calculus one and several variables 10e salas solutions manual ch11
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Calculus one and several variables 10E Salas solutions manualTRANSCRIPT
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
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588 SECTION 11.1
CHAPTER 11
SECTION 11.1
1. lub = 2; glb = 0 2. lub = 2; glb = 0
3. no lub; glb = 0 4. lub = 1, no glb
5. lub = 2; glb = −2 6. lub = 3; glb = −1
7. no lub; glb = 2 8. lub = 2; glb = −2
9. lub = 212 ; glb = 2 10. lub = 0; glb = −1
11. lub = 1; glb = 0.9 12. lub = 219 , glb = −2 1
9
13. lub = e; glb = 0 14. no lub, glb = 1
15. lub = 12 (−1 +
√5); glb = 1
2 (−1 −√
5) 16. no lub, no glb
17. no lub; no glb 18. no lub; no glb
19. no lub; no glb 20. lub = 0; no glb
21. glb S = 0, 0 ≤(
111
)3< 0 + 0.001 22. glb = 1; 1 ≤ 1 < 1 + 0.0001
23. glb S = 0, 0 ≤(
110
)2n−1
< 0 +(
110
)k (n >
k + 12
)
24. glb = 0; 0 ≤(
12
)n
< 0 +(
14
)k
for n > 2k
25. Let ε > 0. The condition m ≤ s is satisfied by all numbers s in S. All we have to show therefore is
that there is some number s in S such that
s < m + ε.
Suppose on the contrary that there is no such number in S. We then have
m + ε ≤ x for all x ∈ S.
This makes m + ε a lower bound for S. But this cannot be, for then m + ε is a lower bound for S that
is greater than m, and by assumption, m is the greatest lower bound.
26. (a) Let M = |a1| + · · · + |an|. Then for any i, |ai| < M , so S is bounded
(b) lub S = max {a1, a2, . . . , an} ∈ S
glb S = min {a1, a2, . . . , an} ∈ S
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SECTION 11.1 589
27. Let c = lub S. Since b ∈ S, b ≤ c. Since b is an upper bound for S, c ≤ b. Thus, b = c.
28. S consists of a single element, equal to lub S.
29. (a) Suppose that K is an upper bound for S and k is a lower bound. Let t be any element of T . Then
t ∈ S which implies that k ≤ t ≤ K. Thus K is an upper bound for T and k is a lower bound, and
T is bounded.
(b) Let a = glb S. Then a ≤ t for all t ∈ T. Therefore, a ≤ glb T. Similarly, if b = lub S, then
t ≤ b for all t ∈ T, so lub T ≤ b. It now follows that glb S ≤ glb T ≤ lub T ≤ lub S.
30. (a) Let S = {r : r <√
2, r rational}.(b) Let T = {t : t < 2, t irrational}.
31. Let c be a positive number and let S = {c, 2c, 3c, . . .}. Choose any positive number M and consider the
positive number M/c. Since the set of positive integers is not bounded above, there exists a positive
integer k such that k ≥ M/c. This implies that kc ≥ M . Since kc ∈ S, it follows that S is not
bounded above.
32. (a) If S is a set of negative numbers, then 0 is an upper bound for S. It follows that α = lub S ≤ 0.
(b) If T is a set of positive numbers, then 0 is a lower bound for T . It follows that β = glb T ≥ 0.
33. (a) See Exercise 75 in Section 1.2.
(b) Suppose x20 > 2. Choose a positive integer n such that
2x0
n− 1
n2< x2
0 − 2.
Then,2x0
n− 1
n2< x2
0 − 2 =⇒ 2 < x20 −
2x0
n+
1n2
=(x0 − 1
n
)2
(c) If x20 < 2, then choose a positive integer n such that
2x0
n+
1n2
< 2 − x20.
Then
x20 +
2x0
n+
1n2
< 2 =⇒(x0 + 1
n
)2< 2
34. Assume that there are only finitely many primes, p1, p2, · · · , pn and let Q = p1 · p2 · · · pn + 1. Q has
a prime divisor p. But Q is not divisible by any of the pi, so p �= p1 for all i a contradiction.
35. (a) n = 5 : 2.48832; n = 10 : 2.59374; n = 100 : 2.70481; n = 1000 : 2.71692; n = 10, 000 : 2.71815
(b) lub = e; glb = 2
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590 SECTION 11.2
36. a2 a3 a4 a5 a6 a7 a8 a9 a10
94
53
65
12 −3 4 9
453
65
(b) a20 = 94 , a30 = −3, a40 = 6
5 , a50 = 94
(c) lubS = 4, glbS = −3
37. (a) a1 a2 a3 a4 a5
1.4142 1.6818 1.8340 1.9152 1.9571
a6 a7 a8 a9 a10
1.9785 1.9892 1.9946 1.9973 1.9986
(b) Let S be the set of positive integers for which an < 2. Then 1 ∈ S since
a1 =√
2 ∼= 1.4142 < 2.
Assume that k ∈ S. Since a2k+1 = 2ak < 4, it follows that ak+1 < 2. Thus k + 1 ∈ S and S is
the set of positive integers.
(c) 2 is the least upper bound.
(d) Let c be a positive number. Then c is the least upper bound of the set
S =
{√c,
√c√c,
√c
√c√c, . . .
}.
38. (a) a1∼= 1.4142136 a2
∼= 1.8477591 a3∼= 1.9615706 a4
∼= 1.9903695 a5∼= 1.9975909
a6∼= 1.9993976 a7
∼= 1.9998494 a8∼= 1.9999624 a9
∼= 1.9999906 a10∼= 1.9999976
(b) a1 =√
2 < 2. Assume true for an. Then an+1 =√
2 + an <√
2 + 2 = 2.
(c) lub S = 2.
(d) For any positive number c, lub S is the positive number satisfying
x =√c + x, that is, x = (1 +
√1 + 4c)/2
SECTION 11.2
1. an = 2 + 3(n− 1) = 3n− 1, n = 1, 2, 3, . . . 2. an = 1 − (−1)n, n = 1, 2, 3, . . .
3. an =(−1)n−1
2n− 1, n = 1, 2, 3, . . . 4. an =
2n − 12n
, n = 1, 2, 3, . . .
5. an =n2 + 1
n, n = 1, 2, 3, . . . 6. an = (−1)n
n
(n + 1)2, n = 1, 2, 3, . . .
7. an =
{n
1/n,
if n = 2k − 1
if n = 2k,where k = 1, 2, 3, . . .
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SECTION 11.2 591
8. an =
{n
1/n2,
if n = 2k
if n = 2k − 1,where k = 1, 2, 3, . . .
9. decreasing; bounded below by 0 and above by 2
10. not monotonic; bounded below by −1 and above by 12 .
11.n + (−1)n
n= 1 + (−1)n
1n
: not monotonic; bounded below by 0 and above by32
12. increasing; bounded below by 1.001 but not bounded above.
13. decreasing; bounded below by 0 and above by 0.9
14. increasing; bounded below by 0 and above by 1
15.n2
n + 1= n− 1 +
1n + 1
: increasing; bounded below by12
but not bounded above
16. increasing; bounded below by√
2, but not bounded above:√n2 + 1 > n
17.4n√
4n2 + 1=
2√1 + 1/4n2
and1
4n2decreases to 0: increasing;
bounded below by 45
√5 and above by 2
18.2n
4n + 1=
2n
(2n)2 + 1=
12n + 1
2n
. decreasing; bounded below by 0 and above by 25 .
19. increasing; bounded below by 251 but not bounded above
20.n2
√n3 + 1
=1√
1n + 1
n4
and1n
+1n4
decreases to 0 =⇒ increasing; bounded below by√
22
,
but not bounded above.
21.2n
n + 1= 2 − 2
n + 1increases toward 2: increasing; bounded below by 0 and above by ln 2.
22. increasing (n ≥ 2); bounded below by2√
2310
, but not bounded above.
23. decreasing; bounded below by 1 and above by 4
24. not monotonic; not bounded below and not bounded above
25. increasing; bounded below by√
3 and above by 2
26. decreasing (sincen + 1n
decreases to 1); bounded below by 0 and above by ln 2.
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592 SECTION 11.2
27. (−1)2n+1√n = −√
n: decreasing; bounded above by −1 but not bounded below
28.√n + 1√n
=
√1 +
1n
decreasing; bounded below by 1 and above by√
2
29.2n − 1
2n= 1 − 1
2n: increasing; bounded below by
12
and above by 1
30.12n
− 12n + 3
=3
2n(2n + 3)decreasing; bounded below by 0 and above by 3
10 .
31. consider sinx as x → 0+: decreasing; bounded below by 0 and above by 1
32. not monotonic; bounded below by − 12 and above by 1
4
33. decreasing; bounded below by 0 and above by 56
34. increasing (becausex
lnxis an increasing function on [4,∞).); bounded below by
4ln 4
but not bounded above.
35.1n− 1
n + 1=
1n (n + 1)
: decreasing; bounded below by 0 and above by12
36. not monotonic; bounded below by −1 and above by 1
37. Set f(x) =lnx
x. Then, f ′(x) =
1 − lnx
x2< 0 for x > e: decreasing;
bounded below by 0 and above by 13 ln 3.
38. not monotonic; not bounded below nor above (because exponentials grow faster than polynomials).
39. Set an =3n
(n + 1)2. Then,
an+1
an= 3
(n + 1n + 2
)2
> 1: increasing;
bounded below by 34 but not bounded above.
40.1 − ( 1
2 )n
( 12 )n
= 2n − 1 increasing; bounded below by 1 but not bounded above.
41. For n ≥ 5
an+1
an=
5n+1
(n + 1)!· n!5n
=5
n + 1< 1 and thus an+1 < an.
Sequence is not nonincreasing: a1 = 5 < 252 = a2.
42. For n ≥ M,an+1
an=
Mn+1
(n + 1)!· n!Mn
=M
n + 1< 1, so the sequence decreases for n ≥ M .
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SECTION 11.2 593
43. boundedness: 0 < (cn + dn)1/n < (2dn)1/n = 21/nd ≤ 2d
monotonicity : an+1n+1 = cn+1 + d n+1 = ccn + ddn
< (cn + dn)1/ncn + (cn + dn)1/ndn
= (cn + dn)1+1/n
= (cn + dn)(n+1)/n
= an+1n
Taking the (n + 1)st root of each side we have an+1 < an. The sequence is decreasing.
44. If for all n
|an| ≤ M and |bn| ≤ N,
then for all n
|αan + βbn| ≤ |α||an| + |β||bn| ≤ |α|M + |β|N
and
|anbn| = |an||bn| ≤ MN
45. a1 = 1, a2 = 12 , a3 = 1
6 , a4 = 124 , a5 = 1
120 , a6 = 1720 ; an = 1
n!
46. a1 = 1, a2 = 8, a3 = 27, a4 = 64, a5 = 125, a6 = 216; an = n3
47. a1 = a2 = a3 = a4 = a5 = a6 = 1; an = 1
48. a1 = 1, a2 = 32 , a3 = 7
4 , a4 = 158 , a5 = 31
16 , a6 = 6332 ; an = (2n − 1)/2n−1
49. a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9, a6 = 11; an = 2n− 1
50. a1 = 1, a2 = 12 , a3 = 1
3 , a4 = 14 , a5 = 1
5 , a6 = 16 ; an = 1/n
51. a1 = 1, a2 = 4, a3 = 9, a4 = 16, a5 = 25, a6 = 36; an = n2
52. a1 = 1, a2 = 3, a3 = 7, a4 = 15, a5 = 31, a6 = 63; an = 2n − 1
53. a1 = 1, a2 = 1, a3 = 2, a4 = 4, a5 = 8, a6 = 16; an = 2n−2 (n ≥ 3)
54. a1 = 3, a2 = 1, a3 = 3, a4 = 1, a5 = 3, a6 = 1; an = 2 − (−1)n
55. a1 = 1, a2 = 3, a3 = 5, a4 = 7, a5 = 9, a6 = 11; an = 2n− 1
56. a1 = 1, a2 = 3, a3 = 4, a4 = 5, a5 = 6, a6 = 7; an = n + 1 (n ≥ 2)
57. First a1 = 21 − 1 = 1. Next suppose ak = 2k − 1 for some k ≥ 1. Then
ak+1 = 2ak + 1 = 2(2k − 1
)+ 1 = 2k+1 − 1.
