Is $\exp(-1/z^2)$ differentiable at $0$? Let $f: \mathbb{C} \rightarrow \mathbb{C}: \begin{cases} \exp(-1/z^2) & z \neq 0 \\ 0 & z=0 \end{cases}$ be a function. Is $f$ differentiable in $0$?
Suppose $f$ is differentiable in $a$, then $\lim_{z \rightarrow 0} \frac{f(z)-f(0)}{z-0}=\lim_{z \rightarrow 0} \frac{\exp(-1/z^2)}{z}$ has to exist in $\mathbb{C}$. I'm not sure how to evaluate this limit. Can I use L'Hopital's rule here or does that only work for real functions?

It is infinitely differentiable($\mathcal{C}^\infty$), but not analytic($\mathcal{C}^\omega$).
The first bit is easy to prove by induction, the second bit is easy to prove by looking at that Taylor series at $0$(which is the zero polynomial), which clearly does not converge to the function in any neighbourhood of $0$.
