

n=1(−1)n−1

na conditionally convergent

series.Conclusion: This power series is absolutely convergent on |x − 1| < 2

e, con-

ditionally convergent at x = 2e

and divergent everywhere else

Exercise 33.10Suppose that the power series

∑∞n=0 anx

n converges if x = −3 and divergesif x = 7. Indicate which of the following statements must be true, cannot betrue, or may be true.(a) The power series converges if x = −10.

204

(b) The power series diverges if x = 3.(c) The power series converges if x = 6.(d) The power series diverges if x = 2.(e) The power series diverges if x = −7.(f) The power series converges if x = −4.

Solution.The thing we know for sure is that the series converges for |x| < 3 anddiverges for |x| > 7.(a) Cannot be true.(b) May be true.(c) May be true.(d) Cannot be true.(e) May be true.(f) May be true

Exercise 33.11Give an example of a power series that converges on the interval [−11,−3).


∑∞n=1

(x+7)n

n4n

Exercise 33.12Determine all the values of the real number x for which the series

∞∑n=1

xn

3nn(log (3n))3

converges

Solution.The radius of convergence is∣∣∣∣ an

an+1

∣∣∣∣ = 3 · n + 1

n

(log (3n + 3)

log 3n

)3

→ 3 as n →∞.

So the series converges for all x satisfyin |x| < 3. If x = −3 the series becomes∑∞n=1

(−1)n

n(log 3n)3which converges by the alernating series test. If x = 3 the

series becomes∑∞

n=11

n(log 3n)3which converges by the integral test. Thus, the

given series converges on the interval [−3, 3] and diverges elsewhere

205

Exercise 34.1Find the Taylor series of f(x) = 1

1−x, where −1 < x < 1.

Solution.Finding successive derivatives we obtain

f ′(x) =(1− x)−2 f ′(0) = 1

f ′′(x) =2(1− x)−3 f ′′(0) = 2 = 2!

f ′′′(x) =3 · 2(1− x)−4 f ′′′(0) = 3!

...

f (n)(x) =n · (n− 1) · · · 3 · 2(1− x)−(n+1) f (n)(0) = n!

Thus,

Pn(x) = 1 + x + x2 + x3 + · · ·+ xn =1− xn+1

1− x.

This is the nth partial sum of a geometric series that converges for |x| < 1.Moreover,

1 + x + x2 + x3 + · · · = 1

1− x.

This shows that1

1− x= 1 + x + x2 + · · ·

for all −1 < x < 1


f(x) =

{0 if x = 0

e−2

x2 if x 6= 0

(a) Find the Taylor polynomial of order n of f at x = 0.(b) Show that f(x) 6= limn→∞ Pn(x) for all x near 0. That is, the Taylorseries of f about x = 0 does not converge to f(x) for number very close to 0.

206

Solution.(a) Using a graphing calculator we see that the derivative of f of any orderis 0 at x = 0. Hence, for all n ≥ 0 we have Pn(x) = 0 for x close to 0. Thus,

for x close to 0 we have f(x) = e−2

x2 6= 0 = limn→∞ Pn(x)

Exercise 34.3(a) Show that the above result holds for n = 0. Hint: Apply the FundamentalTheorem of Caculus on the interval [a, x].(b) Suppose that the result holds for up to n. That is, for any x ∈ [a, a + h]we can estimate f(x) by Pn(x) for x near a :

f(x) = f(a)+f ′(a)

1!(x−a)+

f ′′(a)

2!(x−a)2 + · · ·+ f (n)(a)

n!(x−a)n +Rn+1(x)

Suppose that f has continuous derivatives up to order n+2. Use integrationby parts to show that

Rn+1(x) =f (n+1)(a)

(n + 1)!(x− a)n+1 + Rn+2(x).

Hence,

f(x) = f(a)+f ′(a)

1!(x−a)+

f ′′(a)

2!(x−a)2+· · ·+f (n+1)(a)

(n + 1)!(x−a)n+1+R)n + 2(x).

Solution.(a) By the Fundamental Theorem of Calculus we have f(x) = f(a)+

∫ x

af ′(t)dt.

Thus, the result is true for n = 0. (b) Suppose that the result holds up n.Suppose that f (n+2) exists and continuous. Using integration by parts wefind

1

n!

∫ x

a

f (n+1)(t)(x− t)ndt =−[

f (n+1)(t)

(n + 1)n!(x− t)n+1

]x

a

+1

n!

∫ x

a

f (n+2)(t)

n + 1(x− t)n+1dt

=f (n+1)(a)

(n + 1)!(x− a)n+1 +

1

(n + 1)!

∫ x

a

f (n+2)(t)(x− t)n+1dt

=f (n+1)(a)

(n + 1)!(x− a)n+1 + Rn+2(x)

207

Exercise 34.4 (Lagrange’s Form of Remainder)(a) Show that there exist x1, x2 ∈ [a, x] such that f (n+1)(x1) ≤ f (n+1)(t) ≤f (n+1)(x2) for all t ∈ [a, x].(b) Use (a) to show that

f (n+1)(x1)

(n + 1)!(x− a)n+1 ≤ Rn+1(x) ≤ f (n+1)(x2)

(n + 1)!(x− a)n+1

where

Rn+1(x) =1

n!

