# Calculate derivative of multicase function involving exponentials as $x \to 0$ by definition

While this seemed (and probably does seem to some of you) like a simple question a first it stumbled me a bit.

We were asked to calculate the derivative of:

$$f(x) = \left\{ \begin{array}{lr} e^{\frac{-1}{x^2}} & x \ne0\\ 0 & x =0 \end{array} \right.$$

at $x=0$ by definition, meaning though I can solve it easily with L'Hopital or maybe a taylor expansion of $e$ I cannot use it, only definition of derivatives and limits.

So I started by the defintion of the derivative at $x=0$ to get: $\lim_{x \to 0} \frac{e^{\frac{-1}{x^2}}-0}{x-0} = \lim_{x \to 0} \frac{e^{\frac{-1}{x^2}}}{x}$

Thinking I'd simplify it a bit I got to $\frac{1}{x \cdot e^{\frac{1}{x^2}}}$

Knowing limits around $0$ could be tricky I thought to take the onesided limits of $0^+, 0^-$

Thinking this would be a simple job I set out the prove the limit was $0$ (purely by intuition).Looking for example at the limit as $x \to 0^+$ I started by letting there be an $\epsilon$ I want to prove that $\exists \delta : x<\delta -> |f(x)|<\epsilon$

This is where I got stuck since what I'm left with is (after some algebra) $x \cdot e^{\frac{1}{x^2}} > \epsilon$

Now I'm sure there some algerba involved here to "extract" our variable to allow me to control $x$ by $\delta$ but I can't see it. Also I'm sure there must be some other way to solve it (by definition of course) since I also tried to substitute $u=\frac{1}{x^2}$ to reach a similar limit only to get stuck on the limit of $\frac{1}{\frac{e^u}{\sqrt{u}}}$ as $u \to \infty$ which I also believe to $0$ but also I don't know how to prove.

Any guidance would be appreciated! Thank you!

Note that, for every $x\ne0$, $$\left|\frac{\mathrm e^{-1/x^2}}x\right|=\exp(-a(x^2)),\qquad a(u)=\frac1{u}+\frac12\log u,$$ hence, if ever the limit of $a(u)$ when $u\to0$, $u\gt0$, is $+\infty$, then the limit of the ratio defining the derivative, is $0$. Note that $$a(u)=\frac{2+u\log u}{2u},$$ whose numerator goes to $2$ and the denominator goes to $0$ while being positive, when $u\to0$, $u\gt0$. Can you finish?
• This actually makes it a lot clearer, thought I wouldn't have thought of that myself. Hmm regarding the end, I am not sure what do you mean by finish since this seems finished to me (maybe that's what you meant). Only thing I'm missing is the limit of $u \cdots log(u)$ in your proof which seems to be zero, not sure how to prove that without L'Hopital but I guess that's a topic for another question :) Thank you! – byentropy Aug 25 '14 at 13:07