Young man, in mathematics you don’t understand things. You just get used to them.
John von Neumann
Problem 1: Suppose f\colon \mathbb{R} \to \mathbb{R} is monotone increasing. Prove that (a) if f is bounded from above, then \lim_{x\to\infty}f(x) exists; (b) if f is not bounded from above, then \lim_{x\to\infty}f(x)=\infty.
Idea: (a) Let L be the supremum of the image of f. Argue that \lim_{x\to\infty}f(x) = L. (b) Write down the statements “f is not bounded from above” and “\lim_{x\to\infty}f(x)=\infty” mathematically.
Problem 2: Define mathematically that the sequence \{a_n\}_{n=1}^\infty (a) converges to 5; (b) does not converge to 3; (c) converges to \infty; (d) does not converge to \infty.
Answer: (a) for every \epsilon > 0 there exists N such that |a_n-5|<\epsilon for all n> N. (b) there exists \epsilon > 0 for every N, |a_n-3|\ge \epsilon for some n>N. (c) for every M, there exists N such that a_n > M for all n>N. (d) there exists M, for every N, a_n < M for some n > N.
Problem 3: Prove that if \{a_n\}_{n=1}^\infty converges (in the strict sense), then it is bounded.
Proof: Suppose the sequence converges to L. We know from Problem 2(a) there exists N such that |a_n-L|<1 for all n > N. Therefore the sequence is bounded from below by \min(a_1, \dots, a_N, L-1) and bounded from above by \max(a_1, \dots, a_N, L+1).
Problem 4: Adding, deleting or changing finite number of elements in a sequence does not change its convergence and its limit (in case it converges).
Hint: It is enough to show that adding, deleting or changing one element in a sequence does not change its convergence and its limit.
Fact: Suppose \{a_n\}_{n=1}^\infty converges to L, then \{|a_n|\}_{n=1}^\infty converges to |L|.
Application of problem 4: Prove that \{0, 1, 0, 1, \dots \} does not converge.
Proof: Suppose \{a_n\}_{n=1}^\infty = \{0, 1, 0, 1, \dots \} converges. By problem 4, \{a_{n+1}\}_{n=1}^\infty converges and has the same limit as \{a_n\}_{n=1}^\infty. Therefore \lim_{n\to\infty}{a_n} = \lim_{n\to\infty}{a_{n+1}}, and so \lim_{n\to\infty}|{a_n}-a_{n+1}| = 0. However, |a_n - a_{n+1}| = 1 for all n. A contradiction.
Problem 5: Suppose that \lim_{x\to 0}f(x) = L. Prove that \lim_{n\to\infty}f(1/n) = L.
Proof: Given \epsilon > 0. We know that there is \delta > 0 such that |f(x)-L|<\epsilon for all 0<|x|<\delta. Take N = 1/\delta. For all n > N, we know that 0 < 1/n < 1/N = \delta, thus |f(1/n)-L|<\epsilon.
Application of problem 5: \lim_{n\to\infty}\sin(1/n) = 0, \lim_{n\to\infty}\cos(1/n) = 0, \lim_{n\to\infty}\sin(1/n^2) = 0.
Problem 6: Let \{a_n\}_{n=1}^\infty and \{b_n\}_{n=1}^\infty be two sequences. Prove that if both \{a_n-b_n\}_{n=1}^\infty and \{a_n\}_{n=1}^\infty converge, then so does \{b_n\}_{n=1}^\infty.
Idea: b_n = a_n - (a_n - b_n).
Problem 7: Find (a) \lim_{n\to\infty}\frac{n^5-n^4-7n+4}{3n^5-n^3-2n^2-1}; (b) \lim_{n\to\infty}\sqrt[n]{1^n + 2^n + \dots + 10^n}; (c) \lim_{n\to\infty}\sqrt[n]{1+2+\dots+n}; (d) \lim_{n\to\infty}\sqrt[n]{1^5+2^5+\dots+n^5}; (e) \lim_{n\to\infty}\sqrt[n]{1(n-1)+2(n-2)+\dots+(n-1)1}.
Answers: (a) 1/3; (b) 10; (c) 1; (d) 1; (e) 1.