∫ x

a

f (n+1)(t)(x− t)ndt.

(c) Show that

f (n+1)(x1) ≤ Rn+1(x)(n + 1)!

(x− a)n+1≤ f (n+1)(x2).

(d) Show that there is a c ∈ [a, x] such that

f (n+1)(c) = Rn+1(x)(n + 1)!

(x− a)n+1

and therefore

Rn+1(x) =f (n+1)(c)

(n + 1)!(x− a)n+1.

Solution.(a) Since f (n+1)(t) is continuous on [a, x], by Exercise 17.4 there exist x1, x2 ∈[a, x] such that

f (n+1)(x1) ≤ f (n+1)(t) ≤ f (n+1)(x2) for all t ∈ [a, x].

(b) This follows by multiplying the result in (a) by (x − t)n and then inte-grating from a to x.(c) This follows by multiplying the result in (b) by the ration (n+1)!

(x−a)n+1 .

(d) This follows by applying the Intermediate Value Theorem to the functionf (n+1)(t) on the interval [a, x]

Exercise 34.5 (Estimating Rn+1(x))Suppose that there is M > 0 such that |f (n+1)(x)| ≤ M for all x ∈ [a, a + h].(a) Show that for all x ∈ [a, a + h] we have

|Rn+1(x)| ≤ M

(n + 1)!||x− a|n+1.

208

(b) Show thatlim

n→∞Rn+1(x) = 0.

Hint: Exercise 1.14 and Squeeze rulw.


|Rn+1(x)| =∣∣∣∣f (n+1)(c)

(n + 1)!(x− a)n+1

∣∣∣∣ ≤ M

(n + 1)!||x− a|n+1.

(b) Since

limn→∞

(x− a)n+1

(n + 1)!= 0

and

−M(x− a)n+1

(n + 1)!≤ Rn+1(x) ≤ M

(x− a)n+1

(n + 1)!

the result follows from the squeeze rule

Exercise 34.6Let f : R → R be a function such that f, f ′, f ′′ exist and are continuous.Furthermore, f ≥ 0 and f ′′ ≤ 0. Show that f is a constant function.

Solution.Let a ∈ R. Using Taylor Theorem we can write

f(x) = f(a) + f ′(a)(x− a) +f ′′(c)

2(x− a)2

for some c between a and x. But f ′′ ≤ 0 and f ≥ 0 so that the last equalityimplies that

f(a) + f ′(a)(x− a) = f(x)− f ′′(c)

2(x− a)2 ≥ f(x) ≥ 0

for all x ∈ R. But this true only when f ′(a) = 0 and f(a) ≥ 0. Since a wasarbitrary, we conclude that f ′(x) = 0 for all x ∈ R. Hence, f is a constantfunction

Exercise 34.7Find the Taylor polynomial of order n about 0 for f(x) = ex, and write downthe corresponding remainder term.

209

Solution.Since the derivative of f of any order is just ex we find

f(x) = 1 +x

1!+

x2

2!+ · · ·+ xn

n!+ Rn+1(x)

where

Rn+1(x) =ec

(n + 1)!xn+1

Exercise 34.8Find the Taylor Polynomial of order 3 for the function f(x) = cos x centeredat x = π

6.

Solution.Simple calculation leads to

P4(x) =

√3

2− 1

2

(x− π

6

)−√

3

4

(x− π

6

)2

+1

12

(x− π

6

)3

Exercise 34.9Find the Lagrange form of the remainder Rn+1(x) for the function f(x) =

11+x

.

Solution.By successive differentiation we find f (n+1)(x) = (−1)n+1(n+1)!(1+x)−(n+2).Hence,

Rn+1(x) =(−1)n+1

(1 + c)n+2xn+1

Exercise 34.10Let g(x) be a function such that g(5) = 3, g′(5) = −1, g′′(5) = 1 and g′′′(5) =−3.(a) What is the Taylor polynomial of degree 3 for g(x) near 5?(b) Use (a) to approximate g(4.9).

Solution.(a) We have: c0 = g(5) = 3, c1 = g′(5) = −1, c2 = g′′(5)

2!= 1

2, and c3 = g′′′(5)

3!=

−12. Thus, P3(x) = 3− (x− 5) + 1

2(x− 5)2 − 1

2(x− 5)3.

(b) g(4.9) = 3− (4.9− 5) + 12(4.9− 5)2 − 1

2(4.9− 5)3 = 3.1675.

210

Exercise 34.11Suppose that the function f(x) is approximated near x = 0 by a sixth degreeTaylor polynomial

P6(x) = 3x− 4x3 + 5x6.

