How do you show that $e^{-1/x^2}$ is differentiable at $0$? Define the real valued function $f(x)$ by $f(x) = e^{-1/x^2}$ if $x \ne 0$ and $0$ if $x = 0$.

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*How do you show that $e^{-1/x^2}$ is differentiable at $0$?

*How do you show that this function is infinitely differentiable?


You want to compute $$\lim_{h \to 0}\frac{\exp(-1/h^2)}{h}$$ Show this is equivalent to computing 
$$
\lim_{k \to +\infty}k{\exp(-k^2)}
\qquad \text{and} \qquad
\lim_{k \to -\infty}k{\exp(-k^2)}.
$$
and show they are equal. Then things get easy. 
For your second question, see this.
