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?


Hint: 
You can prove by induction that there exists polynomials $P_n, Q_n$ such that
$$\frac{d^n}{d x^n}  e^{-1/x^2} = \frac{P_n(x)}{Q_n(x)} e^{-\frac{1}{x^2}}$$
You can also show by the change of variables $t=\frac{1}{x}$ that, for each $m$ you have 
$$\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^m}=0$$
From here you can deduce both claims.