Find the value of the following:(a) f(0) (b) f ′(0) (c) f ′′′(0) (d) f (5)(0) (e) f (6)(0)

Solution.If

P6(x) = c0 + c1x + c2x2 + c3x

3 + c4x4 + c5x

5 + c6x6

then c0 = 0, c1 = 3, c2 = 0, c3 = −4, c4 = c5 = 0, and c6 = 5.

(a) f(0) = c0 = 0, (b) f ′(0) = c1 = 3, (c) f ′′′(0) = 3!c3 = −24, (d)f (5)(0) = 5!c5 = 0, (e) f (6)(0) = 6!c6 = 3600.

Exercise 34.12Find the third degree Taylor polynomial approximating

f(x) = arctan x,

near a = 0.

Solution.We have

f(x) = arctan x, c0 = f(0) = 0

f ′(x) =1

1 + x2, c1 = f ′(0) = 1

f ′′(x) =− (1 + x2)−2(2x), c2 =f ′′(0)

2!= 0

f ′′′(x) =− 2(1 + x2)−3(4x2)− 2(1 + x2)−2, c3 =f ′′′(0)

3!= −1

3

Thus,

P3(x) = x− 1

3x3.

211

Exercise 34.13Find the fifth degree Taylor polynomial approximating

f(x) = ln (1 + x),

near a = 0.

Solution.We have

f(x) = ln (1 + x), c0 = f(0) = 0f ′(x) = 1

1+x, c1 = f ′(0) = 1

f ′′(x) = − 1(1+x)2

, c2 = f ′′(0)2!

= −12

f ′′′(x) = 2(1+x)3

, c3 = f ′′′(0)3!

= 13

f (4)(x) = − 6(1+x)4

, c4 = f (4)(0)4!

= −14

f (5)(x) = 24(1+x)5

, c5 = f (5)(0)5!

= 15

Thus,

P5(x) = x− 1

2x2 +

1

3x3 − 1

4x4 +

1

5x5.

212


Exercise 35.1Let f(x) = cos x.(a) Using successive differentiation find a formula for f (n)(0).(b) Show that

P2n(x) = P2n+1(x) = 1− 1

2!x2 +

1

4!x4 − · · ·+ (−1)n x2n

(2n)!=

n∑k=0

(−1)k x2k

(2k)!.

(c) Find the radius of convergence of the series

∞∑n=0

(−1)n x2n

(2n)!= 1− x2

2!+

x4

4!− · · ·

(d) Show that

|Rn+1(x)| ≤ |x|n+1

(n + 1)!.

(e) Show that limn→∞ Rn+1(x) = 0. Hence, conclude that

cos x = 1− x2

2!+

x4

4!− · · ·+ (−1)n x2n

(2n)!+ · · ·

Solution.(a) By successive differentiation we find

f ′(x) = − sin xf ′′(x) = − cos xf ′′′(x) = sin xf (4)(x) = cos x

...

We see that the derivatives go through a cycle of length 4 and then repeatthat cycle forever. It follows that

f (k)(0) =

{0 if k is odd

(−1)k2 if k is even.

213

(b) Hence,

P2n(x) = P2n+1(x) = 1− 1

2!x2 +

1

4!x4 − · · ·+ (−1)n x2n

(2n)!=

n∑k=0

(−1)k x2k

(2k)!.

(c) Now, consider the series

∞∑n=0

(−1)n x2n

(2n)!= 1− x2

2!+

x4

4!− · · ·

and let an = (−1)n

(2n)!. Then

1

R= lim

n→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

1

(2n + 2)(2n + 1)= 0 < 1.

This shows that the series is convergent for all values of x.(d) Since |f (n+1)(c)| ≤ 1 we find that

|Rn+1(x)| |x|n+1

(n + 1)!.

(e) But

limn→∞

xn+1

(n + 1)!= 0

so that limn→∞ Rn+1(x) = 0. This implies that

cos x = 1− x2

2!+

x4

4!− · · ·+ (−1)n x2n

(2n)!+ · · ·

Exercise 35.2Let f(x) = ex.(a) Find f (n)(0) for all n ≥ 0.(b) Find an expression for Pn(x).(c) Consider the series

∞∑n=0

xn

n!= 1 +

x

1!+

x2

2!+ · · · .

214

Find the radius of convergence.(d) Find an expression for Rn+1(x) and show that

limn→∞

Rn+1(x) = 0.

Hence, conclude that

ex = 1 +x

1!+

x2

2!+ · · · =

∞∑n=0

xn

n!.

Solution.(a) For all nonnegative integer k we have f (k)(x) = ex and f (k)(0) = 1.(b) Thus, the nth Taylor polynomial is given by the expression

Pn(x) = 1 +x

1!+

x2

2!+ · · ·+ xn

n!.

(c) Let an = 1n!

. Then by the ratio test we have

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

1

n + 1= 0 < 1.

Thus, the series∑∞

n=0xn

n!converges for all values of x.

