Show that Cauchy's function is infinitely differentiable Show that 
$$f(x)=
\begin{cases}
exp(-\frac{1}{x^2}), & \text{if $x\gt 0$} \\[2ex]
0, & \text{if $x\le 0$ }
\end{cases}$$
is infinitely differentiable. 
Clearly $f^{(n)}(x)=0$ for all $x\lt 0$ and $f_{-}^{(n)}(0)=0$. Also since the derivatives of $exp(-1/x^2)$ produce $exp(-1/x^2)$ and a polynomial in $1/x$, using the rules for differentiation we can evaluate $f^{(n)}$ if $x\gt 0$ with any $n$. It remains to show that $f_{+}^{(n)}(0)=0$. 
Now this final part is where I'm struggling. How can I show this part? I would greatly appreciate any help.

Hint: Use induction and l'Hopital.  Consider $\lim_{x\rightarrow 0^+}\frac{e^{-1/x^2}}{x^2}$.  Let $y=1/x$, so this becomes $\lim_{y\rightarrow\infty}\frac{e^{-y^2}}{y^{-2}}=\lim_{y\rightarrow\infty}\frac{y^2}{e^{y^2}}$.  After one step of l'Hopital, you have your answer.  This approach can be generalized for any $n\geq 0$.
