**Problem Statement:** Use the definition of the derivative to show that if and if is differentiable at .

**Proof:** By definition

To evaluate this limit we will apply the Squeeze Theorem. Since for all it follows that as and that as . Either way,

So, by the Squeeze Theorem it follows that . Thus, is differentiable at since 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!

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.

Oh, you’re right! I think I’ve fixed it now. Thanks, Jason!