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594 SECTION 11.2
58. True for n = 1. Assume true for n. Then an+1 = an + 5 = 5n− 2 + 5 = 5(n + 1) − 2.
59. First a1 =120
= 1. Next suppose ak =k
2k−1for some k ≥ 1. Then
ak+1 =k + 12k
ak =k + 12k
k
2k−1=
k + 12k
.
60. True for n = 1. Assume true for n. Then an+1 = an − 1n(n + 1)
=1n− 1
n(n + 1)=
n + 1 − 1n(n + 1)
=1
n + 1
61. (a) If r = 1 then Sn = n for n = 1, 2, 3, . . .
(b) Sn = 1 + r + r2 + · · · + rn−1
rSn = r + r2 + · · · + rn
Sn − rSn = 1 − rn
Sn =1 − rn
1 − r, r �= 1.
62.1
k(k + 1)=
1k− 1
k + 1, so
Sn = a1 + a2 + · · · + an−1 + an
=1
1 · 2 +1
2 · 3 + · · · + 1(n− 1)n
+1
n(n + 1)
=(
1 − 12
)+(
12− 1
3
)+ · · · +
(1
n− 1− 1
n
)+(
1n− 1
n + 1
)= 1 − 1
n + 1=
n
n + 1since all middle terms cancel out.
63. (a) Let Sn denote the distance traveled between the nth and (n + 1)st bounce. Then
S1 = 75 + 75 = 150, S2 = 34 (75) + 3
4 (75) = 150(
34
), . . . , Sn = 150
(34
)n−1
.
(b) An object dropped from rest from a height h feet above the ground will hit the ground in 14
√h
seconds. Therefore it follows that the ball will be in the air
Tn = 2(
14
)√Sn
2=
5√
32
(34
)(n−1)/2
seconds.
64. P0 = 5, P12 = 5 · 2, P24 = 5 · 22; so Pn = 5 2n/12
65. {an} is an increasing sequence; an → 12 as n → ∞.
66. {an} is a decreasing sequence; an → 2 as n → ∞.
67. (a) Let S be the set of positive integers for which an+1 > an. Since a2 = 1 +√a1 = 2 > 1, 1 ∈ S.
Assume that ak = 1 +√ak−1 > ak−1. Then
ak+1 = 1 +√ak > 1 +
√ak−1 = ak.
Thus, k ∈ S implies k + 1 ∈ S. It now follows that {an} is an increasing sequence.
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SECTION 11.3 595
(b) Since {an} is an increasing sequence,
an = 1 +√an−1 < 1 +
√an, or an −√
an − 1 < 0.
Rewriting the second inequality as
(√an)2 −√
an − 1 < 0
and solving for√an it follows that
√an < 1
2 (1 +√
5). Hence, an < 12 (3 +
√5) for all n.
(c) a2 = 2, a3∼= 2.4142, a4
∼= 2.5538, a5∼= 2.5981, . . . , a9
∼= 2.6179, . . . , a15∼= 2.6180;
(e) lub {an} = 12 (3 +
√5) ∼= 2.6180
68. (a) We show that an < an+1 for all n. True for n = 1 since a1 = 1 <√
3 = a2. Assume true for
n, that is, an < an+1; we need to show that an+1 < an+2.
But an+1 =√
3an <√
3an+1 = an+2, as required.
(b) True since a1 < 3, and an < 3 =⇒ an+1 <√
3 · 3 = 3
(c) a1 = 1, a2 =√
3, a3∼= 2.2795, . . . , a14
∼= 2.9996, a15∼= 2.9998
(d) lub = 3
SECTION 11.3
1. diverges 2. converges to 0
3. converges to 0 4. diverges
5. converges to 1:n− 1n
= 1 − 1n→ 1 6. converges to 1:
n + (−1)n
n= 1 +
(−1)n
n→ 1
7. converges to 0:n + 1n2
=1n
+1n2
→ 0
8. converges to 0:π
2n→ 0, so sin
( π
2n
)→ sin 0 = 0
9. converges to 0: 0 <2n
4n + 1<
2n
4n=
12n
→ 0
10. diverges:n2
n + 1≥ n2
2n=
n
211. diverges
12. diverges:4n√n2 + 1
≈ 4n
n→ ∞
13. converges to 0
14. converges to 0:4n
2n + 106<
4n2n
=n
2n−2→ 0
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596 SECTION 11.3
15. converges to 1:nπ
4n + 1→ π
4so tan
nπ
4n + 1→ tan
π
4= 1
16. converges to 0:1010
√n
n + 1=
1010
√n + 1/
√n→ 0
17. converges to49
:(2n + 1)2
(3n− 1)2=
4 + 4/n + 1/n2
9 − 6/n + 1/n2→ 4
9
18. converges to ln 2 :2n
n + 1→ 2, so ln
(2n
n + 1
)→ ln 2
19. converges to12
√2:
n2
√2n4 + 1
=1√
2 + 1/n4→ 1√
2
20. converges to 1:n4 − 1
n4 + n− 6=
1 − 1n4
1 + 1n3 − 6
n4
→ 1
21. diverges: cosnπ = (−1)n 22. diverges:n5
17n4 + 12= n
(1
17 + 12n4
)
23. converges to 1:1√n→ 0 so e1/
√n → e0 = 1
24. converges to√
4 = 2
25. converges to 0 : lnn− ln (n + 1) = ln(
n
n + 1
)→ ln 1 = 0
26. converges to 1:2n − 1
2n= 1 − 1
2n=⇒ 1
27. converges to12:
√n + 12√n
=12
√1 +
1n→ 1
228. converges to 0:
1n− 1
n + 1=
1n(n + 1)
→ 0
29. converges to e2:(
1 +1n
)2n
=[(
1 +1n
)n]2
→ e2
30. converges to√e :
(1 +
1n
)n/2
=
√(1 +
1n
)n
→√e
31. diverges; since 2n > n3 for n ≥ 10,2n
n2>
n3
n2= n
32. converges to ln 9 : 2 ln 3n− ln(n2 + 1) = ln(
9n2
n2 + 1
)→ ln 9
33. converges to 0:| sinn|√
n≤ 1√
n
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SECTION 11.3 597
34. converges to π/4:n
n + 1→ 1, arctan 1 = π/4
35. converges to 1/2:√n2 + n− n =
(√n2 + n− n
) √n2 + n + n√n2 + n + n
=n√
n2 + n + n→ 1
2
36. converges to 2:√
4n2 + n
n=√
4 + (1/n) →√
4 = 2.
37. converges to π/2
38. converges to 31/9: let s = 3.444 · · ·. Then 10s− s = 31 and s = 31/9.
39. converges to −π/2:1 − n
n→ −1; arcsin (−1) = π/2.
40. converges to 1:(n + 1)(n + 4)(n + 2)(n + 3)
=n2 + 5n + 5n2 + 5n + 6
→ 1.
41. (a) n√n → 1 (b)
3n
n!→ 0
42. (a)n
n√n!
→ e (b) does not converge
43. b < n√an + bn = b n
√(a/b)n + 1 < b n
√2. Since 21/n → 1 as n → ∞, it follows that n
√an + bn → b
by the pinching theorem.
44. (a) −1 < r ≤ 1 (b) −1 < r < 1
45. Set ε > 0. Since an → L, there exists N1 such that
if n ≥ N1, then |an − L| < ε/2.
Since bn → M , there exists N2 such that
if n ≥ N2, then |bn −M | < ε/2.
Now set N = max {N1, N2}. Then, for n ≥ N ,
|(an + bn) − (L + M)| ≤ |an − L| + |bn −M | < ε
2+
ε
2= ε.
46. Let ε > 0, choose k such that n ≥ k =⇒ |an − L| < ε
|α| + 1.
Then for n ≥ k, |αan − αL| = |α||an − L| ≤ (|α| + 1)|an − L| < ε
Therefore αan → αL.
47. Since(
1 +1n
)→ 1 and
(1 +
1n
)n
→ e,
(1 +
1n
)n+1
=(
1 +1n
)n(1 +
1n
)→ (e)(1) = e.
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598 SECTION 11.3
48. (a) If k = j, an =αk + αk−1 ·
1n
+ · · · + α0 ·1nk
βk + βk−1 ·1n
+ · · · + β0 ·1nk
→ αk
βk
(b) If k < j, an =αk + αk−1 ·
1n
+ · · · + α0 ·1nk
βj · nj−k + βj−1 · nj−1−k + · · · + β0 ·1nk
→ 0
(c) If k > j, an =αk · nk−j + αk−1 · nk−1−j + · · · + α0 ·
1nj
βj + βj−1 ·1n
+ · · · + β0 ·1nj
diverges
49. Suppose that {an} is bounded and non-increasing. If L is the greatest lower bound of the range of this
sequence, then an ≥ L for all n. Set ε > 0. By Theorem 11.1.4 there exists ak such that ak < L + ε.
Since the sequence is non-increasing, an ≤ ak for all n ≥ k. Thus,
L ≤ an < L + ε or |an − L| < ε for all n ≥ k
and an → L.
50. Let ε > 0. If an → L, then there exists a positive integer k such that
|an − L| < ε for all n ≥ k
If n ≥ k, then 2n ≥ k and 2n− 1 ≥ k, and thus
|en − L| = |a2n − L| < ε and |on − L| = |a2n−1 − L| < ε
It follows that en → L and on → L. If en → L and on → L, then there exist k1 and k2 such
that
if m ≥ k1, then |em − L| = |a2m − L| < ε
and
if m ≥ k2, then |om − L| = |a2m−1 − L| < ε
Let k = max{2k1, 2k2 − 1}. If n ≥ k then
either an = a2m with m > k1 or an = a2m−1 with m ≥ k2
In either case, |an − L| < ε. This shows that an → L.
51. Let ε > 0. Choose k so that, for n ≥ k,
L− ε < an < L + ε, L− ε < cn < L + ε and an ≤ bn ≤ cn.
For such n,
L− ε < bn < L + ε.
52. Let M be a bound for {bn}. Then |anbn| ≤ |an|M.
Given ε > 0, choose k such that |an| < ε/M for n ≥ k. Then |anbn| < ε for n ≥ k.
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SECTION 11.3 599
53. Let ε > 0. Since an → L, there exists a positive integer N such that L− ε < an < L + ε for all n ≥ N.
Now an ≤ M for all n, so L− ε < M, or L < M + ε. Since ε is arbitrary, L ≤ M.
54. The converse is false. For example, let an = (−1)n. Then |an| → 1, but {an} diverges.
55. Assume an → 0 as n → ∞. Let ε > 0. There exists a positive integer N such that |an − 0| < ε for
all n ≥ N . Since | |an| − 0| ≤ |an − 0|, it follows that |an| → 0. Now assume that |an| → 0. Since
−|an| ≤ an ≤ |an|, an → 0 by the pinching theorem.
56. Let ε > 0. There exists a positive integer N1 such that |an − L| < ε for all n > 2N1 − 1, and there
exists a positive integer N2 such that |bn − L| < ε for all n > 2N2. The sequence a1, b1, a2, b2, . . .
can be represented by the sequence c1, c2, c3, . . ., where
cn =
{a(n+1)/2, if n is odd
bn/2, if n is even.
}
Let N = max{2N1 − 1, 2N2}. Then n > N =⇒ |cn − L| < ε =⇒ cn → L.
57. By the continuity of f, f(L) = f(
limn→∞
an
)= lim
n→∞f(an) = lim
n→∞an+1 = L.
58.2n
n!=
21· 22· 23· · · 2
n= 2 · 2
n· (terms that are ≤ 1) ≤ 4
n.
Since4n→ 0 and 0 <
2n
n!≤ 4
n,
2n
n!→ 0 as well.
59. Set f(x) = x1/p. Since1n→ 0 and f is continuous at 0, it follows by Theorem 11.3.12 that(
1n
)1/p
→ 0.
60. Since |an − L| = |(an − L) − 0| = ||an − L| − 0|,|an − L| < ε iff |(an − L) − 0| < ε iff ||an − L| − 0| < ε,
So an → L iff an − L → 0 iff |an − L| → 0.
