I know that one must use the right- and left-hand limit definition of a derivative to calculate $f'(0)$ for the function
$$f(x) = \begin{cases} e^{-1/{x^2}} & \text{if } x≠0\\ 0 & \text{if } x = 0 \end{cases}$$
since we cannot be sure that $f(x)$ is differentiable at $x = 0$. However, when trying to evaluate this derivative limit expression, I encountered a problem:
$$f'(0) = \lim_{h \to 0^+} \frac{f(x+h) - f(h)}{h} = \lim_{h \to 0^+} \frac {e^{-1/h^2}-0}h = \lim_{h \to 0^+} \frac {e^{-1/h^2}}{h}$$
There does not seem to be a way to evaluate the limit $$\lim_{h \to 0^+} \frac {e^{-1/h^2}}{h}$$ without running into an indeterminate form. One could set up a table of values to calculate it and the left-hand version, which would reveal that they both approach $0$ (so $f'(0)$ is then $0$). However, I was curious if this is the only path to take? Is there a way to algebraically calculate this limit? I tried applying L'hospital's rule as well, but that also just ended up resulting in an indeterminate form.