(d) It remains to show that the series converges to ex. For that, we need touse Taylor Theorem. Write f(x) = Pn(x) + En(x) where

|En(x)| =∣∣∣∣f (n+1)(c)

(n + 1)!xn+1

∣∣∣∣ =ec

(n + 1)!|x|n+1

(e) Since limn→∞xn

n!= 0 we find

limn→∞

Rn+1(x) = 0.

Hence,

ex = 1 +x

1!+

x2

2!+ · · · =

∞∑n=0

xn

n!

Exercise 35.3Let f(x) = ln (1 + x).(a) Find f (n)(0) for all n ≥ 0.

215

(b) Find an expression for Pn(x).(c) Consider the series

∞∑n=0

(−1)n−1xn

n= 1 +

x

1!+

x2

2!+ · · · .

Find the radius of convergence.(d) Show that

|Rn+1(x)| ≤ 1

|1 + c|n+1· |x|n+1

(n + 1)!.

(e) Show thatlim

n→∞Rn+1(x) = 0.

Hence, conclude that

ln (1 + x) =∞∑

n=1

(−1)n−1xn

n, − 1 < x ≤ 1.

Solution.(a) Taking derivatives:

f ′(x) =(1 + x)−1 f ′(0) = 1 = 0!

f ′′(x) =− (1 + x)−2 f ′′(0) = −1!

f ′′′(x) =2(1 + x)−3 f ′′′(0) = 2!

f (4)(x) =− 6(1 + x)−4 f (4)(0) = −3!

...

f (n)(x) =(−1)n−1(n− 1)!(1 + x)−n f (n)(0) = (−1)n−1(n− 1)!

(b) Hence,

Pn(x) = x− x2

2+

x3

3− · · ·+ (−1)n−1xn

n=

n∑k=1

(−1)k−1xk

k.

(c) Letting an = (−1)n−1 1n

and applying the ratio test we find that

1

R= lim

n→∞

∣∣∣∣an+1

an

∣∣∣∣ = 1.

216

Hence, the series∑∞

n=1(−1)n−1 xn

nconverges for all −1 < x < 1. By the

alternating series test we know that∑∞

n=1(−1)n−1

nis convergent so the interval

of convergence of the previous series is −1 < x ≤ 1.(d) It remains to show that the series converges to ln (1 + x). Using Taylortheorem, we can write f(x) = Pn(x) + Rn+1(x), where

|Rn+1(x)| =∣∣∣∣f (n+1)(c)

(n + 1)!xn+1

∣∣∣∣ ≤ 1

|1 + c|n+1· |x|

n+1

(n + 1).

(e) Since limn→∞xn

n!= 0 we find

limn→∞

Rn+1(x) = 0.

Hence,

ln (1 + x) =∞∑

n=1

(−1)n−1xn

n, − 1 < x ≤ 1

Exercise 35.4Find the Taylor series of x

ex about x = 0.

Solution.Replacing x by −x in the expansion of ex we find

e−x = 1− x +x2

2!− · · ·+ (−1)n xn

n!+ · · ·

Now, multiplying both sides of this equality by x to obtain

x

ex= xe−x = x− x2 +

x3

2!+ · · ·+ (−1)n xn+1

n!+ · · ·

This series converges for all x

Exercise 35.5Find the Taylor series of f(x) = 1

1+x2 about x = 0.

Solution.Replacing x by −x2 in Formula (1− x)−1 we can write

1

1 + x2= 1− x2 + x4 − x6 + · · ·+ (−1)nx2n + · · ·

This series converges for −1 < x < 1.

217

Exercise 35.6Let f(x) = sin x.(a) Using successive differentiation find a formula for f (n)(0).(b) Show that

P2n(x) = P2n+1(x) = x− x3

3!+

x5

5!−· · ·+ (−1)n

(2n + 1)!x2n+1 =

n∑k=0

(−1)k x2k+1

(2k + 1)!.

(c) Find the radius of convergence of the series

∞∑n=0

(−1)n

(2n + 1)!x2n+1 = x− x3

3!+

x5

5!+ · · · .

(d) Show that

|Rn+1(x)| ≤ |x|n+1

(n + 1)!.

(e) Show that limn→∞ Rn+1(x) = 0. Hence, conclude that

sin x =n∑

k=0

(−1)k x2k+1

(2k + 1)!.

Solution.(a) By successive differentiation we find

f ′(x) = cos xf ′′(x) = − sin xf ′′′(x) = − cos xf (4)(x) = sin x

...

We see that the derivatives go through a cycle of length 4 and then repeatthat cycle forever. It follows that

f (k)(0) =

{0 if k is even

(−1)k−12 if k is odd.

(b) Hence, for n ≥ 1,

P2n(x) = P2n1(x) = x− x3

3!+

x5

5!−· · ·+ (−1)n

(2n + 1)!x2n+1 =

n∑k=0

(−1)k x2k+1

(2k + 1)!.

218

(c) Let an = (−1)n

(2n+1)!. Then

1

R= lim

n→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

1

(2n + 3)(2n + 2)= 0 < 1.

This shows that the series∑∞

n=0(−1)n

(2n+1)!x2n+1 is convergent for all values of x.

