# Recitation 6

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 $x$?