Mathematics 242 – Principles of Analysis Selected Solutions – Exam 1 October 11, 2000

I. Let A = {x ∈ R : 1 < x^{2} < 9} and let B be the interval [−2, 4).

A) What is the set A ∩ B?

Solution: 1 < x^{2} < 9 is equivalent to either 1 < x < 3, or −3 < x < −1. So A is the union of two intervals: A = (−3, −1) ∪ (1, 3). The intersection with B is

A ∩ B = [−2, −1) ∪ (1, 3)

B) Explain why the set C = {|x + 1/2| : x ∈ A} is bounded, and give upper and lower bounds for it.

Solution: Since C is a set of numbers obtained by taking absolute values, it is bounded below by 0. Since x ∈ A implies |x| < 3, |x + 1/2| ≤ |x| + |1/2| < 3 + 1/2 for all x ∈ A. Hence C is also bounded above.

C) What are the least upper bound and greatest lower bound for the set C from part B? Explain your reasoning.

Solution: |x + 1/2| = |x − (−1/2)| is the distance from x to −1/2 along the number line. By part A, we know that A = (−3, −1)∪(1, 3). Hence the closest points to −1/2 in A are x near −1, and we can get arbitrarily close to |x + 1/2| = 1/2 for x close to −1 in A. The greatest lower bound for C is 1/2. Similarly, the farthest points from −1/2 are the x near 3 in A. Hence the least lower bound for C is 7/2.

II. For parts A, B see the text or class notes. Comments: There will be questions like this on the next midterm and final. You will be throwing away lots of hard-to-replace points if you do not know the definitions, statements of results, and proofs that I put on the review sheet for an exam. Also, part C of this question was the only problem on this exam intended to be slightly “tricky”!

C) Suppose that x_{n} is a sequence of negative numbers and x_{n+1}/x_{n} ≤ 1 for all n ≥ 1. Show that x_{n} converges to some a ∈ R.

Solution: Since all the x_{n} are negative, when we multiply both sides of the inequality x_{n+1}/x_{n} ≤ 1 by x_{n} to clear denominators, the inequality reverses: x_{n+1} ≥ x_{n} for all n. Also since all of the x_{n} are negative, the number M = 0 is an upper bound for the sequence. By the result of part B, every increasing sequence that is bounded above converges to some a ∈ R. The most we can say here is that a ≤ 0.

A) Identify lim_{n→∞} x_{n} for the sequence

x_{n} = \frac{3n}{2n + 5},

and prove that x_{n} converges to your limit using the ϵ - n_{0} definition of convergence.

Solution: By our usual methods, the limit should be \frac{3}{2}. To prove this, note that

|\frac{3n}{2n + 5} − \frac{3}{2}| = |\frac{−15}{4n + 10}|

= \frac{15}{4n + 10}

< \frac{15}{4n}

Hence, given ϵ > 0, if n ≥ n_{0} > \frac{15}{4ϵ}, then \frac{1}{n} < \frac{4ϵ}{15}, so we will have

|\frac{3n}{2n + 5} − \frac{3}{2}| < \frac{15}{4n} < \frac{15}{4} · \frac{4ϵ}{15} = ϵ.

By the definition of convergence, this shows that lim_{n→∞} x_{n} = \frac{3}{2}.

B) Use the “Big Limit Theorem” for sequences to determine

lim_{n→∞} \begin{matrix}√ & \overline{4n^{2} + 3n + 1} \\ ^{3} & \sqrt{27n^{3} + 3} \end{matrix}

Solution: The fastest growing terms on the top and bottom both are roughly constant times n as n → ∞ (square root of n^{2} and cube root of n^{3}). Hence, if we divide both top and bottom by \frac{1}{n}, we get

lim_{n→∞} \begin{matrix}√ & \overline{4n^{2} + 3n + 1} \\ ^{3} & \sqrt{27n^{3} + 3} \end{matrix} = lim_{n→∞} \frac{\sqrt{4n^{2} + 3n + 1} · \frac{1}{n}}{\sqrt[3]{27n^{3} + 3} · \frac{1}{n}}

= lim_{n→∞} \begin{matrix}√ & \overline{4 + \frac{3}{n} + \frac{1}{n^{2}}} \\ ^{3} & \sqrt{27 + \frac{3}{n^{3}}} \end{matrix}

= \frac{2}{3}

C) Consider the sequence

x_{n} = sin (\frac{π\sqrt{n}}{2})

Find three convergent subsequences of x_{n} but converging to three different limits. (For each one, give a formula for the index sequence n_{k} for the subsequence). What does this say about lim_{n→∞} x_{n}?

Solution: The simplest way to find the required subsequences is to try to find n_{k} → ∞ where the sine function takes on some constant value. For instance, we know that sin(x) = 0 for all integer multiplies of π, so if

\frac{π\sqrt{n}}{2} = kπ

for any integer k ≥ 1, then we get a zero term in the sequence. From the equation above, n_{k} = (2k)^{2} gives one such subsequence, with x_{n_{k}} = 0 for all k ≥ 1. Hence lim_{k→∞} x_{n_{k}} = 0. Similarly, to get sin(x) = 1, we can take

\frac{π\sqrt{n}}{2} = \frac{(4k + 1)π}{2}

for all integers k ≥ 1. Hence n_{k} = (4k + 1)^{2} gives a subsequence converging to +1. Finally, to get sin(x) = −1, we can take

\frac{π\sqrt{n}}{2} = \frac{(4k + 3)π}{2}

for all integers k ≥ 1. Hence n_{k} = (4k + 3)^{2} gives a subsequence converging to −1. The fact that we can find these subsequences says that the whole x_{n} sequence does not converge.