61. an = e1−n → 0 62. diverges
63. an =1n!
→ 0 64. an = 1 · 12· 23· · · n− 1
n=
1n
converges to 0
65. an =12[1 − (−1)n] diverges 66. an =
2n − 12n−1
→ 2
67. L = 0, n = 32 68.1√n→ 0.
1√n< 0.001 for n ≥ 10002 + 1
69. L = 0, n = 4 70.n10
10n→ 0.
n10
10n< 0.001 for n ≥ 15
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600 SECTION 11.3
71. L = 0, n = 7 72.2n
n!→ 0.
2n
n!< 0.001 for n ≥ 10
73. L = 0, n = 65 74.lnn
n→ 0.
lnn
n< 0.001 for n ≥ 9119
75. (a) an+1 = 1 +√an Suppose that an → L as n → ∞. Then an+1 → L as n → ∞. Therefore
L = 1 +√L which, since L > 1, implies L = 1
2 (3 +√
5).
(b) an+1 =√
3an Suppose that an → L as n → ∞. Then an+1 → L as n → ∞. Therefore
L =√
3L which, since L > 1, implies L = 3.
76. (a) a2∼= 2.6458, a3
∼= 2.9404, a4∼= 2.9900, a5
∼= 2.9983, a6∼= 2.9997
(b) True for n = 1. Assume true for n. Then an+1 =√
6 + an−1 ≤√
6 + 3 = 3
(c) an+12 − an
2 = 6 + an − an2 = (3 − an)(2 + an) ≥ 0 since 0 ≤ an ≤ 3.
Since ak ≥ 0 for all k, this implies an+1 ≥ an
(d) an → 3
77. (a) a2 a3 a4 a5 a6 a7 a8 a9 a10
0.5403 0.8576 0.6543 0.7935 0.7014 0.7640 0.7221 0.7504 0.7314
(b) L is a fixed point of f(x) = cosx, that is, cosL = L; L ∼= 0.739085.
78. (a) a2∼= 1.5403, a3
∼= 1.5708, a4∼= a5
∼= · · · ∼= a10∼= 1.5708.
(b) L ∼= 1.570796 Let f(x) = x + cosx. L must satisfy L = f(L), so L = L + cosL,
and cosL = 0. Indeed, the L we found is justπ
2∼= 1.570796327
PROJECT 11.3
1. (a)a2 a3 a4 a5 a6 a7 a8
2.000000 1.750000 1.732143 1.732051 1.732051 1.732051 1.732051
(b) L =12
(L +
3L
)which implies L2 = 3 or L =
√3.
2. The Newton-Raphson method applied to the function f(x) = x2 −R gives
an = an−1 −f(an−1)f ′(an−1)
= an−1 −a2n−1 −R
2an−1
=12an−1 +
12
R
an−1=
12
(an−1 +
R
an−1
), n = 2, 3, . . . .
3 & 4. (a) f(x) = x3 − 8, so xn → 2 (b) f(x) = sinx− 12 , so xn → π
6
(c) f(x) = lnx− 1, so xn → e
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SECTION 11.4 601
SECTION 11.4
1. converges to 1: 22/n = (21/n)2 → 12 = 1 2. converges to 1: e−α/n → e0 = 1
3. converges to 0: for n > 3, 0 <
(2n
)n
<
(23
)n
→ 0
4. converges to 0:log10 n
n=
1ln 10
· lnn
n→ 0
5. converges to 0:ln (n + 1)
n=[ln (n + 1)n + 1
](n + 1n
)→ (0)(1) = 0
6. converges to 0:3n
4n=(
34
)n
→ 0 7. converges to 0:x100n
n!=
(x100)n
n!→ 0
8. converges to 1: n1/(n+2) =(n1/n
)n/(n+2)
→ 1 9. converges to 1: nα/n = (n1/n)α → 1α = 1
10. converges to 0: ln(n + 1n
)→ ln(1) = 0
11. converges to 0:3n+1
4n−1= 12
(3n
4n
)= 12
(34
)n
→ 12(0) = 0
12. converges to12
:∫ 0
−n
e2x dx =12− e−2n
2→ 1
2
13. converges to 1: (n + 2)1/n = e1n ln(n+2) and, since
1n
ln (n + 2) =[ln (n + 2)n + 2
](n + 2n
)→ (0)(1) = 0,
it follows that (n + 2)1/n → e0 = 1.
14. converges to e−1 :(
1 − 1n
)n
=(
1 +(−1)n
)n
→ e−1 (by (11.4.7)
15. converges to 1:∫ n
0
e−x dx = 1 − 1en
→ 1 16. diverges.
17. converges to π: integral = 2∫ n
0
dx
1 + x2= 2 tan−1 n → 2
(π2
)= π
18. converges to 0:∫ n
0
e−nx dx = −e−n2
n+
1n→ 0
19. converges to 1: recall (11.4.6) 20. converges to 0: n2 sinnπ = 0 for all n
21. converges to 0:ln (n2)
n= 2
ln n
n→ 2(0) = 0
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602 SECTION 11.4
22. converges to π :∫ 1−1/n
−1+1/n
dx√1 − x2
= sin−1
(1 − 1
n
)− sin−1
(−1 +
1n
)→ sin−1(1) − sin−1(−1) = π
23. diverges: Since limx→0
sinx
x= 1,
n
πsin
π
n=
sin (π/n)π/n
→ 1. Therefore,
n2 sinπ
n= nπ
(nπ
sinπ
n
)→ nπ.
24. diverges 25. converges to 0:5n+1
42n−1= 20
(516
)n
→ 0
26. converges to e3x :(1 +
x
n
)3n
=[(
1 +x
n
)n]3→ (ex)3 = e3x
27. converges to e−1:(n + 1n + 2
)n
=(
1 − 1n + 2
)n
=
(1 +
(−1)n + 2
)n+2
(1 +
(−1)n + 2
)2 → e−1
1= e−1
28. converges to 2:∫ 1
1/n
dx√x
= 2 − 2√n→ 2.
29. converges to 0: 0 <
∫ n+1
n
e−x2dx ≤ e−n2
[(n + 1) − n] = e−n2 → 0
30. converges to 1:(
1 +1n2
)n
=
[(1 +
1n2
)n2]1/n
→ (e1)0 = 1
31. converges to 0:nn
2n2 =( n
2n)n
→ 0 sincen
2n→ 0
32. converges to 0:∫ 1/n
0
cos ex dx →∫ 0
0
cos ex dx = 0
33. converges to ex: use (11.4.7)
34. diverges:(
1 +1n
)n2
=[(
1 +1n
)n]n> 2n,
[(1 +
1n
)n
≈ e > 2]
35. converges to 0:
∣∣∣∣∣∫ 1/n
−1/n
sinx2dx
∣∣∣∣∣ ≤∫ 1/n
−1/n
| sin x2| dx ≤∫ 1/n
−1/n
1 dx =2n→ 0
36.(t +
x
n
)n= tn
(1 +
x/t
n
)n
; converges to 0 if t < 1, converges to ex if t = 1, diverges if t > 1.
37. converges:sin(6/n)sin(3/n)
→ 2 38. converges:arctann
n→ 0
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SECTION 11.4 603
39.√n + 1 −
√n =
√n + 1 −√
n√n + 1 +
√n
(√n + 1 +
√n)
=1√
n + 1 +√n→ 0
40.√n2 + n− n =
√n2 + n− n√n2 + n + n
(√n2 + n + n) =
n√n2 + n + n
=1
1 +√
1 + 1/n→ 1
2
41. (a) The length of each side of the polygon is 2r sin(π/n). Therefore the perimeter, pn, of the polygon
is given by: pn = 2rn sin(π/n).
(b) 2rn sin(π/n) → 2πr as n → ∞ : The number 2rn sin (π/n) is the perimeter of a regular polygon
of n sides inscribed in a circle of radius r. As n tends to ∞, the perimeter of the polygon tends
to the circumference of the circle.
42. Since 0 < c < d, d < (cn + dn)1/n < (2dn)1/n = 21/nd → d, so by the pinching theorem
(cn + dn)1/n → d.
43. By the hint, limn→∞
1 + 2 + . . . + n
n2= lim
n→∞n(n + 1)
2n2= lim
n→∞1 + 1/n
2=
12.
44. diverges:12 + 22 + · · · + n2
(1 + n)(2 + n)=
n(n + 1)(2n + 1)6(1 + n)(2 + n)
=2n3 + 3n2 + n
6n2 + 18n + 12→ ∞
45. By the hint, limn→∞
13 + 23 + . . . + n3
2n4 + n− 1= lim
n→∞n2(n + 1)2
4(2n4 + n− 1)= lim
n→∞1 + 2/n + 1/n2
8 + 4/n3 − 4/n4=
18.
46. Here we show that every convergent sequence is a Cauchy sequence. Let ε > 0. If an → L, then there
exists a positive integer k such that
|ap − L| < ε
2for all p ≥ k
With m,n ≥ k we have
|am − an| ≤ |am − L| + |L− an| = |am − L| + |an − L| < ε
2+
ε
2= ε.
47. (a) mn+1 −mn =1
n + 1(a1 + · · · + an + an+1) −
1n
(a1 + · · · + an)
=1
n(n + 1)[nan+1 −
n
(
︷ ︸︸ ︷a1 + · · · + an)
]> 0 since {an} is increasing.
(b) We begin with the hint
mn <|a1 + · · · + aj |
n+
ε
2
(n− j
n
).
Since j is fixed,
|a1 + · · · + aj |n
→ 0
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604 SECTION 11.4
and therefore for n sufficiently large
|a1 + · · · + aj |n
<ε
2.
Sinceε
2
(n− j
n
)<
ε
2,
we see that, for n sufficiently large, |mn| < ε. This shows that mn → 0.
48. (a) Since {an} converges, it is a Cauchy sequence (see Exercise 46), so given ε > 0 we can find k
such that |an − am| < ε for n,m ≥ k.
In particular, |an+1 − an| < ε, so limn→∞
(an − an−1) = 0.
(b) {an} does not necessarily converge. For example let an = lnn. This diverges, but
an − an−1 = lnn− ln(n− 1) = ln(
n
n− 1
)→ ln 1 = 0
49. (a) Let S be the set of positive integers n (n ≥ 2) for which the inequalities hold. Since(√b)2
− 2√ab +
(√a)2 =
(√b−
√a)2
> 0,
it follows thata + b
2>
√ab and so a1 > b1. Now,
a2 =a1 + b1
2< a1 and b2 =
√a1b1 > b1.
Also, by the argument above,
a2 =a1 + b1
2>√
a1b1 = b2,
and so a1 > a2 > b2 > b1. Thus 2 ∈ S. Assume that k ∈ S. Then
ak+1 =ak + bk
2<
ak + ak2
= ak, bk+1 =√
akbk >√b2k = bk,
and
ak+1 =ak + bk
2>√
akbk = bk+1.
Thus k + 1 ∈ S. Therefore, the inequalities hold for all n ≥ 2.
(b) {an} is a decreasing sequence which is bounded below.
{bn} is an increasing sequence which is bounded above.
Let La = limn→∞
an, Lb = limn→∞
bn. Then
an =an−1 + bn−1
2implies La =
La + Lb
2and La = Lb.
50.e−
(1 + 1
100
)100e
∼= 0.004995 : within 0.01%;e5 −
(1 + 5
100
)100e5
∼= 0.11395 : within 12%
e−(1 + 1
1000
)1000e
∼= 0.0004995 : within 0.05%;e5 −
(1 + 5
1000
)1000e5
∼= 0.01238 : within 1.3%
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SECTION 11.5 605
51. The numerical work suggests L ∼= 1. Justification: Set f(x) = sinx− x2. Note that f(0) = 0 and
f ′(x) = cosx− 2x > 0 for x close to 0. Therefore sinx− x2 > 0 for x close to 0 and sin 1/n− 1/n2 > 0
for n large. Thus, for n large,
1n2
< sin1n<
1n
(| sinx| ≤ |x| for all x)
(1n2
)1/n
<
(sin
1n
)1/n
<
(1n
)1/n
(1
n1/n
)2
<
(sin
1n
)1/n
<1
n1/n.