(d) Since |f (n+1)(c)| ≤ 1 we find that

|Rn+1(x)| |x|n+1

(n + 1)!.

(e) But

limn→∞

xn+1

(n + 1)!= 0

so thatlim

n→∞Rn+1(x) = 0.

This implies that

sin x =∞∑

n=0

(−1)n

(2n + 1)!x2n+1 = x− x3

3!+

x5

5!+ · · ·

Exercise 35.7Find the MacLaurin series of x

1−2x.

Solution.We have

x

1− 2x= x

∞∑n=0

(2x)n =∞∑

n=1

4n−1xn

which is valid for all −12

< x < 12

Exercise 35.8Find the coefficient of (x − 2)2 in the Taylor series expansion of f(x) = 1

x

about x = 2.

Solution.The coefficient is f ′′(2)

2!= 1

8

219

Exercise 35.9Find the Maclaurin series for the function f(x) = x6e−x2

. Give your answerin sigma notation.

Solution.We have

x6e−x2

= x6

∞∑n=0

(−x2)n

n!=

∞∑n=0

(−1)n x2n+6

n!

Exercise 35.10Compute each of the following sums in terms of known functions:

(a)∑∞

n=0(−1)nx4n+1

n!

(b)∑∞

n=1(−1)nx4n+1

(2n+1)!

(c)∑∞

n=0(−1)nx6n

(2n+2)!

Solution.(a)∑∞

n=0(−1)nx4n+1

n!=∑∞

n=0(−1)nx4n

n!= xe−x4

.

(b)∑∞

n=1(−1)nx4n+1

(2n+1)!= 1

x

∑∞n=1

(−1)n(x2)2n+1

(2n+1)!= sin x2−x2

x.

(c)∑∞

n=0(−1)nx6n

(2n+2)!= − 1

x6

∑∞n=1

(−1)n(x3)2n

(2n)!= 1−cos x3

x6

Exercise 35.11The hyperbolic cosine of x is defined to be the function cosh x = ex+e−x

2. Find

the MacLaurin series of cosh x.

Solution.Using the Taylor expansion of ex and e−x one can easily see that

cosh x =∞∑

n=0

x2n

(2n)!

Exercise 35.12The hyperbolic sine of x is defined to be the function sinh x = ex−e−x

2. Find

the MacLaurin series of sinh x.

Solution.Using the Taylor expansion of ex and e−x one can easily see that

sinh x =∞∑

n=0

x2n+1

(2n + 1)!

220

Exercise 35.13 (Binomial Series)Consider the function f(x) = (1 + x)n where n ∈ R.(a) Using successive differentiation show that f (k)(0) = k(k−1) · · · (k−n+1).

Thus, f (k)(0)k!

= C(n, k) where

C(n, k) = n!k!(n−k)!

and C(n, 0) = 1.

(b) Find the interval of convergence of the binomial series (1+x)n =∑∞

k=0 C(n, k)xk.

Solution.(a) This is done by successive differentiation.(b) Using the absolute ration test we find∣∣∣∣an+1

an

∣∣∣∣ =|n− k|k + 1

|x| → |x| as k →∞.

Thus, the interval of convergence if |x| < 1. Convergence at the endpointsdepends on the values of n and needs to be checked every time

Exercise 35.14Find the MacLaurin series of f(x) = 1√

x+1.

Solution.We have

(1 + x)−12 =

∞∑n=0

C(−1

2, k)xk = 1− 1

2x +

(1)(3)

222!x2 − (1)(3)(5)

233!x3 + · · ·

221

∑∞n=1 fn(x) converges uniformly on D. For each x ∈ D let

f(x) =∑∞

n=1 fn(x). That is, {Sn}∞n=1 converges uniformly to f.(a) Let ε > 0 be given. Show that there is a positive integer N such that ifn ≥ N we have ∣∣∣∣∣

n∑k=1

fk(x)− f(x)

∣∣∣∣∣ < ε

2

for all x ∈ D.(b) Show that for n > m ≥ N we have∣∣∣∣∣

n∑k=m+1

fk(x)

∣∣∣∣∣ =

∣∣∣∣∣n∑

k=1

fk(x)−m∑

k=1

fk(x)

∣∣∣∣∣ < ε

for all x ∈ D.

Solution.(a) Since the series

∑∞n=1 fn(x) is uniformly convergent, the sequence of par-

tial sums {Sn}∞n=1 is uniformly convergent, say to a function f. Thus for agiven ε > 0, there is a positive integer N such that for n ≥ N we have

|∑n

k=1 fk(x)− f(x)| < ε2

for all x ∈ D.

