**Fact **If \lim_{n\to\infty} a_n = \infty, then \lim_{n\to \infty}(1 + 1/a_n)^{a_n} = e.

**Remark** Using Heine’s theorem, we can show that if \lim_{x\to \infty}f(x) = \infty, then \lim_{x\to \infty}(1 + 1/f(x))^{f(x)} = e.

**Corollary of fact **If \lim_{n\to\infty} |a_n| = \infty, then \lim_{n\to\infty}(1 + 1/a_n)^{a_n} = e.

**Proof of corollary **Define the sequence \{b_n\}_{n=1}^\infty by b_n = |a_n| if a_n > 0 and b_n = |a_n|-1 if a_n < 0. By the pizza rule, \lim_{n\to\infty}b_n = \infty. Moreover, if a_n > 0, then (1+1/a_n)^{a_n}=(1+1/|a_n|)^{|a_n|} = (1+1/b_n)^{b_n}; if a_n < 0, then (1+1/a_n)^{a_n} = (1-1/|a_n|)^{-|a_n|} = [1+1/(|a_n|-1)]^{|a_n|} = (1+1/b_n)^{b_n+1}. Therefore, we always have (1+1/b_n)^{b_n} \le (1+1/a_n)^{a_n} \le (1+1/b_n)^{b_n+1}. By the sandwich rule, \lim_{n\to\infty}(1 + 1/a_n)^{a_n} = e.

**Problem 1:** Suppose \lim_{x\to x_0} f(x) = 0 and f(x) \neq 0 for all x. Show that \lim_{x\to x_0}(1+f(x))^{1/f(x)}=e.

**Proof** By the Heine’s theorem, it suffices to show that for every sequence \{a_n\}_{n=1}^\infty such that \lim_{n\to\infty}a_n = x_0 (and a_n \neq x_0 for all n), \lim_{n\to \infty}(1+f(a_n))^{1/f(a_n)}=e. Given such sequence \{a_n\}_{n=1}^\infty. Since \lim_{x\to x_0} f(x) = 0, by the Heine’s theorem, \lim_{n\to\infty} f(a_n) = 0. Define b_n = 1/f(a_n). We know that \lim_{n\to\infty} |b_n| = \infty. By corollary of fact, \lim_{n\to \infty}(1+f(a_n))^{1/f(a_n)}=\lim_{n\to\infty}(1+1/b_n)^{b_n}=e.

**Remark **Problem 1 is still true if x_0 = \infty. The proof goes through line by line.

**Problem 2:** Find the limits (a) \lim_{x\to 0}(\cos x + \sin x^2)^{1/x^2}; (b) \lim_{x\to 0}(1+\sin x)^{-1/x}.

**Idea:** (a) Take f(x) = \cos x + \sin x^2 - 1 = -2\sin^2(x/2) + \sin x^2 and find \lim_{x\to 0} f(x)/x^2. (b) Take f(x) = \sin x and find \lim_{x\to 0} f(x)/x.

**Problem 3:** Suppose that f\colon \mathbb{R}\to \mathbb{R} is continuous and that for every x we have f(x) = f(x^2). show that f is constant.

**Proof **Take any x > 0. Define a sequence a_0 = x, a_{n+1} = \sqrt{a_n}. By mathematical induction, we know that a_n = x^{1/2^n} and f(a_n) = f(a_0) = f(x). Because \lim_{n\to\infty} a_n = 1 and f is continuous, f(x) = \lim_{n\to\infty}f(a_n) = f(1). For x<0, since x^2 > 0, we know that f(x) = f(x^2) = f(1). Finally, by continuity, f(0) = \lim_{x\to 0} f(x) = f(1).

**Problem 4:** Suppose f\colon [0,1]\to \mathbb{R} is continuous and f(0) = f(1). Show that there exist x, y\in [0,1] such that |x-y|=1/27 and f(x) = f(y).

**Proof **Define the function g\colon [0, 26/27]\to \mathbb{R} by g(x) = f(x) - f(x+1/27). Notice that g(0) + g(1/27) + g(2/27) + \dots + g(25/27) + g(26/27) = f(0) - f(1) = 0. Therefore we can find two distinct numbers a, b in 0, 1/27, 2/27, \dots, 25/27, 26/27 such that g(a) \le 0, g(b) \ge 0. By the intermediate value theorem, there is some x \in [a,b] such that g(x) = 0, and so f(x) = f(x + 1/27).

**Problem 5:** Suppose that f\colon \mathbb{R}\to \mathbb{R} is continuous. Show that if f(f(3)) = 3, then there is x_0 such that f(x_0) = x_0.

**Proof** If f(3) = 3, we are done already. Otherwise, let a, b be respectively the smaller and bigger numbers in 3, f(3). Notice that f(a) = b, f(b) = a because f(f(3)) = 3. Define g(x) = f(x) - x. Since g(a) = f(a) - a = b - a > 0, g(b) = f(b) - b = a - b < 0, by the intermediate value theorem, there exists a \le x_0 \le b such that g(x_0) = 0 that is f(x_0) = x_0.

**Problem 6:** Show that the function f\colon \mathbb{R}\to \mathbb{R} defined by f(x) = x^3 - 2x + 1 - 5\sin(x^8+x-9) is onto \mathbb{R}.

**Proof **Because x^3 - 2x + 1 - 5 \le f(x) \le x^3 -2x + 1 + 5, by the pizza rule \lim_{x\to -\infty}f(x) = -\infty and \lim_{x\to\infty}f(x) = \infty. Take any y. We can find a < 0 < b such that f(a) \le y and f(b) \ge y. By the intermediate value theorem, there exists a \le x \le b such that f(x) = y.

**Puzzle** Is the converse of the intermediate theorem true? We say a function f\colon \mathbb{R}\to \mathbb{R} has the “intermediate value property” if it satisfies the conclusion of the intermediate value theorem: for any two values a and b, and any y between f(a) and f(b), there is some c between a and b with f(c) = y. Take the function f\colon \mathbb{R}\to \mathbb{R}defined by f(x) = \sin(1/x) for x\neq 0 and f(0) = 0. One can show that the function has the intermediate value property, however it is not continuous at 0. This example is not satisfactory since f is continuous at all x but 0. Can you find a function f with the intermediate value property such that it is discontinuous at several x? Or maybe discontinuous at every x?