# Harrison Chapman

## Math 317: Homework 7

Due: Friday, April 5, 2019
1. (28.8) Let $f(x) = x^2$ when $x$ is rational and $f(x) = 0$ otherwise.

1. Show that $f$ is continuous at $0$.

2. Show that $f$ is discontinuous at all $x \ne 0$.

3. Prove that $f$ is differentiable at $0$. (It is insufficient to simply claim $f'(x) = 2x$)

1. Proof. Let $$(x_n)$$ be any sequence that converges to 0. Let $$\epsilon > 0$$ be arbitrary. Convergence of $$(x_n)$$ to 0 means that there exists $$N$$ so that for all natural numbers $$n > N$$, $$\vert x_n \vert < \sqrt \epsilon$$ Then $$\vert f(x_n) \vert \le x_n^2 < \epsilon$$. So $$(f(x_n))$$ converges to $$f(0) = 0$$, and so $$f$$ is continuous at 0.

2. Proof. First, suppose $c \ne 0$ is rational. Let $(x_n)$ be a sequence of irrationals converging to $c$, which must exist by density of irrationals (or, if you like, by example, we have $x_n = c + \pi/n$). Then $f(x_n) = 0$ for all $n$, so $(f(x_n))$ converges to 0, but $f(c) \ne 0$.

On the other hand, if $c \ne 0$ is irrational, let $(x_n)$ be a sequence of rationals converging to $c$ (if you like, $x_n$ is the sequence of decimal approximations to $c$), then $f(c) = 0$ but $f(x_n) = x_n^2$ so $(f(x_n))$ converges to $c^2 \ne 0$. So $f$ is not continuous at any $c \ne 0$.

3. Proof. The definition of the derivative of $f$ at 0 is $f'(0) = \lim_{x \to 0} \frac{f(x)}x$. Let $% $. Notice that for $% $, $x \ne 0$ we have that $% $, proving that $f'(0) = 0$.

2. (28.14) Suppose $f$ is differentiable at $a$.

1. Prove that $\lim_{h\to 0}\frac{f(a+h) - f(a)}{h} = f'(a)$.

2. Prove that $\lim_{h\to 0}\frac{f(a+h) - f(a-h)}{2h} = f'(a)$.

1. Proof. Let $\epsilon > 0$, so that there exists $\delta > 0$ for which $% $ implies that $% $. But with $h = x-a$, this means that $% $ implies that $% $. This holds for all $\epsilon > 0$, so $\lim_{h\to 0}\frac{f(a+h) - f(a)}{h} = f'(a)$.

2. Proof. Notice that (a) actually also proves that $\lim_{h\to 0}\frac{f(a) - f(a-h)}{h} = f'(a)$. We have that

3. (29.5) Let $f$ be a real-valued function which is differentiable on $\mathbb R$. Prove that if $% $ for all $x, y \in \mathbb R$, then $f$ is a constant function.

Proof. If $% $, then for all $x \ne y$, $% $. But taking limits as $x \to y$ of both sides, we realize that $f'(y) \le 0$. But this means that $f$ is constant.

4. (29.9) Show that $ex \le e^x$ for all $x \in \mathbb R$.

Proof. Notice that $ex \le e^x$ is true if and only if $0 \le e^x - ex$. Let $g(x) = e^x - ex$. Then $g(1) = 0$. Notice that $g'(x) = e^x - e$. When $% $, $% $ and when $x > 1$, $g'(x) > 0$.

Let $% $. Then $g(x) \ge 0$, as it must be decreasing always until $x = 1$, when it is its minimum value of $0$. On the other hand, when $x > 1$, the function is always increasing from its minimum value of $0$, so $g(x) \ge 0$ as well. So in all cases $g(x) \ge 0$.

5. Let $f$ be real-valued and differentiable on $\mathbb R$, and let $g(x) = f(x+1) - f(x)$. Suppose additionally that $\lim_{x\to \infty} f'(x) = 0$.

Prove that $\lim_{x\to\infty} g(x) = 0$.

Hint. Use the Mean Value Theorem.

Proof. Let $\epsilon > 0$, so there exists $M$ so that for all $c > M$ we have that $% $. Let $x > M$, and consider the interval $[x, x+1]$. By the Mean Value Theorem, there exists an input $c$ between $x$ and $x+1$ for which $f'(c) = (f(x+1) - f(x))/1 = g(x)$. But then $% $ as $c > x > M$. This holds for all $x > M$ and all $\epsilon > 0$, so $\lim_{x \to \infty} g(x) = 0$.