Mathematics 242 – Principles of Analysis Solutions for Exam 2 – April 5, 2013

I. A) (15) Show using the ε, n_{0} definition that

lim_{n→∞} \frac{5n + 1}{3n + 3} = \frac{5}{3}.

Solution: Given ε > 0, let n_{0} > \frac{4}{3ε}. Then for all n ≥ n_{0},

|\frac{5n + 1}{3n + 3} − \frac{5}{3}| = \frac{4}{3n + 3} < \frac{4}{3n} ≤ \frac{4}{3n_{0}} < ε.

This shows the limit is \frac{5}{3}.

B) (15) Show using the ε, δ definition that

lim_{x→1} \frac{5x + 1}{3x + 3} = 1.

Solution: Given ε > 0, let δ = min (1, \frac{3ε}{2}). Then for all x in the deleted neighborhood of 1 defined by 0 < |x − 1| < δ we have 0 < x < 2, so \frac{1}{|x+1|} < 1 and |x − 1| < \frac{3ε}{2}. Therefore

|\frac{5x + 1}{3x + 3} − 1| = \frac{2|x − 1|}{3|x + 1|} = \frac{1}{|x + 1|} · \frac{2}{3} · |x − 1| < 1 · \frac{2}{3} · \frac{3ε}{2} =< ε.

This shows the limit is 1.

II. A) (20) State and prove the Monotone Convergence Theorem for sequences. (You may

give the proof in the case that the sequence is monotone increasing.)

Solution: The statement is that any monotone bounded sequence of real numbers converges. If the sequence is monotone increasing, let a = lub{x_{n} | n ∈ N}. Then for all ε > 0, a − ε is not an upper bound for {x_{n} | n ∈ N}, so there exist n_{0} such that a − ε < x_{n_{0}} ≤ a. But then since {x_{n}} is monotone increasing, we have a − ε < x_{n_{0}} ≤ x_{n} ≤ a for all n ≥ n_{0}. This shows x_{n} → a, since it implies |x_{n} − a| < ε for all n ≥ n_{0}.

B) (10) Let {x_{n}} be the sequence defined by x_{1} = 1 and

x_{n+1} = \frac{5x_{n} + 3}{16}

for all n ≥ 1. Does this sequence converge? Why? If it does, what is the limit?

Solution: We have x_{1} = 1 and x_{2} = \frac{5·1+3}{16} = \frac{1}{2}. Assuming x_{k+1} < x_{k}, it follows that

x_{k+2} = \frac{5x_{k+1} + 3}{16} < \frac{5x_{k} + 3}{16} = x_{k+1}.

Therefore, the sequence is monotone (strictly) decreasing. The terms are clearly bounded below by 0, so the sequence converges by the Monotone Convergence Theorem. The limit is found by letting n → ∞ in the recurrence. If a denotes the limit then a = \frac{5a+3}{16}, so a = \frac{3}{11}.

III. (10) Let x_{n} = sin(n) − cos(n) (the sequence of values of sin(x) − cos(x) at the angles x given by a whole number n ≥ 1 in radian measure). Show that there exists a subsequence x_{n_{k}} = sin(n_{k}) − cos(n_{k}) that converges. State any “big theorems” that you are using.

Solution 1: We have

|x_{n}| = | sin(n) − cos(n)| ≤ | sin(n)| + | cos(n)| ≤ 2

for all n by the triangle inequality. This shows the sequence is bounded. Hence the Bolzano-Weierstrass theorem implies it has a convergent subsequence. (Note: This is not the tightest possible upper bound. In fact, by trig identities,

sin(n) − cos(n) = \sqrt{2} sin (n − \frac{π}{4} )

so | sin(n)−cos(n)| ≤ \sqrt{2} is also true for all n. There are n for which | sin(n)−cos(n)| > 1, though, so that is not a valid upper bound.)

It is also possible to argue like this:

Solution 2: Since | sin(n)| ≤ 1 for all n, by the Bolzano-Weierstrass theorem there exists an index sequence n_{k} such that sin(n_{k}) converges to some a ∈ R. There is no guarantee that cos(n_{k}) also converges, but we can apply the Bolzano-Weierstrass theorem again to that sequence and get a subsequence of the index sequence n_{k}, say n_{k_{ℓ}} for some strictly increasing sequence k_{ℓ} of k-values, such that cos(n_{k_{ℓ}}) converges to some b ∈ R. But then sin(n_{k_{ℓ}}) is a subsequence of a convergent sequence, so it also converges to a. Then by the limit sum theorem, as ℓ → ∞,

sin(n_{k_{ℓ}}) − cos(n_{k_{ℓ}}) → a − b.

(Note: Solution 1 is certainly more direct(!))

IV. Give an example, or give a reason why there can be no such examples:

A) (10) A function f that is continuous at all real c, with f(x) = 3 for all x ∈ Q, but f(\sqrt{2}) = 4.

Solution: There are no such examples because we can find sequences x_{n} → \sqrt{2} where all the x_{n} are rational. If f is continuous at c = \sqrt{2}, then f(x_{n}) → f(\sqrt{2}) = 4 but f(x_{n}) = 3 for all n if x_{n} is rational, so this is impossible.

B) (10) A function f that is continuous and bounded on an open interval (a, b) with M = lub{f(x) | x ∈ (a, b)}, but such that f(x) ≠ M for all x ∈ (a, b).

Solution: An example is f(x) = x on (0, 1). We have M = 1, but f(x) ≠ 1 for any x ∈ (0, 1). (The Extreme Value Theorem does not apply on open intervals.)

C) (10) A continuous function on [0, 1] with f(0) = 4, f(1) = −2, but f(x) ≠ 0 for all x ∈ (0, 1). Solution: There are no such examples because the Intermediate Value Theorem says that there must be solutions of all equations f(c) = k for k between f(0) = 4 and f(1) = −2, including k = 0.

Extra Credit. (10) A function is said to be right-continuous at c if the one-sided limit lim_{x→c^{+}} f(x) exists and equals f(c). True or False (and give a proof or a counterexample): If f is right-continuous at all c ∈ [a, b] for some closed interval [a, b], then {f(x) | x ∈ [a, b]} is a bounded subset of R.

Solution: This is not true. An example is

f(x) = \{ \begin{matrix}0 & if x ≥ 0 \\ 1/x & if x < 0 \end{matrix}

for x ∈ [−1, 1]. Then lim_{x→c^{+}} f(x) = f(c) for all c ∈ [−1, 1], but the set of values is not bounded, since lim_{x→0^{−}} f(x) = −∞.