As n → ∞ both bounds tend to 1 and therefore the middle term also tends to 1.
52. Numerical work suggests L ∼= 1/3. Conjecture: L = 1/k.
Proof: (nk + nk−1)1/k − n = n(1 + 1/n)1/k − n =(1 + 1/n)1/k − 1
1/n;
limn→∞
(1 + 1/n)1/k − 11/n
= limh→0
(1 + h)1/k − 1h
= f ′(1),
where f(x) = x1/k. Since f ′(x) = (1/k)x(1/k)−1, f ′(1) = 1/k.
53. (a) a3 a4 a5 a6 a7 a8 a9 a10
2 3 5 8 13 21 34 55
(b) r1 r2 r3 r4 r5 r6
1 2 1.5 1.667 1.600 1.625
(c) Following the hint,
1 +1
rn−1= 1 +
1anan−1
= 1 +an−1
an=
an + an−1
an=
an+1
an= rn.
Now, if rn → L, then rn−1 → L and
1 +1L
= L which, since L > 1, implies L =1 +
√5
2∼= 1.618034.
54. With the partition {0, 1n ,
2n , . . . ,
nn} and f(x) = x, we have
an =1n
(1n
+2n
+ · · · + n
n
)=
1n
[f
(1n
)+ f
(2n
)+ · · · + f
(nn
)]=
n∑i=1
f(xi)Δxi,
so it is a Riemann sum for∫ 1
0
x dx, and therefore limn→∞
an =∫ 1
0
x dx =12
SECTION 11.5
(We’ll use � to indicate differentiation of numerator and denominator.)
1. limx→0+
sin x√x
�= limx→0+
2√x cos x = 0 2. lim
x→1
lnx
1 − x
�= limx→1
1/x−1
= −1
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606 SECTION 11.5
3. limx→0
ex − 1ln (1 + x)
�= limx→0
(1 + x)ex = 1 4. limx→4
√x− 2x− 4
�= limx→4
12√x
1=
14
5. limx→π/2
cos x
sin 2x�= lim
x→π/2
− sin x
2 cos 2x=
12
6. limx→a
x− a
xn − an�= lim
x→a
1nxn−1
=1
nan−1
7. limx→0
2x − 1x
�= limx→0
2x ln 2 = ln 2 8. limx→0
tan−1 x
x
�= limx→0
11+x2
1= 1
9. limx→1
x1/2 − x1/4
x− 1�= lim
x→1
(12x−1/2 − 1
4x−3/4
)=
14
10. limx→0
ex − 1x(1 + x)
�= limx→0
ex
1 + 2x= 1
11. limx→0
ex − e−x
sin x
�= limx→0
ex + e−x
cos x= 2 12. lim
x→0
1 − cosx3x
�= limx→0
sinx
3= 0
13. limx→0
x + sinπx
x− sinπx
�= limx→0
1 + π cosπx1 − π cosπx
=1 + π
1 − π
14. limx→0
ax − (a + 1)x
x
�= limx→0
ax ln a− (a + 1)x ln(a + 1)1
= ln(
a
a + 1
)
15. limx→0
ex + e−x − 21 − cos 2x
�= limx→0
ex − e−x
2 sin 2x�= lim
x→0
ex + e−x
4 cos 2x=
12
16. limx→0
x− ln(x + 1)1 − cos 2x
�= limx→0
1 − 1x+1
2 sin 2x�= lim
x→0
1(x+1)2
4 cos 2x=
14
17. limx→0
tanπx
ex − 1�= lim
x→0
π sec2 πx
ex= π
18. limx→0
cosx− 1 + x2/2x4
�= limx→0
− sinx + x
4x3
�= limx→0
− cosx + 112x2
�= limx→0
sinx
24x=
124
19. limx→0
1 + x− ex
x(ex − 1)�= lim
x→0
1 − ex
xex + ex − 1�= lim
x→0
−ex
xex + 2ex= −1
2
20. limx→0
ln(secx)x2
�= limx→0
tanx
2x�= lim
x→0
sec2 x
2=
12
21. limx→0
x− tan x
x− sin x
�= limx→0
1 − sec2 x
1 − cos x
�= limx→0
−2 sec2 x tan x
sin x= lim
x→0
−2 sec2 x
cos x= −2
22. limx→0
xenx − x
1 − cosnx�= lim
x→0
enx + nxenx − 1n sinnx
�= limx→0
enx(2n + n2x)n2 cosnx
=2n
23. limx→1−
√1 − x2
√1 − x3
= limx→1−
√1 − x2
1 − x3=
√23
=13
√6 since lim
x→1−
1 − x2
1 − x3
�= limx→1−
2x3x2
=23
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SECTION 11.5 607
24. limx→0
2x− sinπx
4x2 − 1= 0
25. limx→π/2
ln (sin x)(π − 2x)2
�= limx→π/2
− cot x
4(π − 2x)�= lim
x→π/2
csc2 x
−8= −1
8
26. limx→0+
√x√
x + sin√x
�= limx→0+
12√x
12√x
+ cos√x
2√x
= limx→0+
11 + cos
√x
=12
27. limx→0
cosx− cos 3xsin(x2)
�= limx→0
− sinx + 3 sin 3x2x cos(x2)
�= limx→0
− cosx + 9 cos 3x2 cos(x2) − 4x2 sin(x2)
= 4
28. limx→0
√a + x−√
a− x
x
�= limx→0
12√a+x
+ 12√a−x
1=
1√a
=√a
a
29. limx→π/4
sec2 x− 2 tan x
1 + cos 4x�= lim
x→π/4
2 sec2 x tan x− 2 sec2 x
−4 sin 4x
�= limx→π/4
2 sec4 x + 4 sec2 x tan2 x− 4 sec2 x tan x
−16 cos 4x=
12
30. limx→0
x− arcsinx
sin3 x
�= limx→0
1 − 1√1−x2
3 sin2 x cosx�= lim
x→0
−x(1−x2)3/2
6 sinx cos2 x− 3 sin3 x
�= limx→0
−1−2x2
(1−x2)5/2
6 cos3 x− 21 sin2 x cosx= −1
6
31. limx→0
tan−1 x
tan−1 2x�= lim
x→0
11 + x2
˙1 + 4x2
2=
12
32. limx→0
sin−1 x
x
�= limx→0
1√1−x2
1= 1
33. 1 : limx→∞
π/2 − tan−1 x
1/x�= lim
x→∞x2
1 + x2= 1
34. −1 : limn→∞
ln(1 − 1n )
sin( 1n )
= limx→0+
ln(1 − x)sinx
�= limx→0+
1(1 − x) cosx
= −1
35. 1 : limx→∞
1x[ ln (x + 1) − ln x ]
= limx→∞
1/xln (1 + 1/x)
= limt→0+
t
ln (1 + t)�= lim
t→0+(1 + t) = 1
36.13
: limn→∞
sinh(π/n) − sin(π/n)sin3(π/n)
= limx→∞
sinh(π/x) − sin(π/x)sin3(π/x)
�= limu→0+
sinhu− sinu
sin3 u
�= limu→0+
coshu− cosu3 sin2 u cosu
�= limu→0+
sinhu + sinu
6 sinu cos2 u− 3 sin3 u
�= limu→0+
coshu + cosu6 cos3 u− 21 sin2 u cosu
=26
=13
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
608 SECTION 11.5
37. limx→0
x3
4x − 1= 0 38. lim
x→0
4xsin2 x
does not exist
39. limx→0
xπ2 − arccosx
= 1 40. limx→2+
√2x− 2√x− 2
= 0
41. limx→0
tanhx
x= 1 42. lim
x→π/2
1 + cos 2x1 − sinx
= 4
43. limx→0
(2 + x + sin x) �= 0, limx→0
(x3 + x− cos x) �= 0
44. limn→∞
n(a1/n − 1
)= lim
x→∞a1/x − 1
1/x�= lim
x→∞
a1/x ln a(− 1
x2
)(− 1
x2
) = ln a
45. The limit does not exist if b �= 1. Therefore, b = 1.
limx→0
cos ax− 12x2
�= limx→0
−a sin ax
4x�= lim
x→0
−a2 cos ax4
= − a2
4
Now, − a2
4= − 4 implies a = ±4.
46. limx→0
sin 2x + ax + bx3
x3
�= limx→0
2 cos 2x + a + 3bx2
3x2need a = −2 to keep numerator 0
�= limx→0
−4 sin 2x + 6bx6x
�= limx→0
−8 cos 2x + 6b6
= 0 if 6b = 8
=⇒ a = −2, b =43
47. Recall limx→0+
(1 + x)1/x = e:
limx→0+
(1 + x)1/x − e
x
�= limx→0+
[(1 + x)1/x
] [x− (1 + x) ln (1 + x)x2 + x3
]
= e limx→0+
x− (1 + x) ln (1 + x)x2 + x3
�= e limx→0+
− ln (1 + x)2x + 3x2
�= e limx→0+
−1/(1 + x)2 + 6x
= −e
2
48. (a) limh→0
f(x + h) − f(x− h)2h
�= limh→0
f ′(x + h) − f ′(x− h)(−1)2
= f ′(x)
(note that here we differentiated with respect to h, not x. )
(b) limh→0
f(x + h) − 2f(x) + f(x− h)h2
�= limh→0
f ′(x + h) − f ′(x− h)2h
�= limh→0
f ′′(x + h) + f ′′(x− h)2
= f ′′(x)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
SECTION 11.5 609
49. limx→0
1x
∫ x
0
f(t) dt �= limx→0
f(x)1
= f(0)
50. (a) limx→0
Si(x)x
�= limx→0
sinx
x= 1
(b) limx→0
Si(x) − x
x3
�= limx→0
sinx/x− 13x2
= limx→0
sinx− x
3x3
�= limx→0
cosx− 19x2
�= limx→0
− sinx
18x= − 1
18
51. (a) limx→0
C(x)x
�= limx→0
cos2 x1
= 1
(b) limx→0
C(x) − x
x3
�= limx→0
cos2 x− 13x2
�= limx→0
−2 cosx sinx
6x�= lim
x→0
−2 cos2 x + 2 sin2 x
6= −1
3
52. (a) limx→a
∫ x
a
f(t) dt
f(x)�= lim
x→a
f(x)f ′(x)
= 0
(b) Similarly, limx→a
∫ x
a
f(t) dt
f(x)�= lim
x→a
f(x)f ′(x)
�= · · · �= limx→a
f (k−1)(x)f (k)(x)
= 0
53. A(b) = 2∫ √
b
0
(b− x2) dx = 2[bx− x3
3
]√b
0
=43b√b and T (b) =
12
(2√b)b = b
√b.
Thus, limb→0
T (b)A(b)
=b√b
43 b
√b
=34.
54. T (θ) =12(1 − cos θ) sin θ; S(θ) =
θ
2− 1
2sin θ:
limθ→0+
T (θ)S(θ)
= limθ→0+
(1 − cos θ) sin θ
θ − sin θ
�= limθ→0+
(1 − cos θ) cos θ + sin2 θ
1 − cos θ= lim
θ→0+
cos θ − cos 2θ1 − cos θ
�=− sin θ + 2 sin 2θ
sin θ= lim
θ→0+
− sin θ + 4 sin θ cos θsin θ
= 3
55. (a) f(x) → ∞ as x → ±∞
-5 5x
5
10
15
y
f
(b) f(x) → 10 as x → 4
Confirmation: limx→4
x2 − 16√x2 + 9 − 5
�= limx→4
2x
x (x2 + 9)−1/2= lim
x→42√x2 + 9 = 10
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
610 SECTION 11.6
56. (a) limx→±∞
f(x) = 0
−10 10x
0.17
y
(b) f(x) → 16
as x → 0; limx→0
x− sinx
x3
�= limx→0
1 − cosx3x2
�= limx→0
sinx
6x=
16
57. (a) f(x) → 0.7 as x → 0
-2 -1 1 2x
0.6
0.8
y
f
(b) Confirmation: limx→0
2sinx − 1x
�= limx→0
ln(2) 2sinx cosx1
= ln 2 ∼= 0.6931
58. (a) g(x) → −1.6 as x → 0x
− 1
− 2
y
(b) Confirmation: limx→0
3cosx − 3x2
�= limx→0
3cosx(− sinx) ln 32x
= − limx→0
3cosx sinx
2xln 3
= − 3 ln 32
∼= −1.6479
SECTION 11.6
(We’ll use � to indicate differentiation of numerator and denominator.)
