## Fall 2003 #1

Problem Statement: Use the definition of the derivative to show that $f(x)=x^2\sin(\dfrac{1}{x})+5$ if $x\neq 0$ and $f(x)=5$ if $x=0$ is differentiable at $x=0$

Proof: By definition

$f'(0)=\lim_{h\rightarrow 0}\dfrac{f(0+h)-f(0)}{h}$

$=\lim_{h\rightarrow 0}\dfrac{h^2\sin(\dfrac{1}{h})+5-5}{h}$

$=\lim_{h\rightarrow 0}\dfrac{h^2\sin(\dfrac{1}{h})}{h}$

$=\lim_{h\rightarrow 0}h\sin(\dfrac{1}{h})$

To evaluate this limit we will apply the Squeeze Theorem. Since $-1\leq\sin(\dfrac{1}{h})\leq 1$ for all $h\neq 0$ it follows that $-h\leq h\sin(\dfrac{1}{h})\leq h$ as $h\rightarrow 0^+$ and that $h\leq h\sin(\dfrac{1}{h})\leq -h$ as $h\rightarrow 0^-$. Either way,

$\lim_{h\rightarrow 0}-h=0=\lim_{h\rightarrow 0}h$

So, by the Squeeze Theorem it follows that $\lim_{h\rightarrow 0}h\sin(\dfrac{1}{h})=0$. Thus, $f(x)$ is differentiable at $x=0$ since $f'(0)$ exists and is finite.

$\Box$

Reflection: This one was very straight forward. When I first did the problem I didn’t go through the Squeeze Theorem argument though. After re-reading the proof and trying to defend it to Jeremy I realized that the Squeeze Theorem argument is kind of necessary. I told him, “well, on the qual I would write that out” and he reminded me of what my high school volleyball coach constantly told our team, “you practice how you play.” So, in the future I will be making sure to write out every single detail every time I take a practice exam.

On another note, I’ve been timing these exams and I am having enough time, yay!