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.


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!

This entry was posted in Analysis, Differentiable, Squeeze Theorem. Bookmark the permalink.

2 Responses to Fall 2003 #1

  1. Jason says:

    Couple corrections: In the first line of the limit definition of the derivative, you need to replace the x with zero. Also, when you note the inequality and multiply both sides by h, you can’t do that because h could be negative (since h goes to zero). You need to break it up into two cases (h approaches 0 from the left and then h approaches 0 from the right). Overall, the solution is on the right track.

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