1. limx→−∞
x2 + 11 − x
�= limx→−∞
2x−1
= ∞ 2. limx→∞
20xx2 + 1
= 0
3. limx→∞
x3
1 − x3= lim
x→∞1
1/x3 − 1= −1 4. lim
x→∞x3 − 12 − x
= −∞
5. limx→∞
x2 sin1x
= limh→0+
[(1h
)(sinh
h
)]= ∞
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
SECTION 11.6 611
6. limx→∞
ln(xk)x
�= limx→∞
k/x
1= 0
7. limx→π
2−
tan 5xtanx
= limx→π
2−
[(sin 5xsinx
)( cosxcos 5x
)]=
15
since
limx→π
2−
sin 5xsinx
= 1 and limx→π
2−
cosxcos 5x
�= limx→π
2−
sinx
5 sin 5x=
15
8. limx→0
(x ln | sinx|) = limx→0
ln | sinx|1/x
�= limx→0
cosxsinx
− 1x2
= limx→0
(− x
sinx
)(x cosx) = 0
9. limx→0+
x2x = limx→0+
(xx)2 = 12 = 1 [see (11.6.4)]
10. limx→∞
x sinπ
x= lim
t→0+
sinπt
t
�= limt→0+
π cosπt1
= π
11. limx→0
x( ln |x| )2 = limx→0
(ln |x|)21/x
�= limx→0
2 ln |x|−1/x
�= limx→0
21/x
= limx→0
2x = 0
12. limx→0+
lnx
cotx�= lim
x→0+
1/x− csc2 x
= limx→0+
− sin2 x
x= lim
x→0+− sinx
x· sinx = 0
13. limx→∞
1x
∫ x
0
et2dt
�= limx→∞
ex2
1= ∞
14. limx→∞
√1 + x2
x= lim
x→∞
√1x2
+ 1 = 1
15. limx→0
[1
sin2 x− 1
x2
]= lim
x→0
x2 − sin2 x
x2 sin2 x
�= limx→0
2x− 2 sinx cosx2x2 sinx cosx + 2x sin2 x
= limx→0
2x− sin 2xx2 sin 2x + 2x sin2 x
�= limx→0
2 − 2 cos 2x2x2 cos 2x + 4x sin 2x + 2 sin2 x
�= limx→0
4 sin 2x−4x2 sin 2x + 12x cos 2x + 6 sin 2x
�= limx→0
8 cos 2x−8x2 cos 2x− 32x sin 2x + 24 cos 2x
=13
16. Since limx→0
ln(| sinx|x) = limx→0
(x ln | sinx|) = 0 by Exercise 8, limx→0
| sinx|x = e0 = 1
17. limx→1
x1/(x−1) = e since limx→1
ln[x1/(x−1)
]= lim
x→1
ln x
x− 1�= lim
x→1
1x
= 1
18. Take log:
limx→0+
ln(xsinx
)= lim
x→0+(sinx lnx) = lim
x→0+
(lnx
cscx
)�= lim
x→0+
1/x− cscx cotx
= limx→0+
− sin2 x
x cosx= 0, so lim
x→0+xsinx = e0 = 1
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
612 SECTION 11.6
19. limx→∞
(cos
1x
)x
= 1 since limx→∞
ln[(
cos1x
)x]= lim
x→∞
ln(
cos1x
)(1/x)
�= limx→∞
(− sin (1/x)
cos (1/x)
)= 0
20. Take log:
limx→π/2
ln (| secx|cosx) = limx→π/2
cosx ln | secx| = limx→π/2
ln | secx|secx
�= limx→π/2
tanx
secx tanx= lim
x→π/2cosx = 0, so lim
x→π/2| secx|cosx = e0 = 1
21. limx→0
[1
ln (1 + x)− 1
x
]= lim
x→0
x− ln (1 + x)x ln (1 + x)
�= limx→0
x
x + (1 + x) ln (1 + x)
�= limx→0
11 + 1 + ln (1 + x)
=12
22. Take log: limx→∞
ln(x2 + a2)(1/x)2 = limx→∞
ln(x2 + a2)x2
�= limx→∞
2xx2+a2
2x= 0,
so limx→∞
(x2 + a2)(1/x)2 = e0 = 1
23. limx→0
[1x− cotx
]= lim
x→0
sinx− x cosxx sinx
�= limx→0
x sinx
sinx + x cosx
�= limx→0
sinx + x cosx2 cosx− x sinx
= 0
24. limx→∞
ln(x2 − 1x2 + 1
)3
= 3 limx→∞
ln(x2 − 1x2 + 1
)= 0
25. limx→∞
(√x2 + 2x− x
)= lim
x→∞
[(√x2 + 2x− x
)(√x2 + 2x + x√x2 + 2x + x
)]
= limx→∞
2x√x2 + 2x + x
= limx→∞
2√1 + 2/x + 1
= 1
26. limx→∞
(1 +
a
x
)bx= lim
x→∞
[(1 +
a
x
)x]b= (ea)b = eab.
27. limx→∞
(x3 + 1
)1/ lnx = e3 since
limx→∞
ln[(x3 + 1
)1/ lnx]
= limx→∞
ln(x3 + 1
)lnx
�= limx→∞
(3x2
x3 + 1
)1/x
= limx→∞
31 + 1/x3
= 3.
28. Take log: limx→∞
ln(ex + 1)x
�= limx→∞
ex
ex+1
1�= lim
x→∞ex
ex= 1,
so limx→∞
(ex + 1)1/x = e
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
SECTION 11.6 613
29. limx→∞
(coshx)1/x = e since
limx→∞
ln [(coshx)1/x] = limx→∞
ln (coshx)x
�= limx→∞
sinhx
coshx= 1.
30. Take log: limx→∞
3x ln(
1 +1x
)= 3 lim
x→∞
ln(1 + 1
x
)1x
�= 3 limx→∞
−1/x2
1+1/x
− 1x2
= 3,
so limx→∞
(1 +
1x
)3x
= e3
31. limx→0
(1
sinx− 1
x
)= lim
x→0
x− sinx
x sinx
�= limx→0
1 − cosxsinx + x cosx
�= limx→0
sinx
2 cosx− x sinx= 0
32. Take log: limx→0
ln(ex + 3x)x
�= limx→0
ex+3ex+3x
1= 4, so lim
x→0(ex + 3x)1/x = e4
33. limx→1
(1
lnx− x
x− 1
)= lim
x→1
x− 1 − x lnx
(x− 1) lnx
�= limx→1
− lnx
(x− 1)(1/x) + lnx
= limx→1
−x lnx
x− 1 + x lnx
�= limx→1
− lnx− 12 + lnx
= − 12
34.√
2: take log: limx→0
ln(1 + 2x) − ln 2x
�= limx→0
2x ln 21 + 2x
1=
ln 22
; limx→0
(1 + 2x
2
)1/x
= e12 ln 2 =
√2
35. 0 :1n
ln1n
= − lnn
n→ 0 36. 0: lim
n→∞nk
2n→ 0
37. 1 : ln[(lnn)1/n
]=
1n
ln (lnn) → 0
38. 0: limn→∞
lnn
np
�= limn→∞
1/npnp−1
= limn→∞
1p np
= 0
39. 1 : ln[(n2 + n
)1/n] =1n
ln [n(n + 1)] =lnn
n+
ln (n + 1)n
→ 0
40. 1: limn→∞
ln(nsin(π/n)
)= lim
n→∞[sin(π/n) lnn] = lim
n→∞
(sin(π/n)
1/n
) (lnn
n
)= 0, so nsin(π/n) → 1
41. 0 : 0 ≤ n2 lnn
en<
n3
en, lim
x→∞x3
ex= 0
42. 1: take log: ln(√n− 1)1/
√n =
ln(√n− 1)√n
→ 0, so (√n− 1)1/
√n → 1
43. limx→0
(sinx)x = 1 44. limx→π/4
(tanx)tan2x =1e
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
614 SECTION 11.6
45. limx→0
(1
sinx− 1
tanx
)= 0 46. lim
x→0+(sinhx)−x = 1
47.
vertical asymptote y-axis
48.
vertical asymptote x = 1
horizontal asymptote y = 1
49.
horizontal asymptote x-axis
50.
horizontal asymptote x-axis
51.
horizontal asymptote x-axis
52.
vertical asymptote y-axis
horizontal asymptote x-axis
53.b
a
√x2 − a2 − b
ax =
√x2 − a2 + x√x2 − a2 + x
(b
a
)(√x2 − a2 − x
)=
−ab√x2 − a2 + x
→ 0 as x → ∞
54. coshx− sinhx =12(ex + e−x) − 1
2(ex − e−x) = e−x → 0, as x → ∞
55. for instance, f(x) = x2 +(x− 1)(x− 2)
x3
56. for instance, F (x) = x +sinx
1 + x2
57. limx→0+
− 2xcos x
�= limx→0+
2− sin x
. L’Hospital’s rule does not apply here since limx→0+
cos x = 1.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
SECTION 11.6 615
58. (a) Let S be the set of positive integers for which the statement is true. Since limx→∞
lnx
x= 0, 1 ∈ S.
Assume that k ∈ S. By L’Hospital’s rule,
limx→∞
(lnx)k+1
x
�= limx→∞
(k + 1)(lnx)k
x= 0 (since k ∈ S).
Thus k + 1 ∈ S, and S is the set of positive integers.
(b) Choose any positive number α. Choose a positive integer k > α.
Then, for x > e,
0 <(lnx)α
x<
(lnx)k
x
and the result follows by the pinching theorem and part (a).
59. Let y =(a1/x + b1/x
2
)x
. Then ln y = x ln[a1/x + b1/x
2
]=
ln(a1/x + b1/x
2
)1/x
.
Now,
limx→∞
ln(a1/x + b1/x
2
)1/x
�= limx→∞
2a1/x + b1/x
· −a1/x ln a− b1/x ln b
2x2
−1/x2
= limx→∞
a1/x ln a + b1/x ln b
a1/x + b1/x= 1
2 ln ab = ln√ab
Thus, limx→∞
ln y = ln√ab =⇒ lim
x→∞y =
√ab.
60. (a) limk→0+
v(t) = limk→0+
mg(1 − e−(k/m)t
)k
�= limk→0+
gte−(k/m)t
1= gt
(b)dv
dt= g =⇒ v(t) = gt + C; v(0) = 0 =⇒ C = 0 and v(t) = gt.