(b) If n > m ≥ N then we have∣∣∣∣∣n∑

k=m+1

fk(x)

∣∣∣∣∣ =

∣∣∣∣∣n∑

k=1

fk(x)−m∑

k=1

fk(x)

∣∣∣∣∣≤

∣∣∣∣∣n∑

k=1

fk(x)− f(x)

∣∣∣∣∣+∣∣∣∣∣

m∑k=1

fk(x)− f(x)

∣∣∣∣∣<

ε

2+

ε

2= ε

Exercise 36.2 (Weierstrass)For each integer n ≥, let fn : D → R be a continuous function that isbounded on D with |fn(x)| ≤ Mn for all x ∈ D. Suppose that the series

222

of numbers∑∞

n=1 Mn is convergent. For each positive integer n define thepartial sum

Sn(x) =n∑

k=1

fk(x).

(a) Let ε > 0 be given. Show that there is a positive integer N such that forall m, n ≥ N we have ∣∣∣∣∣

n∑k=1

Mk −m∑

k=1

Mk

∣∣∣∣∣ < ε.

Hint: The sequence {∑n

k=1 Mk}∞n=1 is Cauchy.(b) Suppose that n > m ≥ N. By (a) we have |

∑nk=m+1 Mk| < ε. Show that

for all x ∈ D we have|Sn(x)− Sm(x)| < ε

Hence, the sequence {Sn}∞n=1 is uniformly Cauchy.(c) Conclude that the series

∑∞n=1 fn is uniformly convergent. Hint: Exercise

32.15

Solution.(a) Since the series

∑∞n=1 Mn is convergent, the sequence {

∑nk=1 Mk}∞n=1 is

convergent and hence it is Cauchy. Thus, for a fixed ε we can find a positiveinteger N such that for m,n ≥ N we have∣∣∣∣∣

n∑k=1

Mk −m∑

k=1

Mk

∣∣∣∣∣ < ε.

(b) We have |Sn(x)−Sm(x)| =∣∣∑n

k=m+1 fk(x)∣∣ ≤∑n

k=m+1 |fk(x)| ≤∑n

k=m+1 Mk =|∑n

k=m+1 Mk| < ε. Hence, the sequence {Sn}∞n=1 is uniformly Cauchy.(c) This follows from Exercise 32.15

Exercise 36.3Use Weierstrass M test to show that the series

∑∞n=0

xn

3n converges uniformlyon [−2, 2].

Solution.Note first that for all x ∈ [−2, 2] we have

∣∣xn

3n

∣∣ =∣∣x3

∣∣n ≤ (23

)n. Let fn(x) = xn

3n

and Mn =(

23

).The series

∑∞n=0

(23

)nis a convergent geometric series. Hence,

by Weierstrass M test the given series is uniformly convergent in [−2, 2]

223

Exercise 36.4Let

∑∞n=0 anx

n be a power series with radius of convergence R. Let 0 < c < Rand D = [−c, c].(a) Define fn(x) = anx

n and Mn = |ancn|. Clearly, fn is continuous in D and

Mn > 0 for all integer n ≥ 0. Show that∑∞

n=0 Mn converges. Hint: Exercise33.1(f)(b) Let x ∈ D. Show that if x ∈ [0, c] then |gn(x)| ≤ Mn. Hint: xn isincreasing for x ≥ 0.(c) Answer the same question if x ∈ [−c, 0]. (d) Conclude that the series isuniformly convergent on D.

Solution.(a) Since 0 < c < R the series

∑∞n=0 anc

n is absolutely convergent by Exercise33.1(f). That is, the series

∑∞n=0 Mn is convergent.

(b) If x ∈ [0, c] then x ≥ 0 and (xn)′ ≥ 0 so that xn is increasing in [0, c].Since 0 ≤ x ≤ c w conclude that |fn(x)| = |an||x|n ≤ |an|cn = Mn.(c) If x ∈ [−c, 0] then −x ∈ [0, c] so that |fn(x)| = |an||x|n = |an|| − x|n ≤|an|cn = Mn.(d) This follows from Weiestrass M test

Exercise 36.5Show that the following series converges uniformly.

∞∑n=0

x2

3n(x2 + 1).

Solution.We have

0 ≤ x2

3n(x2 + 1)≤ 1

3n.


n=013n converges, the given series converges uniformly by

the Weierstrass M test

Exercise 36.6Let {an}∞n=1 be a bounded sequence with |an| ≤ M for all n ∈ N. Show thatther series

∑∞n=1

an

nx converges uniformly for all x ≥ c > 1.

224

Solution.We use the Weierstrass M-test. For x ≥ c we have

∣∣an

nx

∣∣ ≤ Mnc . Because c > 1,

the series∑∞

n=1Mnc converges (p-series with p > 1). Thus, by the Weierstrass

M=test, the given series converges uniformly and absolutely for x ≥ c


∑∞n=1

sin nxn2 converges uniformly for all x ∈ R.

Solution.We have

0 ≤∣∣∣∣sin nx

n2

∣∣∣∣ ≤ 1

n2


n=11n2 is convergent, the Weierstrass M-test asserts that

the given series converges uniformly for all x ∈ R.

Exercise 36.8Suppose that {fn}∞n=1 is a sequence of functions defined on a set D such that|fn+1(x) − fn(x)| ≤ Mn for all x ∈ D and n ∈ N. Assume that

∑∞n=1 Mn is

convergent. Show that the series∑∞

n=1 fn(x) is uniformly convergent on D.

