Problem 1: Consider the sequence \{a_n\}_{n=1}^\infty defined by a_1 = 1 and a_{n+1} = a_n + 1/a_n for every n \ge 1. Show that \{a_n\}_{n=1}^\infty is monotone increasing and \lim_{n\to\infty} a_n = \infty.
Proof: One can see that each a_n is positive (a rigorous proof requires mathematical induction). Thus a_{n+1} = a_n + 1/a_n > a_n, that is, the sequence is monotone increasing. Suppose, for a moment, that the sequence is bounded from above, by Problem 1(a) from Recitation 4, the sequence has a limit, say L. Because a_na_{n+1} = a_n^2 + 1, L^2 = \lim_{n\to\infty}a_na_{n+1} = \lim_{n\to\infty}a_n^2+1 = L^2+1, which is impossible. Therefore the sequence is not bounded from above. By Problem 1(b) from Recitation 4, \lim_{n\to\infty} a_n = \infty.
Remark: Alternatively, one can prove by mathematical induction that a_n \le n. Again, by induction, we can show that a_n \ge 1 + 1/1 + 1/2 + \dots + 1/(n-1). Denote by H_n := 1/1 + 1/2 + \dots + 1/n, which is the Harmonic series. It can be shown that H_n \ge 1 + \lfloor \log_2 n \rfloor / 2, which diverges to infinity.
Problem 2: Suppose that \lim_{n\to\infty} a_n = 0. Prove that \lim_{n\to\infty} 5^{a_n} = 1.
Proof: We know that \lim_{n\to\infty}5^{1/n} = 1. Given \epsilon > 0, find M such that 5^{1/M} < 1+\epsilon. Since \lim_{n\to\infty} a_n = 0, we can find N such that -1/M < a_n < 1/M for all n > N. Now, for all n > N, 1 - \epsilon < \frac{1}{1+\epsilon} < 5^{-1/M} < 5^{a_n} < 5^{1/M} < 1 + \epsilon, that is, |5^{a_n} - 1| < \epsilon.
Problem 3: Prove that if \lim_{n\to\infty} a_n = A > 0 and \lim_{n\to\infty} b_n = B, then \lim_{n\to\infty} a_n^{b_n} = A^B.
Exercise 1: Use mathematical induction to prove the following. For all \epsilon\in(0,1) and M\in\mathbb{N}, (1-\epsilon)^M \ge 1-M\epsilon.
Exercise 2: Suppose that \epsilon \in (0, 1), M\in\mathbb{N}. If 1-\epsilon/M<a<1+\epsilon/M and -M < b < M, then 1 - \epsilon < a^b < 1+\epsilon.
Proof: We first prove that case when A = 1. Since \lim_{n\to\infty} b_n = B, by Problem 3 from Recitation 4, the sequence \{b_n\}_{n=1}^\infty is bounded. Let N be a natural number such that -M < b_n < M for all n. Suppose that \epsilon \in (0,1) is given. Since \lim_{n\to\infty} a_n = 1, there exists N such that 1-\epsilon/M<a_n<1+\epsilon/M. For all n > N, by Exercise 2, 1 - \epsilon < a_n^{b_n} < 1 + \epsilon, and so \lim_{n\to\infty} a_n^{b_n} = 1.
Finally, we prove the general case. Let a'_n = a_n / A. Because \lim_{n\to\infty} a_n' = \lim_{n\to\infty} a_n/A = 1, we know that \lim_{n\to\infty} a_n'^{b_n} = 1. Moreover, from problem 2, we know that \lim_{n\to\infty} A^{b_n} = A^B. Therefore \lim_{n\to\infty} a_n^{b_n} = \lim_{n\to\infty} (a_n' \cdot A)^{b_n} = \lim_{n\to\infty} a_n'^{b_n} \lim_{n\to\infty} A^{b_n} = A^B.
Facts: If \lim_{n\to\infty}a_n=\infty, then \lim_{n\to\infty}(1+1/a_n)^{a_n}=e. If \lim_{x\to x_0}f(x)=\infty, then \lim_{n\to\infty}(1+1/f(x))^{f(x)}=e.
Corollary 1: If \lim_{n\to\infty}a_n=\infty and \lim_{n\to\infty}b_n/a_n = L, then (a) \lim_{n\to\infty}(1+1/a_n)^{b_n}=e^L and (b) \lim_{n\to\infty}(1-1/a_n)^{b_n}=e^{-L}.
Corollary 2: If \lim_{x\to x_0}f(x)=\infty and \lim_{x\to 0}g(x)/f(x) = L, then (a) \lim_{n\to\infty}(1+1/f(x))^{g(x)}=e^{L} and (b) \lim_{n\to\infty}(1-1/f(x))^{g(x)}=e^{-L}.
Problem 4: Find the limits (a) \lim_{n\to\infty} \left(\sin(1/n)+\cos(1/n)\right)^{n^2}; (b) \lim_{n\to\infty} \left(1+\tan(1/n)\right)^{n}; (c) \lim_{n\to\infty} \left(\frac{n^2+n+1}{n^2-n-1}\right)^{n}; (d) \lim_{x\to 0} \left(\cos x\right)^{1/x^2}.
Answers: (a) \infty; (b) e; (c) e^2; (d) e^{-1/2}.
Idea: (a) In Corollary (1a), take a_n = 1/(\sin(1/n)+\cos(1/n)-1)=1/(2\sin(1/2n)(\cos(1/2n)-\sin(1/2n))) and b_n = n^2. (b) In Corollary (1a), take a_n = 1/\tan(1/n) and b_n = n. (c) In Corollary (1a), take a_n = (n^2-n-1)/(2n+2) and b_n = n. (d) In Corollary (2b), take x_0 = 0, f(x) = 1/(1-\cos x) = 1/(2\sin^2(x/2)) and g(x) = 1/x^2.
Problem 5: Let f\colon \mathbb{R} \to \mathbb{R} be a continuous function such that f(x) = f(2x) for every x. Show that f is constant.
Idea: In Problem 6 from Homework 2, we actually proved the following. Let f\colon \mathbb{R} \to \mathbb{R} be a function such that f(x) = f(2x) for every x. If \lim_{x\to 0}f(x) = L, then f(x) = L for all x \neq 0. In particular, if f is continuous, L must be f(0).