61. (a) Ab = 1 − (1 + b)e−b
(b) xb =2 −
(2 + 2b + b2
)e−b
1 − (1 + b)e−b; yb =
14 − 1
4
(1 + 2b + 2b2
)e−2b
2 [1 − (1 + b)e−b]
(c) limb→∞
Ab = 1; limb→∞
xb = 2; limb→∞
yb =18
62. (a) Vx = π
[14− 1
4(1 + 2b + 2b2
)e−2b
]
(b) Vy = 2π[2 −
(2 + 2b + b2
)e−b]
(c) limb→∞
Vx =π
4, lim
b→∞Vy = 4π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
616 SECTION 11.6
63. (a) limx→0+
(1 + x2
)1/x = 1
1 2x
1
2
3
y
f
g
(b) limx→0+
(1 + x2
)1/x = 1 since
limx→0+
ln[(
1 + x2)1/x] = lim
x→0+
ln(1 + x2
)x
�= limx→0+
2x(1 + x2)
= 0
64. (a) limx→∞
f(x) ∼= 1.5
5 10 15 20x
1.5
y
(b) limx→∞
f(x) = limx→∞
[√x2 + 3x + 1 − x
]= lim
x→∞3x + 1√
x2 + 3x + 1 + x= lim
x→∞3x + 1
x√
1 + (3/x) + 1/x2 + x=
32
65. (a) limx→∞
g(x) ∼= − 1.7.10 50
x
-2
y
g
(b) limx→∞
g(x) = limx→∞
[3√x3 − 5x2 + 2x + 1 − x
]= lim
x→∞−5x2 + 2x + 1(
3√x3 − 5x2 + 2x + 1
)2+ x 3
√x3 − 5x2 + 2x + 1 + x2
= − 53∼= −1.667
66. [P (x)]1/n − x =([P (x)]1/n − x
)· [P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1
[P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1
=P (x) − xn
[P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1
=b1x
n−1 + b2xn−2 + · · · + bn
[P (x)](n−1)/n + x[P (x)](n−2)/n + · · · + xn−2[P (x)]1/n + xn−1→ b1
nas x → ∞
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49
SECTION 11.7 617
SECTION 11.7
1. 1 :∫ ∞
1
dx
x2= lim
b→∞
∫ b
1
dx
x2= lim
b→∞
[− 1x
]b1
= limb→∞
[1 − 1
b
]= 1
2.π
2:
∫ ∞
0
dx
1 + x2= lim
b→∞tan−1 b =
π
2
3.π
4:
∫ ∞
0
dx
4 + x2= lim
b→∞
∫ b
0
dx
4 + x2= lim
b→∞
[12
tan−1 x
2
]b0
= limb→∞
12
tan−1
(b
2
)=
π
4
4.1p
:∫ ∞
0
e−px dx = limb→0
(−e−pb
p+
1p
)=
1p
5. diverges:∫ ∞
0
epxdx = limb→∞
∫ b
0
epx dx = limb→∞
[1pepx]b0
= limb→∞
1p
(epb − 1
)= ∞
6. 2 :∫ 1
0
dx√x
= lima→0+
∫ 1
a
dx√x
= lima→0+
(2 − 2√a) = 2
7. 6:∫ 8
0
dx
x2/3= lim
a→0+
∫ 8
a
x−2/3 dx = lima→0+
[3x1/3
]8a
= lima→0+
[6 − 3a1/3
]= 6
8. diverges:∫ 1
0
dx
x2= lim
a→0+
∫ 1
a
dx
x2= lim
a→0+
(−1 +
1a
)= ∞
9.π
2:
∫ 1
0
dx√1 − x2
= limb→1−
∫ b
0
dx√1 − x2
= limb→1−
sin−1 b =π
2
10. 2:∫ 1
0
dx√1 − x
= lima→0+
∫ 1
a
dx√1 − x
= lima→0+
[−2
√1 − x
]1a
= lima→0+
(0 + 2√
1 − a) = 2
11. 2:∫ 2
0
x√4 − x2
dx = limb→2−
∫ b
0
x(4 − x2
)−1/2dx = lim
b→2−
[−(4 − x2
)1/2]b0
= limb→2−
(2 −
√4 − b2
)= 2
12.π
2:
∫ a
0
dx√a2 − x2
= limb→a−
∫ b
0
dx√a2 − x2
= limb→a−
[sin−1
(b
a
)− sin−1(0)
]=
π
2
13. diverges:∫ ∞
e
lnx
xdx = lim
b→∞
∫ b
e
lnx
xdx = lim
b→∞
[12
(lnx)2]be
= limb→∞
[12
(ln b)2 − 12
]= ∞
14. diverges:∫ ∞
e
dx
x lnx= lim
b→∞
∫ b
e
dx
x lnx= lim
b→∞[ln(lnx)]be = ∞
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618 SECTION 11.7
15. −14
: ∫ 1
0
x lnx dx = lima→0+
∫ 1
a
x lnx dx = lima→0+
[12x2 lnx− 1
4x2
]1
a
(by parts)∧
= lima→0+
[14a2 − 1
2a2 ln a− 1
4
]= −1
4
Note: limt→0+
t2 ln t = limt→0+
ln t
1/t2�= lim
t→0+
1/t−2/t3
= −12
limt→0+
t2 = 0.
16. 1:∫ ∞
e
dx
x(lnx)2= lim
b→∞
∫ b
e
dx
x(lnx)2= lim
b→∞
[− 1
lnx
]be
= limb→∞
[− 1
ln b+ 1]
= 1
17. π:∫ ∞
−∞
dx
1 + x2= lim
a→−∞
∫ 0
a
dx
1 + x2+ lim
b→∞
∫ b
0
dx
1 + x2
= lima→−∞
[tan−1 x
]0a
+ limb→∞
[tan−1 x
]b0
= −(−π
2
)+
π
2= π
18.ln 32
:∫ ∞
2
dx
x2 − 1= lim
b→∞
∫ b
2
dx
x2 − 1= lim
b→∞
[12
ln∣∣∣∣x− 1x + 1
∣∣∣∣]b2
= limb→∞
[12
ln(b− 1b + 1
)+
12
ln 3]
=12
ln 3
19. diverges:∫ ∞
−∞
dx
x2= lim
a→−∞
∫ −1
a
dx
x2+ lim
b→0−
∫ b
−1
dx
x2+ lim
c→0+
∫ 1
c
dx
x2+ lim
d→∞
∫ d
1
dx
x2;
and, limc→0+
∫ 1
c
dx
x2= lim
c→0+
[− 1x
]1
c
= limc→0+
[1c− 1]
= ∞
20. 2:∫ 3
1/3
dx3√
3x− 1= lim
a→ 13+
∫ 3
a
dx
(3x− 1)1/3= lim
a→ 13+
[3(3x− 1)2/3
2 · 3
]3
a
= lima→ 1
3+
[82/3
2− (3a− 1)2/3
2
]= 2
21. ln 2: ∫ ∞
1
dx
x(x + 1)= lim
b→∞
∫ b
1
[1x− 1
x + 1
]dx
= limb→∞
[ln(
x
x + 1
)]b1
= limb→∞
[ln(
b
b + 1
)− ln
(12
)]= ln 2
22. −1 :∫ 0
−∞xex dx = lim
a→−∞
∫ 0
a
xex dx = lima→−∞
[xex − ex]0a = lima→−∞
[−1 − aea + ea] = −1
23. 4:
∫ 5
3
x√x2 − 9
dx = lima→3−
∫ 5
a
x(x2 − 9
)−1/2dx
= lima→3−
[(x2 − 9
)1/2]5a
= lima→3−
[4 −
(a2 − 9
)1/2] = 4
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SECTION 11.7 619
24. ∫ 4
1
dx
x2 − 4= lim
b→2−
∫ b
1
dx
x2 − 4+ lim
a→2+
∫ 4
a
dx
x2 − 4= lim
b→2−
[14
ln∣∣∣∣x− 2x + 2
∣∣∣∣]b1
+ lima→2+
[14
ln∣∣∣∣x− 2x + 2
∣∣∣∣]4
a
= limb→2−
(14
ln∣∣∣∣b− 2b + 2
∣∣∣∣− 14
ln13
)+ lim
a→2+
(14
ln26− 1
4ln∣∣∣∣a− 2a + 2
∣∣∣∣)
= ∞ diverges
25.∫ 3
−3
dx
x(x + 1)diverges since
∫ 3
0
dx
x(x + 1)diverges:
∫ 3
0
dx
x(x + 1)= lim
a→0+
∫ 3
a
(1x− 1
x + 1
)dx = lim
a→0+
[ln |x| − ln |x + 1|
]3a
= lima→0+
[ ln 3 − ln 4 − ln a + ln (a + 1)] = ∞.
26.14
:∫ ∞
1
x
(1 + x2)2dx = lim
b→∞
∫ b
1
x
(1 + x2)2dx = lim
b→∞
[ −12(1 + x2)
]b1
= limb→∞
[ −12(1 + b2)
+14
]=
14
27.∫ 1
−3
dx
x2 − 4diverges since
∫ 1
−2
dx
x2 − 4diverges:
∫ 1
−2
dx
x2 − 4= lim
a→−2+
∫ 1
a
14
[1
x− 2− 1
x + 2
]dx
= lima→−2+
[14
(ln |x− 2| − ln |x + 2|)]1
a
= lima→−2+
14
[− ln 3 − ln |a− 2| + ln |a + 2| ] = −∞.
28.π
2: ∫ ∞
0
1ex + e−x
dx = limb→∞
∫ b
0
1ex + e−x
dx = limb→∞
[tan−1 ex
]b0
=π
2− π
4
∫ 0
−∞
1ex + e−x
dx = limb→−∞
∫ 0
b
1ex + e−x
dx = limb→−∞
[tan−1 ex
]0b
=π
4
29. diverges:∫ ∞
0
coshx dx = limb→∞
∫ b
0
coshx dx = limb→∞
[sinhx]b0 = ∞
30. Since∫ 2
1
dx
x2 − 5x + 6= lim
b→2−
∫ b
1
dx
(x− 2)(x− 3)= lim
b→2−
[ln∣∣∣∣x− 3x− 2
∣∣∣∣]b1
= limb→2−
(ln∣∣∣∣b− 3b− 2
∣∣∣∣− ln 2)
diverges, so does∫ 4
1
dx
x2 − 5x + 6
31.12
:∫ ∞
0
e−x sinx dx = limb→∞
∫ b
0
e−x sinx dx = limb→∞
− 12[e−x cosx + e−x sinx
]b0
(by parts)∧
= limb→∞
12[1 − e−b cos b− e−b sin b
]=
12
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620 SECTION 11.7
32. diverges:∫ ∞
0
cos2 x dx = limb→∞
[x
2+
sin 2x4
]b0
= limb→∞
(b
2+
sin 2b4
)= ∞
33. 2e− 2 :∫ 1
0
e√x
√xdx = lim
a→0+
∫ 1
a
e√x
√xdx = lim
a→0+
[2 e
√x]1a
= 2(e− 1)
34. 2 :∫ π/2
0
cosx√sinx
dx = lima→0+
∫ π/2
a
cosx√sinx
dx = lima→0+
[2√
sinx]π/2a
= 2
35. (a) converges:∫ ∞
0
x
(16 + x2)2dx =
132
(b) converges:∫ ∞
0
x2
(16 + x2)2dx =
π
16
(c) converges:∫ ∞
0
x
16 + x4dx =
π
16(d) diverges
36. (a) converges:∫ 2
0
x3
3√
2 − xdx =
243 3√
455
(b) converges:∫ 2
0
1√2 − x
dx = 2√
2
(c) converges:∫ 2
0
x√2 − x
dx =8√
23
(d) converges:∫ 2
0
1√2x− x2
dx = π
37.∫ 1
0
sin−1 x dx =[x sin−1 x
]10−∫ 1
0
x√1 − x2
dx =π
2− lim
a→1−
∫ a
0
x√1 − x2
dx
(by parts)
Now,∫ a
0
x√1 − x2
dx = − 12
∫ 1−a2
1
1√udu =
[−√u]1−a2
1= 1 −
√1 − a2
u = 1 − x2
Thus,∫ 1
0
sin−1 x dx =π
2− lim
a→1−
(1 −
√1 − a2
)=
π
2− 1.
38. (a)∫ ∞
0
xre−x dx diverges if r ≤ −1:
∫ ∞
0
xre−x dx =∫ 1
0
xre−xdx +∫ ∞
1
xre−x dx and∫ 1
0
xre−x dx diverges.
For any r > −1, we can find k such that xr < ex/2 for x ≥ k (ex/2 grows faster than any power
of x). Then∫ ∞
0
xre−xdx <
∫ k
0
xre−xdx +∫ ∞
k
e−x/2dx, which converges. Thus∫ ∞
0
xre−xdx
converges for all r > −1.
(b) For n = 1 :∫ ∞
0
xe−x dx = limb→∞
[−xe−x − e−x
]b0
= limb→∞
[−be−b − e−b + 1
]= 1
Assume true for n.∫ ∞
0
xn+1e−x dx = limb→∞
([−xn+1e−x
]b0
+ (n + 1)∫ b
0
xne−x dx
)
= limb→∞
(−bn+1e−b) + (n + 1)∫ ∞
0
xne−x dx = 0 + (n + 1)n! = (n + 1)!
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SECTION 11.7 621
39.
∫ ∞
0
1√x (1 + x)
dx =∫ 1
0
1√x (1 + x)
dx +∫ ∞
1
1√x (1 + x)
dx
= lima→0+
∫ 1
a
1√x (1 + x)
dx + limb→∞
∫ b
1
1√x (1 + x)
dx
Now,∫
1√x (1 + x)
dx =∫
21 + u2
du = 2 arctanu + C = 2 arctan√x + C.
u =√x
Therefore, lima→0+
∫ 1
a
1√x (1 + x)
dx = lima→0+
[2 arctan
√x]1a
= lima→0+
2[π/4 − arctan
√a)] =
π
2
and limb→∞
∫ b
1
1√x (1 + x)
dx = limb→∞
[2 arctan
√x]b1
= limb→∞
2[arctan
√b− π/4
]=
π
2.
