# Recitation 7

Using the chain rule is like peeling an onion. You have to deal with every layer at a time and if it’s too big you’ll start crying. – A calculus professor

Problem 1 Suppose $f\colon [a,b] \to \mathbb{R}$ is a function that is not bounded from above. Prove that there exists $x_{0} \in [a,b]$ such that $f$ is not bounded from above in any neighborhood of $x_0$.

Proof Because $f$ is not bounded from above, for every $n$, there is $x_n \in [a,b]$ such that $f(x_n) > n$. Since the sequence $\{x_n\}_{n=1}^{\infty}$ is bounded, by the Bolzano–Weirstrass theorem, $\{x_n\}_{n=1}^{\infty}$ has a partial limit, say $x_0$. Hence in any neighborhood of $x_0$ there are infinitely many elements of the sequence, say $\{x_{n_k}\}_{k=1}^\infty$. Because $f(x_{n_k}) > n_k$, $f$ is not bounded from above in that neighborhood.

Problem 2 Let $\{a_{n}\}_{n=1}^{\infty}$ be a sequence that is not bounded from above. Prove that it has a subsequence $\{a_{n_{k}}\}_{k=1}^{\infty}$ such that $\lim_{k \to\infty}a_{n_{k}}=\infty$.

Idea Because $\{a_{n}\}_{n=1}^{\infty}$ is not bounded from above, for every $k$ there is $n_k$ such that $a_{n_k} > k$. The pizza rule applies and $\lim_{k \to\infty}a_{n_{k}}=\infty$. This construction may suffer from the problem that $\{n_k\}_{k=1}^\infty$ is not increasing.

Proof We inductively define an increasing sequence of indices $\{n_k\}_{k=1}^\infty$. As $\{a_{n}\}_{n=1}^{\infty}$ is not bounded from above, there is $n_1$ such that $a_{n_1} > 1$. As $\{a_{n}\}_{n=n_1+1}^{\infty}$ is not bounded from above, there is $n_2 > n_1$ such that $a_{n_2} > 2$. As $\{a_{n}\}_{n=n_2+1}^{\infty}$ is not bounded from above, there is $n_3 > n_2$ such that $a_{n_3} > 3$. And so on. In general, suppose $n_{k-1}$ is already defined. Since $\{a_{n}\}_{n=n_{k-1}+1}^{\infty}$ is not bounded from above, there is $n_k > n_{k-1}$ such that $a_{n_k} > k$. Therefore we obtain an increasing sequence of indices $\{n_k\}_{k=1}^\infty$ such that $a_{n_k} > k$.

Problem 3 Find an example of a one-to-one and onto continuous function $f$ such that $f^{-1}$ is not continuous. (Hint: Find such a function whose domain is $[0,1] \cup (2,3]$.)

Proof sketch Define $f\colon [0,1]\cup(2,3] \to [0,2]$ by $f(x) = x$ for $x\in[0,1]$ and $f(x) = x-1$ for $x\in(2,3]$. The inverse function, say $g(x) = f^{-1}(x)$, is $g(x) = x$ for $x\in[0,1]$ and $g(x) = x+1$ for $x\in(1,2]$, and so $g$ is not continuous at 1.

Problem 4 Suppose $f\colon [a,b] \to \mathbb{R}$ is a continuous function that is one-to-one. Prove that either $f$ is monotone increasing, or $f$ is monotone decreasing.

Proof sketch Assume, for the sake of contradiction, that $f$ is neither monotone increasing nor monotone decreasing. There is $u, v, w\in [a,b]$ such that $u < v < w$ and $f(u) < f(v) > f(w)$ or $f(u) > f(v) < f(w)$. Case 1: Suppose $f(u) < f(v) > f(w)$. We can find such that $f(u) < y < f(v) > y > f(w)$ (why?). By the intermediate value theorem, there is $u < u' < v$ and $v < w' < w$ such that $f(u') = y = f(w')$ contradicting that $f$ is one-to-one. Case 2: Suppose $f(u) > f(v) < f(w)$. This case can be dealt similarly.

Problem 5 Suppose $f(x)>0$ and assume that both $f(x)$ and $g(x)$ are differentiable functions. Find the derivative of $f(x)^{g(x)}$.

Proof Because $f(x)^{g(x)} = e^{g(x)\ln f(x)}$, by the chain rule, $(f(x)^{g(x)})' = e^{g(x)\ln f(x)}(g'(x)\ln f(x) + g(x)f'(x)/f(x)) = f(x)^{g(x)}(g'(x)\ln f(x) + g(x)f'(x)/f(x))$.

Fact The function $f\colon [-\pi/2, \pi/2] \to [-1,1]$ defined by $f(x) = \sin x$ is one-to-one and onto. Its inverse function $g\colon [-1,1]\to [-\pi/2, \pi/2]$ is denoted by $g(x) = \arcsin x$. It is one of the inverse trigonometric functions.

Problem 6 Find the derivatives of the following functions: (a) $f(x) = \sqrt{\sin(x^2+x+1)}$; (b) $f(x) = 2^{x^2+\sin(x^2)}$; (c) $f(x) = \arcsin x$.

Solution (a) $\frac{1}{2}(\sin(x^2+x+1))^{-1/2}\cos(x^2+x+1)(2x+1)$; (b) $2^{x^2+\sin(x^2)}\ln 2 (2x+\cos(x^2)2x)$; (c) $y = \arcsin x$ implies that $\sin y = x$, and so $y' \cos y = 1$. Because $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1-x^2}$, $y' = 1/\cos y = 1/\sqrt{1 - x^2}$.

Problem 7 Suppose $f\colon \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x\sin (1/x)$ when $x\neq 0$. (a) Define $f(0)$ so that $f$ is continuous at 0; (b) Show that $f$ is not differentiable at 0.

Proof (a) $f(0) = \lim_{x\to 0}f(x) = \lim_{x\to 0}x\sin (1/x) = 0$; (b) Notice that $[f(h) - f(0)] / h = h\sin (1/h) / h = \sin(1/h)$ and $\lim_{h\to 0}\sin(1/h)$ does not exist.

Problem 8 Suppose $f\colon \mathbb{R} \to \mathbb{R}$ define by $f(x) = x^2\sin (1/x)$ when $x\neq 0$ and $f(0) = 0$. (a) Show that $f$ is differentiable at 0. (b) Show that $f'(x)$ is not continuous at 0.

Proof (a) $f'(0) = \lim_{h\to 0} h^2\sin(1/h) / h = \lim_{h\to 0} h\sin(1/h) = 0$. (b) When $x\neq 0$, $f'(x) = 2x\sin(1/x) + x^2\cos(1/x)(-1/x^2) = 2x\sin (1/x) - \cos(1/x)$. Now $\lim_{x\to 0}f'(x) = \lim_{x\to 0}2x\sin (1/x) - \cos(1/x)$ does not exist, and so $f'(x)$ is not continuous at 0.

Problem 9 Find the following limits: (a) $\lim_{x \to 0}\frac{e^x-1}{x}$; (b) $\lim_{x \to 1}\frac{\sqrt{x^2+3}-2}{x^2-1}$.

Solution (a) The limit is exactly the definition of $(e^x)'(0) = 1$; (b) 1/4.