Solution.By the Weierstrass M-Test, we know that the series

∑∞n=1(fn+1(x)− fn(x))

is uniformly convergent on D say to a function f. The nth partial sum ofthis series is fn+1(x)− f1(x). Hence, fn → f + f1 uniformly on D


∑∞n=1

x(1+x)n converges uniformly on [1, 2].

Solution.We have x ≤ 2 → 3x ≤ 2 + 2x → x

1+x≤ 2

3. Thus,

x

(1 + x)n≤ xn

(1 + x)n≤(

2

3

)n

.


n=1

(23

)nconverges, by the Weierstrass M-test the given

series is uniformly convergent on [1, 2]

Exercise 36.10Prove that

∑∞n=1 sin

(xn2

)converges uniformly on any bounded interval [a, b].

225

Solution.We have ∣∣∣sin( x

n2

)∣∣∣ ≤ |x|n2

≤ M

n2

where M = |a|+|b|. Since the series∑∞

n=1Mn2 is convergent, by the Weierstrass

M-test the given series is uniformly convergent


∑∞n=1

13n cos

(x3n

)converges uniformly on R.

Solution.We have

1

3ncos( x

3n

)≤ 1

3n.

The result now follows from Weierstrass M-test

226

Exercise 37.1Let c ∈ D. Let R0 > 0 be a number such that |c− a| < R0 < R. By Exercise36.4, the power series

∑∞n=0 an(x − a)n converges uniformly on the interval

[a−R0, a + R0].(a) Let ε > 0 be given. Show that there is a positive integer N such that forall n > m ≥ N we have∣∣∑n

k=0 ak(x− a)k −∑n

k=0 ak(x− a)k∣∣ =

∣∣∑nk=m+1 ak(x− a)k

∣∣ < ε3

for allx ∈ [a−R0, a + R0].

Hint: Exercise 36.1(b) Show that there is a δ1 > 0 such that if |x− a| < δ1 then∣∣∣∣∣

N∑k=0

ak(x− a)k −N∑

k=0

ak(c− a)k

∣∣∣∣∣ < ε

3.

(c) Let δ = min{δ1, R0 − |c− a|}. Show that for |x− a| < δ we have

|f(x)− f(c)| < ε.

Hence, the function f(x) =∑∞

n=0 an(x− a)n is continuous on D.

Solution.(a) This follows from Exercise 36.1.(b) Since

∑Nk=0 ak(x− a)k is a polynomial, it is continuous at c.

(c) Suppose that |x−a| < δ. Then |f(x)−f(c)| =∣∣∣(∑N

k=0 ak(x− a)k +∑∞

k=N+1 ak(x− a)k)−(∑N

k=0 ak(c− a)k +∑∞

k=N+1 ak(c− a)k)∣∣∣ ≤∣∣∣∑N

k=0 ak(x− a)k −∑N

k=0 ak(c− a)k∣∣∣+∣∣∑∞

k=N+1 ak(x− a)k∣∣+∣∣∑∞

k=N+1 ak(c− a)k∣∣ <

ε3

+ ε3

+ ε3

= ε. Hence, the function f(x) =∑∞

n=0 an(x− a)n is continuous atc. Since c was arbitrary, f(x) is continuous in D


∑∞n=0 an(x−a)n where the power series converges for |x−a| < R

and diverges for |x − a| > R. Let F (x) =∫ x

af(t)dt. Suppose that a − R <

x ≤ a. A similar result holds for a ≤ x < a + R. (a) Show that {Sn}∞n=1

converges uniformly to f on [x, a].(b) Evaluate

∫ a

xSn(t)dt.

(c) Show that the power series∑∞

n=0an(x−a)n+1

n+1has radius of convergence R.

(d) Show that F (x) =∑∞

n=0an(x−a)n+1

n+1. Hint: Exercise 32.11

227

Solution.(a) See Remark ?? (2).(b) By integration we have∫ a

x

Sn(t)dt =

∫ a

x

(n∑

k=0

ak(t− a)k

)dt =

[n∑

k=0

ak(t− a)k+1

k + 1

]a

x

= −n∑

k=0

ak(x− a)k+1

k + 1.