Thus,∫ ∞
0
1√x (1 + x)
dx = π.
40.∫ ∞
1
1x√x2 − 1
dx = lima→1+
∫ 2
a
1x√x2 − 1
dx + limb→∞
∫ b
2
1x√x2 − 1
dx
= lima→1+
[sec−1 x
]2a
+ limb→∞
[sec−1 x
]∞2
= lima→1+
(sec−1 2 − sec−1 a) + limb→∞
(sec−1 b− sec−1 2)
= (sec−1 2 − 0) + (π
2− sec−1 2) =
π
2
41. surface area S =∫ ∞
1
2π(
1x
)√1 +
1x4
dx = 2π∫ ∞
1
√x4 + 1x3
dx = ∞ by comparison with∫ ∞
1
1xdx
42. A =∫ π/2
0
(secx− tanx) dx = limb→π/2−
∫ b
0
(secx− tanx) dx = limb→π/2−
[ln(secx− tanx) − ln secx
]b0
= limb→π/2−
[ln(1 + sinx)
]b0
= ln 2
43. (a) (b) A =∫ 1
0
1√xdx = lim
a→0+
∫ 1
a
1√xdx
= lima→0+
[2√x]1a
= 2
(c) V =∫ 1
0
π
(1√x
)2
dx = π
∫ 1
0
1xdx = π lim
a→0+
∫ 1
a
1xdx = π lim
a→0+[lnx]1a diverges
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622 SECTION 11.7
44. (a) (b) A =∫ ∞
0
11 + x2
dx = limb→∞
tan−1 b =π
2
(c) Vx =∫ ∞
0
π · 1(1 + x2)2
dx = limb→∞
π
2
[tan−1 x +
x
1 + x2
]b0
= limb→∞
π
2
(tan−1 b +
b
1 + b2− 0)
=π2
4
(d) Vy =∫ ∞
0
2πx1 + x2
dx = limb→∞
π[ln(1 + x2)
]b0
= ∞
45. (a) (b) A =∫ ∞
0
e−x dx = 1
(c) Vx =∫ ∞
0
πe−2x dx = π/2
(d) Vy =∫ ∞
0
2πxe−x dx = limb→∞
∫ b
0
2πxe−x dx = limb→∞
[2π(−x− 1)e−x
]b0
(by parts)
= 2π(
1 − limb→∞
b + 1eb
)= 2π(1 − 0) = 2π
(e) A =∫ ∞
0
2πe−x√
1 + e−2x dx = limb→∞
∫ b
0
2πe−x√
1 + e−2x dx
∫ b
0
2πe−x√
1 + e−2x dx = −2π∫ e−b
1
√1 + u2 du
u = e−x
= −π[u√
1 + u2 + ln(u +
√1 + u2
)]e−b
1
= π[√
2 + ln(1 +
√2)− e−b
√1 + e−2b − ln
(e−b +
√1 + e−2b
)]Taking the limit of this last expression as b → ∞, we have
A = π[√
2 + ln(1 +
√2)]
.
46. xA =∫ ∞
0
xe−x dx = 1, x =xA
A= 1
yA =∫ ∞
0
12e−2x dx =
14, y =
yA
A=
14; centroid: (1, 1
4 )
Yes: 2πxA = 2π = Vy, 2πyA =12π = Vx
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SECTION 11.7 623
47. (a) The interval [0, 1] causes no problem. For x ≥ 1, e−x2 ≤ e−x and∫ ∞
1
e−x dx is finite.
(b) Vy =∫ ∞
0
2πxe−x2dx = lim
b→∞
∫ b
0
2πxe−x2dx = lim
b→∞π[−e−x2
]b0
= limb→∞
π(1 − e−b2
)= π
48. (a) A =∫ ∞
1
(1x− x
x2 + 1
)dx =
12
ln 2
(b) Vx =∫ ∞
1
[(1x
)2
−(
x
x2 + 1
)2]dx <
∫ ∞
1
dx
x2finite
(c) Vy =∫ ∞
1
2πx(
1x− x
x2 + 1
)dx = 2π
∫ ∞
1
1x2 + 1
dx =12π2
49. (a) (b) A = lima→0+
∫ 1
a
x−1/4 dx = lima→0+
[43x3/4
]1
a
=43
(c) Vx = lima→0+
∫ 1
a
πx−1/2 dx = lima→0+
[2πx1/2
]1a
= 2π
(d) Vy = lima→0+
∫ 1
a
2πx3/4 dx = lima→0+
[8π7x7/4
]1
a
=87π
50. (i) Suppose that∫ ∞
a
g(x) dx = L. Since f(x) ≥ 0 for x ∈ [a,∞),∫ x
a
f(t) dt is increasing.
Therefore it is sufficient to show that∫ x
a
f(t) dt is bounded above. For any number
M ≥ a, we have ∫ M
a
f(x) dx ≤∫ M
a
g(x) dx ≤∫ ∞
a
g(x) dx = L
Therefore,∫ x
a
f(t) dt is bounded and∫ ∞
a
f(x) dx converges.
(ii) If∫ ∞
0
f(x) dx diverges, then∫ ∞
0
g(x) dx can not converge, by (i)
51. converges by comparison with∫ ∞
1
dx
x3/252. converges by comparison with
∫ ∞
2
e−x dx on [2,∞).
53. diverges since for x large the integrand is greater than1x
and∫ ∞
1
dx
xdiverges
54. Converges by comparison with∫ ∞
π
dx
x255. converges by comparison with
∫ ∞
1
dx
x3/2
56. Diverges by comparison with∫ ∞
e
dx
(x + 1) ln(x + 1)
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624 SECTION 11.7
57. (a) limb→∞
∫ b
0
2x1 + x2
dx = limb→∞
[ln(1 + x2
) ]b0
= ∞
Thus, the improper integral∫ ∞
0
2x1 + x2
dx diverges.
(b)limb→∞
∫ b
−b
2x1 + x2
dx = limb→∞
[ln(1 + x2
) ]b−b
= limb→∞
(ln[1 + b2
]− ln
[1 + (−b)2
])= lim
b→∞(0) = 0
58.∫ ∞
0
sinx dx = limb→∞
∫ b
0
sinx dx = limb→∞
[− cosx
]b0
= 1 − limb→∞
cos b; the limit does not exist
limb→∞
∫ b
−b
sinx dx = limb→∞
[− cosx
]b−b
= limb→∞
[− cos b + cos(−b)] = limb→∞
[ 0 ] = 0
59. r(θ) = aecθ, r′(θ) = acecθ
L =∫ θ1
−∞
√a2e2cθ + a2c2e2cθ dθ
(10.7.3)∧
=(a√
1 + c2)(
limb→−∞
∫ θ1
b
ecθ dθ
)
=(a√
1 + c2)(
limb→−∞
[ecθ
c
]θ1b
)
=
(a√
1 + c2
c
)(lim
b→−∞
[ecθ1 − ecb
])=
(a√
1 + c2
c
)ecθ1
60. For all real t, − t2
2< t + 1. Therefore
∫ x
−∞e−t2/2 dt converges by comparison with
∫ x
−∞et+1 dt.
61. F (s) =∫ ∞
0
e−sx · 1 dx = limb→∞
∫ b
0
e−sx dx = limb→∞
[− 1
se−sx
]b0
=1s
provided s > 0.
Thus, F (s) =1s; dom(F ) = (0,∞).
62. F (s) =∫ ∞
0
xe−sx dx = limb→∞
[−xe−sx
s− e−sx
s2
]b0
= limb→∞
(−be−sb
s− e−sb
s2+
1s2
)
=1s2
if s > 0, diverges for s ≤ 0, so dom(F ) = (0,∞).
63. F (s) =∫ ∞
0
e−sx cos 2x dx = limb→∞
∫ b
0
e−sx cos 2x dx
Using integration by parts∫
e−sx cos 2x dx =4
s2 + 4
[12e−sx sin 2x− s
4e−sx cos 2x
]+ C.
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SECTION 11.7 625
Therefore,
F (s) = limb→∞
4s2 + 4
[12e−sx sin 2x− s
4e−sx cos 2x
]b0
=4
s2 + 4limb→∞
[12e−sb sin 2b− s
4e−sb cos 2b +
s
4
]=
4s2 + 4
· s4
=s
s2 + 4provided s > 0.
Thus, F (s) =s
s2 + 4; dom(F ) = (0,∞).
64. F (s) =∫ ∞
0
eaxe−sx dx =∫ ∞
0
e(a−s)x dx = limb→∞
[e(a−s)b
a− s− 1
a− s
]
=1
s− aif s > a, diverges if s ≤ a; so dom F = (a,∞)
65. The function f is nonnegative on (−∞,∞) and∫ ∞
−∞f(x) dx =
∫ 0
−∞0 dx +
∫ ∞
0
6x(1 + 3x2)2
dx =∫ ∞
0
6x(1 + 3x2)2
dx
Now,∫
6x(1 + 3x2)2
dx = − 11 + 3x2
+ C.
Therefore, ∫ ∞
−∞f(x) dx = lim
b→∞
[− 1
1 + 3x2
]b0
= limb→∞
(1 − 1
1 + 3b2
)= 1.
66. f is nonnegative, and∫ ∞
−∞f(x) dx =
∫ ∞
0
ke−kx dx = limb→∞
[−e−kx
]b0
= limb→∞
(−e−kb + 1
)= 1
So, f is a probability density function.
67. μ =∫ ∞
−∞xf(x) dx =
∫ 0
−∞0 dx +
∫ ∞
0
kxe−kx dx = limb→∞
∫ b
0
kxe−kx dx
Using integration by parts,∫
kxe−kx dx = −xe−kx − 1ke−kx + C.
Therefore,
μ =∫ ∞
−∞xf(x) dx = lim
b→∞
[−xe−kx − 1
ke−kx
]b0
= limb→∞
[−be−kb − 1
ke−kb +
1k
]=
1k
68. σ =∫ ∞
−∞(x− μ)2f(x) dx =
∫ ∞
0
(x− 1
k
)2
ke−kx dx
=∫ ∞
0
kx2e−kx dx− 2∫ ∞
0
xe−kx dx +1k
∫ ∞
0
e−kx dx =1k2
69. Observe that F (t) =∫ t
1
f(x) dx is continuous and increasing, that an =∫ n
1
f(x) dx
is increasing, and that (∗) an ≤∫ t
1 f(x) dx ≤ an+1 for t ∈ [n, n + 1 ].
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626 REVIEW EXERCISES
If∫ ∞
1
f(x) dx converges, then F , being continuous, is bounded and, by (∗), {an} is bounded
and therefore convergent. If {an} converges, then {an} is bounded and, by (∗), F is bounded. Being
increasing, F is also convergent; i.e.,∫ ∞
1
f(x) dx converges.
REVIEW EXERCISES
1. |x− 2| ≤ 3 =⇒ −1 ≤ x ≤ 5 : lub = 5, glb = −1.
2. x2 > 3 =⇒ x >√
3 or x < −√
3; no lub, no glb.
3. x2 − x− 2 ≤ 0 =⇒ (x− 2)(x + 1) ≤ 0 =⇒ −1 ≤ x ≤ 2 : lub = 2, glb = −1.
4. cosx ≤ 1 for all x; no lub, no glb.
5. Since e−x2 ≤ 1 for all x, e−x2 ≤ 2 for all x; no lub, no glb.
6. lnx < e =⇒ 0 < x < ee : lub = ee, glb = 0.
7. increasing; bounded below by 12 and above by 2
3 .
8. increasing; bounded below by 0 but not bounded above:n2 − 1
n= n− 1
n→ ∞ as n → ∞.
9. bounded below by 0 and above by 32 ; not monotonic
10. increasing; bounded below by 45 and above by 1.
11.{
2n
n2
}={2, 1, 8
9 , 1, 3225 , . . .
