**Problem 1:** Show \lim_{x\to\infty} [x]/x = 1.

**Idea: **Use x-1<[x]\le x and the sandwich theorem.

**Problem 2:** Given sets A and B. It is known that for every a\in A there exists b\in B such that a < b - 1/10000. Show that \sup A < \sup B.

**Proof: **By the definition of supreme, there is a\in A such that a > \sup A - 1/10000. By the condition in the problem, there is b\in B such that a < b - 1/10000. Thus \sup A < b \le \sup B.

**Remark:** If for every a\in A there exists b\in B such that a < b, we can only conclude that \sup A \le \sup B (why?). In general, we cannot conclude \sup A < \sup B. For example, A = (0, 1) and B = \{1\}.

**Problem 3:** Find two functions f and g such that \lim_{x\to \infty}f(x) = \lim_{x\to \infty}g(x) = \infty, but \lim_{x\to \infty}f(x)/g(x) does not exist (even in the broad sense).

**Idea:** Take h(x) = \sin x + 2 (or any other positive function which does not have a limit as x approaches infinity). Notice that h(x) \ge 1 for all x. Take g(x) = x (or any other function whose limit at infinity is \infty). Now let f(x) = g(x)h(x).

**Problem 4:** Suppose \lim_{x\to \infty}f(x) = \infty and \lim_{x\to \infty}f(x)g(x) = 1. Show that necessarily \lim_{x\to \infty}g(x) = 0.

**Sketch: **Our goal is to prove that for every \epsilon > 0 there is M such that |g(x)|<\epsilon for all x>M. We can find M_1 such that f(x) > 2/\epsilon for all x>M_1 and M_2 such that 0 < f(x)g(x) < 2. Now take M = \max(M_1, M_2).

**Problem 5:** Compute the following limits (a) \lim_{x\to \infty}\sqrt{x+1}-\sqrt{x}; (b) \lim_{x\to \infty}\frac{x+\sin x}{x+\cos x}; (c) \lim_{x\to a}\frac{\sin^2 x - \sin^2 a}{x^2 - a^2}; (d) \lim_{x\to \infty}\sin 2x / \sin 3x; (e) \lim_{x\to 1}\frac{x^n-1}{x^m-1}; (f) \lim_{x\to \infty}\frac{x^4-5}{x^2+4x+7}; (g)\lim_{x\to \infty}\frac{x^3}{x^2+1}-x.

**Answers:** (a) 0; (b) 1; (c) \frac{\sin a}{a} if a\neq 0; 1 if a = 0; (d) 2/3; (e) n/m; (f) \infty; (g) 0.