Looking for help with a proof that n-th derivative of $e^\frac{-1}{x^2} = 0$ for $x=0$.

Given the function

$$f(x) = \left\{\begin{array}{cc} e^{- \frac{1}{x^2}} & x \neq 0 \\ 0 & x = 0 \end{array}\right.$$

show that $\forall_{n\in \Bbb N} f^{(n)}(0) = 0$.

So I have to show that nth derivative is always equal to zero $0$. Now I guess that it is about finding some dependencies between the previous and next differential but I have yet to notice one. Could you be so kind to help me with that?

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Hint: show that all derivatives of $f$ are of the form $R(x)e^{-1/x^2}$ for some rational function $R$. –  Javier Badia Mar 16 '13 at 18:16
Just by using induction? –  darenn Mar 16 '13 at 18:28
BTW 1, 3, 5, ... deriviates are odd continuous functions => equal to zero in 0. You can probably get an easy proof for even deriviates using this. –  Harold Mar 16 '13 at 18:59

Suppose that $R(x)$ is a rational function of $x$. Then \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\left(R(x)\,e^{-1/x^2}\right) &=\left(R'(x)+\frac{2R(x)}{x^3}\right)e^{-1/x^2} \end{align} where $R'(x)+\frac{2R(x)}{x^3}$ is another rational function of $x$.

Inductively, we have that there is some rational $R(x)$, $$\frac{\mathrm{d}^k}{\mathrm{d}x^k}e^{-1/x^2}=R(x)\,e^{-1/x^2}\tag{1}$$

Since $R(1/x)$ is also a rational function of $x$, there is an $n$ so that $$\lim_{x\to\infty}\frac{R(1/x)}{x^{2n}}=0$$ Furthermore, \begin{align} \lim_{x\to+\infty}x^ne^{-x} &=\lim_{x\to+\infty}\frac{x^n}{e^x}\\ &=\lim_{x\to+\infty}\frac{n!}{e^x}&&\text{applying L'Hospital n times}\\[5pt] &=0 \end{align} Now we need to show that \begin{align} \lim_{x\to0}R(x)\,e^{-1/x^2} &=\lim_{u\to\infty}R(1/u)\,e^{-u^2}\\ &=\lim_{u\to\infty}\frac{R(1/u)}{u^{2n}}\,\lim_{u\to\infty}u^{2n}e^{-u^2}\\ &=\lim_{u\to\infty}\frac{R(1/u)}{u^{2n}}\,\lim_{v\to+\infty}v^ne^{-v}\\[7pt] &=0\cdot0\tag{2} \end{align}

Combining $(1)$ and $(2)$ yields $$\lim_{x\to0}\frac{\mathrm{d}^k}{\mathrm{d}x^k}e^{-1/x^2}=0\tag{3}$$

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$$f'(0):=\lim_{x\to 0}\frac{e^{-\frac{1}{x^2}}}{x}=\lim_{x\to 0}\frac{\frac{1}{x}}{e^{\frac{1}{x^2}}}\stackrel{\text{l'Hosp.}}=0$$
$$f''(0):=\lim_{x\to 0}\frac{\frac{2}{x^3}e^{-\frac{1}{x^2}}}{x}=\lim_{x\to 0}\frac{\frac{2}{x^4}}{e^\frac{1}{x^2}}\stackrel{\text{l'Hosp.}\times 2}=0$$