}; the sequence is not monotonic.
However, it is increasing from a3 on. The sequence is bounded below by 89 ; it is not bounded above.
12.{
sin (nπ/2)n2
}={1, 0, − 1
9 , 0, 125 , · · ·
}; bounded below by − 1
9 and above by 1; not monotonic
13. the sequence does not converge; n21/n → ∞ as n → ∞
14. converges to 1:n2 + 3n + 2n2 + 7n + 12
=1 +
3n
+2n2
1 +7n
+12n2
→ 1.
15. converges to 1: limn→∞
1n
ln(
n
n + 1
)= 0 =⇒ lim
n→∞
(n
1 + n
)1/n
= 1
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REVIEW EXERCISES 627
16. converges to 0:4n2 + 5n + 1
n3 + 1=
4n
+5n2
+1n3
1 +1n3
→ 0.
17. converges to 0: cos πn sin π
n → cos 0 sin 0 = 0 as n → ∞
18. diverges: (2 + 1n )n > 2n and 2n diverges.
19. converges to 0: 0 = [ln 1]n ≤ [ln(1 +1n
) ]n ≤ [ln 2]n; [ln 2]n → 0 as n → ∞.
20. converges to ln 8: 3 ln 2n− ln(n3 + 1) = ln8n3
n3 + 1→ ln 8.
21. converges to 32 :
3n2 − 1√4n4 + 2n2 + 3
=3 − 1
n2√4 + 2
n2 + 3n4
→ 32
as n → ∞.
22. converges to 0:(n2 + 4)
13
2n + 1=
(1/n + 4/n3)1/3
2 + 1/n→ 0 as n → ∞.
23. converges to 0:π
ncos
π
n→ 0 cos 0 = 0 as n → ∞.
24. converges 0: (n/π) sin(nπ) = 0 for all positive integers n.
25. converges to 0:∫ n+1
n
e−xdx =[− e−x
]n+1
n= e−n(1 − 1
e) → 0 as n → ∞
26. diverges:∫ n
1
1√xdx =
[2√x]n1
= 2√n− 2 and 2
√n− 2 diverges.
27. Given ε > 0. Since an → L, there exists a positive integer K such that if n ≥ K, then |an − L| < ε.
Now, if n ≥ K − 1, then n + 1 ≥ K and |an+1 − L| < ε. Therefore, an+1 → L.
28. Let ε > 0. Since an → L, there is positive integer K such that if n ≥ K,
|an − L| < ε
2.
The set {|a1 − L|, · · · , |aK − L|} is a finite set so there is a positive integer N such that if n > N ,
|ai − L|n
<ε
2K, i = 1, 2, · · · ,K.
Let M = max {K,N}. Then, if n ≥ M ,∣∣∣∣a1 + · · · + ann
− L
∣∣∣∣ ≤ K∑i=1
|ai − L|n
+n∑
j=K+1
|aj − L|n
< K(ε
2K) + n(
ε
2n) = ε.
Therefore mn → L.
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628 REVIEW EXERCISES
29. As an example, let a = π3 . Then
cos π/3 = 0.5, cos cos 0.5 ∼= 0.87758, · · · .
Using technology (graphing calculator, CAS), we get
cos cos · · · cos π/3 → 0.73910.
and cos (0.73910) ∼= 0.73910.
Hence, numerically, this sequence converges to 0.73910.
30. Let f(x) = sin (cosx) and let a = π/3. Then
f(π/3) ∼= 0.4794, f(f(π/3)) ∼= 0.7753, · · ·
After 14 steps, we get f(f(· · · f(π/3)) ∼= 0.6948 and sin (cos 0.6948) ∼= 0.6948.
31. limx→∞
5x + 2 lnx
x + 3 lnx= lim
x→∞
5 + 2lnx
x
1 + 3lnx
x
= 5;(
limx→∞
lnx
x= 0
)
32. limx→0
ex − 1tan 2x
�= limx→0
ex
2 sec2 2x=
12
33. limx→0
ln(cosx)x2
�= limx→0
− sinx
cosx2x
= limx→0
−12 cosx
· sinx
x= −1
2
34. Set y = x1/(x−1). Then ln y =ln x
x− 1, and lim
x→1
ln x
x− 1�= lim
x→1
1x
= 1
Therefore, limx→1
x1/(x−1) = e.
35. limx→∞
(1 +4x
)2x = limx→∞
[(1 +
4x
)x]2
= e8
36. limx→0
e2x − e−2x
sinx
�= limx→0
2e2x + 2e−2x
cosx= 4
37. limx→0+
x2 lnx = limx→0+
lnx1x2
�= limx→0+
1x−2x3
= limx→0+
−x2
2= 0
38. limx→∞
10x
x10
�= limx→∞
10x ln 1010x9
�= limx→∞
10x(ln 10)2
(10)(9)x8
�= · · · = limx→∞
10x(ln 10)10
10!→ ∞
39. limx→0
ex + e−x − x2 − 2sin2 x− x2
�= limx→0
ex − e−x − 2x2 sinx cosx− 2x
�= limx→0
ex − e−x − 2sin 2x− 2x
�= limx→0
ex + e−x − 22 cos 2x− 2
�= limx→0
ex − e−x
−4 sinx2x�= lim
x→0
ex + e−x
−8 cos 2x= −1
4
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REVIEW EXERCISES 629
40. limx→1
csc(πx) lnx = limx→1
lnx
sinπx
�= limx→1
1/xπ cosπx
= − 1π
41. limx→∞
xe−x2∫ x
0
et2dt = lim
x→∞
∫ x
0
et2dt
ex2
x
�= limx→∞
ex2
2x2ex2 − ex
2
x2
= limx→∞
x2
2x2 − 1=
12
42. Consider lne−1/x2
xn=(− 1x2
− n ln |x|)
.
limx→0
(− 1x2
− n ln |x|)
= limx→0
−1 + nx2 ln |x|x2
= −∞
Therefore limx→0
e−1/x2
xn→ 0.
43.∫ ∞
1
e−√x
√x
dx = limb→∞
∫ b
1
e−√x
√x
dx = limb→∞
[− 2e−
√x]b1
= limb→∞
(−2e−√b + 2e−1) = 2e−1
44.∫
x√1 − x2
dx = −√
1 − x2 + C
∫ 1
0
x√1 − x2
dx = limb→1−
∫ b
0
x√1 − x2
dx = limb→1−
[−√
1 − x2]b0
= limb→1−
(−√
1 − b2 + 1) = 1
45.∫ 1
0
11 − x2
dx = lima→1−
∫ a
0
11 − x2
dx = −12
lima→1−
∫ a
0
(1
x− 1− 1
x + 1
)dx (partial fractions)
=12
lima→1−
[ln(x + 1) − ln(1 − x)
]a0
=12
lima→1−
ln[a + 11 − a
]= ∞;
the integral diverges.
46.∫ π/2
0
sec x dx = limc→π/2−
∫ c
0
sec x dx = limc→π/2−
[ln(secx + tanx)
]c0
= limc→π/2−
ln(sec c + tan c) = ∞;
the integral diverges.
47.∫ ∞
1
sin(π/x)x2
dx = limb→∞
∫ b
1
sin(π/x)x2
dx = limb→∞
[ 1π
cosπ/x]b1
=2π
48.∫ 9
0
1(x− 1)2/3
dx = limc→1−
∫ c
0
1(x− 1)2/3
dx + limc→1+
∫ 9
c
1(x− 1)2/3
dx
limc→1−
∫ c
0
1(x− 1)2/3
dx = limc→1−
3[(x− 1)1/3
]c0
= 3
limc→1+
∫ 9
c
1(x− 1)2/3
dx = limc→1+
3[(x− 1)1/3
]9c
= 6
∫ 9
0
1(x− 1)2/3
= 3 + 6 = 9.
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630 REVIEW EXERCISES
49.∫
1ex + e−x
dx =∫
ex
e2x + 1dx = arctan ex + C
∫ ∞
0
1ex + e−x
dx = limc→∞
∫ c
0
1ex + e−x
dx = limc→∞
arctan ex∣∣∣c0
=π
2
50. Set u = ln x, du =1xdx; u(2) = ln 2. Then∫ ∞
2
1x(lnx)k
dx =∫ ∞
ln 2
1uk
du
The integral converges if k > 1 and diverges otherwise.
For k > 1, ∫ ∞
2
1x(lnx)k
dx = limc→∞
∫ c
ln 2
1uk
du = limc→∞
11 − k
u1−k∣∣∣cln 2
=1
(k − 1) (ln 2)k−1
51.∫ a
0
ln(1/x) dx =∫ a
0
− ln x dx = limc→0+
∫ a
c
− ln x dx = limc→0+
[− x lnx + x
]ac
= limc→0+
[−a ln a + a + c ln c− c] = a ln(1/a) + a
52. y = (a2/3 − x2/3)3/2; y′ = −x−1/3(a2/3 − x2/3)1/2
L =∫ a
0
√1 + (y′)2 dx =
∫ a
0
a1/3x−1/3 dx =3a2
53. For any a ∈ S + T , a = s + t for some s ∈ S and t ∈ T . Hence a ≤ lub(S) + lub(T ).
Therefore, S + T is bounded above and lub(S) + lub(T ) is an upper bound for S + T .
Let M = lub(S + T ) and suppose M < lub(S) + lub(T ). Set ε = (lub(S) + lub(T )) −M . There
exist s ∈ S and t ∈ T such that
lub(S) − s < ε/2, and lub(T ) − t < ε/2
Now,
lub(S) + lub(T ) − (s + t) = lub(S) − s + lub(T ) − t < ε = lub(S) + lub(T ) −M
which implies s + t > M , a contradiction. Therefore lub(S) + lub(T ) = lub(S + T ).
54. (a) Since S is bounded below, there is a number b such that b ≤ s for every s ∈ S. Thus b is a
lower bound for S and B �= ∅.(b) Choose any s ∈ S. Then, for any b ∈ B, b ≤ s. Therefore B is bounded above (each element
s ∈ S is an upper bound for B).
(c) We show first that glb(S) is an upper bound for B. For if not, there is b ∈ B such that b > glb(S).
Then there is an s ∈ S such that glb(S) < s < b, which contradicts the fact that b is a lower
bound of S. It follows that lub(B) ≤ glb(S). If lub(B) < glb(S), then there exists a number
a such that lub(B) < a < glb(S) which implies that a is a lower bound for S and a ∈ B.
Therefore a ≤ lub(B), a contradiction. Thus, lub(B) = glb(S).
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REVIEW EXERCISES 631
55. (a) If∫ ∞
−∞f(x)dx = L, then
limc→∞
∫ c
0
f(x) dx, and limb→−∞
∫ 0
b
f(x) dx
both exist and
limc→∞
∫ c
0
f(x) dx + limb→−∞
∫ 0
b
f(x) dx = L
Let c = −b, then
limc→∞
∫ c
−c
f(x) dx = L
(b) Set f(x) = x. then limc→∞
∫ c
−c
x dx = 0, but∫ ∞
−∞x dx diverges.
56. (a) Assume that f is nonnegative on (−∞,∞).
By Exercise 55,∫ ∞
−∞f(x) dx = L =⇒ lim
c→∞
∫ c
−c
f(x) dx = L
Now assume that limc→∞
∫ c
−c
f(x) dx = L. Since f is nonnegative,
∫ x
0
f(t) dt ≤ L on [0,∞).
Therefore∫ x
0
f(t) dt is a bounded and nondecreasing function, which implies that
limc→∞
∫ c
0
f(x) dx exists. Similarly, limc→∞
∫ 0
−c
f(x) dx exists.
Therefore,∫ ∞
−∞f(x) dx exists, and, by the uniqueness of the limit,
∫ ∞
−∞f(x) dx = L.
57. Let S be a set of integers which is bounded above. Then there is an integer k ∈ S such that k ≥ n
for all n ∈ S, for if not, S is not bounded above. Therefore, k is an upper bound for S.
Let M = lub(S). Then M ≥ k since k ∈ S. Also M ≤ k since k is an upper bound for S. Therefore
M = k; the least upper bound of S is an element of S.
58. lub [Lf (P )] =∫ b
a f(x) dx; glb [Uf (P )] =∫ b
a f(x) dx.