(c) Let R′ be the radius of convergence of∑∞

n=0an(x−a)n+1

n+1. We have

1

R′ = limn→∞

∣∣∣∣∣an+1(x−a)n+2

n+2

an(x−a)n+1

n+1

∣∣∣∣∣= lim

n→∞

n + 1

n + 2· lim

n→∞

∣∣∣∣an+1(x− a)n+1

an(x− a)n

∣∣∣∣ =1

R

Hence, R′ = R.(d) By Exercise 32.11 limit and integration can be interchanged. Hence,we have F (x) =

∫ x

af(t)dt = −

∫ a

xlimn→∞ Sn(t)dt = − limn→∞

∫ a

xSn(t)dt =

limn→∞∑n

k=0ak(x−a)k+1

k+1=∑∞

n=0an(x−a)n+1

n+1


∑∞n=0 an(x−a)n where the power series converges for |x−a| < R

and diverges for |x− a| > R.(a) Show that the power series g(x) =

∑∞n=1 nan(x − a)n−1 has radius of

convergence R.(b) Let G(x) =

∫ x

ag(t)dt. Show that G(x) = f(x)− a0 for |x− a| < R. Hint:

Exercise 37.2.(c) Show that g(x) = f ′(x) for all |x− a| < R. Hint: Exercise 25.2

Solution.(a) Let R′ be the radius of convergence of g(x). We have

1

R′ = limn→∞

∣∣∣∣(n + 1)an+1(x− a)n

nan(x− a)n−1

∣∣∣∣= lim

n→∞

n + 1

n· lim

n→∞

∣∣∣∣an+1(x− a)n+1

an(x− a)n

∣∣∣∣ =1

R

Hence, R′ = R.(b) Integrating term-by-term we find G(x) =

∑∞n=1 an(x− a)n = f(x)− a0.

(c) By Exercise 25.2, we have g(x) = G′(x) = f ′(x)

228

∑∞n=1

xn

n22n has radius of convergence 2 and show that the seriesconverges uniformly to a continuous function on [−2, 2].

Solution.Using the absolute ratio test we find∣∣∣∣an+1

an

∣∣∣∣ =

(n

n + 1

)2 ∣∣∣x2

∣∣∣→ ∣∣∣x2

∣∣∣ as n →∞.

Thus, the series converges for |x| < 2 so that the radius of convergence is 2.Now, ∣∣∣∣ xn

n22n

∣∣∣∣ ≤ 2n

n22n=

1

n2.


n=11n2 is convergent, the given series converges uniformly

on [−2, 2]. Since each function fn(x) = xn

n22n is continuous, by Exercise ??,the series converges to a continuous function

Exercise 37.5Let g(x) =

∑∞n=1

sin (3x)3n . Prove that the series converges for all x ∈ R and

that g(x) is continuous everywhere.

Solution.Since

0 ≤∣∣∣∣sin (3x)

3n

∣∣∣∣ ≤ 1

3n

and the series∑∞

n=113n is convergent, the given series converges uniformly

for all x ∈ R. By Exercise 37.1, g(x) is continuous everywhere


∑∞n=1

1n2+x2 converges to a continuous function for all x ∈ R.

Solution.Notice that x2 ≥ 0 always. So n2 + x2 ≥ n2 + 1 which gives 0 < 1

n2+x2 ≤ 1n2 .

Since∑∞

n=11n2 converges, by Weierstrass M-test, the given series converges

uniformly for all x ∈ R.Recall a theorem saying that if a sequence of continuous function convergesuniformly, the limit function is also continuous (Exercise ??). Using this andobserve that fn(x) = 1

n2+x2 are continuous (as denominator are non-zero), we

have∑∞

n=11

n2+x2 converges to a continuous function for all real number x

229

Exercise 37.7Find the Taylor series about x = 0 of cos x from the series of sin x.

Solution.We know that d

dx(sin x) = cos x, so we start the Taylor series of sin x and

differentiate this series term by term we get the series

cos x = 1− x2

2!+

x4

4!− · · ·+ (−1)n x2n

(2n)!+ · · ·

Exercise 37.8Find the Taylor’s series about x = 0 for arctan x from the series for 1

1+x2 .

SolutionIntegrating term by term of the series 1

1+x2 =∑∞

n=0(−1)nx2n we find

arctan x =

∫dx

1 + x2= C +

∞∑n=0

(−1)n x2n+1

2n + 1

where −1 < x < 1. Since arctan 0 = 0 then C = 0 and therefore

arctan x = x− x3

3+

x5

5− · · · =

∞∑n=0

(−1)n x2n+1

2n + 1

Exercise 37.9Use the first 500 terms of series of arctan x and a calculator to estimate thenumerical value of π.

Solution.Substituting x = 1 in to the series for arctan x gives

π = 4 arctan 1 = 4(1− 1

3+

1

5− 1

7+

1

9− · · · )

By using 500 terms of this series one finds

π ≈ 3.140.

Exercise 37.10Estimate the value of

∫ 1

0sin (x2)dx.

230

Solution.The integrand has no antiderivative expressible in terms of familiar functions.However, we know how to find its Taylor series: we know that

sin t = t− t3

3!+

t5

5!− · · ·

Now if we substitute t = x2, we have

sin (x2) = x2 − x6

3!+

x10

5!− · · ·

In spite of the fact that we cannot antidifferentiate the function, we canantidifferentiate the Taylor series:∫ 1

0sin (x2)dx =

∫ 1

0(x2 − x6

3!+ x5

5!− · · · )dx

= (x3

3− x7

7·3! + x11

11·5! − · · · )|10= 1

3− 1

7·3! + 111·5! −

115·7! + · · ·

≈ 0.31026

231

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