San Jose State University

Math 131A, Fall 2006

Final Exam Solutions

December 13, 2006

9:45 – noon

Name: Erastus Hamm

Score

1 202 203 204 255 206 20

Total 125

Explain your work

1. (20 points) Compute

limn→∞

(√n(n− 1)(n + 1)− n3/2

).

Solution: Multiplying and dividing by√

n(n− 1)(n + 1) + n3/2, we obtain:

√n(n− 1)(n + 1)− n3/2 =

n(n− 1)(n + 1)− n3√n(n− 1)(n + 1) + n3/2

=−n√

n(n− 1)(n + 1) + n3/2

= − 1√(1− 1

n

)(n + 1) +

√n

→ − 1

∞= 0,

as n →∞.

2. (20 points) Determine whether the following series converge or diverge:

(a)∞∑

n=1

[(1 +

1

n

)n

− 2

]; (b)

∞∑n=1

sin2 n3

n3/2.

Solution: (a) The series diverges, since its general term does not go to zero: as n →∞,(1 +

1

n

)n

− 2 → e− 2 6= 0.

(b) Using the fact that |sin| ≤ 1, we obtain

0 ≤ sin2 n3

n3/2≤ 1

n3/2.

for all n ∈ N. Since∑

1/n3/2 is convergent as a p-series with p = 3/2 > 1, by ComparisonTest it follows that the given series also converges.

3. (20 points) A sequence (xn) is defined inductively as follows:

x1 = 5, xn+1 =1 + xn

2.

(a) Show that (xn) is convergent.

(b) Find the limit of (xn).

Solution: (a) Claim 1: (xn) is decreasing.

Proof of Claim 1: By induction. We have x2 = 3 < 5 = x1. Assume xk > xk+1, for somek ≥ 1. Then

xk+1 =1 + xk

2>

1 + xk+1

2= xk+2.

By the Principle of Mathematical Induction (PMI), it follows that (xn) is decreasing.

Claim 2: xn > 1, for all n ∈ N.

Proof of Claim 2: Also by induction. Claim 2 is clearly true for n = 1. Assume xk > 1,for some k ≥ 1. Then

xk+1 =1 + xk

2>

1 + 1

2= 1.

Claim 2 follows by the PMI.

Since (xn) is decreasing and bounded below, it follows that it is convergent.

(b) Let x∗ be the limit of (xn). Letting n →∞ in

xn+1 =1 + xn

2,

we obtain x∗ = (1 + x∗)/2. This implies that x∗ = 1.

4. (20 + 5 points) Define a function f : R → R by

f(x) =

{x2, if x ∈ Q

3x− 2, if x 6∈ Q.

(a) Show that f is continuous at 1.

(b) Show that f is discontinuous at 0.

(c) (extra credit, 5 points) Show that f has exactly two points of continuity.

Proof: (a) Let (xn) be an arbitrary sequence converging to 1. For rational xn, f(xn) = x2n →

12 = 1. For irrational xn, we have f(xn) = 3xn − 2 → 3 · 1− 2 = 1. Therefore, the fullsequence converges and

limn→∞

f(xn) = 1,

which equals f(1). By the Sequential Criterion for continuity, f is continuous at 1.

Remark. A completely rigorous proof would go like this. Let (xn) be an arbitrary se-quence converging to 1. Let (xnk

), (xmk) be the subsequences of rational and irrational

members of (xn), respectively. They also converge to 1, since the entire sequence does.Then f(xnk

) = x2nk→ 12 = 1 and f(xmk

) = 3xmk− 2 → 3 · 1 − 2 = 1, as k → ∞.

Since (f(xnk)) and (f(xmk

)) together make up the whole sequence (f(xn)) and theyboth converge to 1, it follows that f(xn) → 1, as n →∞. Since f(1) = 1, it follows thatf is continuous at 1, by the Sequential Criterion.

(b) Let (yn) be a sequence of irrational numbers converging to 0. Such as sequence existsbecause the irrationals are dense in R. Then

limn→∞

f(yn) = limn→∞

(3yn − 2) = −2 6= 0 = f(0).

By the Sequential Criterion for continuity, it follows that f is discontinuous at 0.

(c) Let a ∈ R be arbitrary and let (zn) be a sequence converging to a. Then f(zn) → a2

for rational zn and f(zn) → 3a − 2, for irrational zn. The limit of the entire sequencef(zn) exists (and equals f(a)) if and only if a2 = 3a−2, i.e., iff a = 1 or a = 2. Therefore,f is continuous at 1 and 2 and discontinuous elsewhere.

5. (20 points) Prove that the equation cos x = x has a solution in the interval[0, π

2

].

Proof: Definef(x) = x− cos x.

The function f is continuous (on R) as a difference of two continuous functions. Sincef(0) = −1 < 0 and f(π/2) = π/2 > 0, by the Intermediate Value Theorem, there existsc ∈ (0, π/2) such that f(c) = 0. Then cos c = c.

6. (20 points) Let f : [0, 1] → R be a continuous function. If f has only integer values,show that it must be constant.

Proof: Since f is continuous and [0, 1] is a compact interval, f([0, 1]) is a compact interval.Denote it by [a, b]. On the other hand, f has only integer values, so [a, b] ⊂ Z. This ispossible only if a = b ∈ Z. Therefore, f(x) ≡ a, i.e., f is constant.