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I know this is a duplicate, but none of the questions I found explained the induction. I proved the $n$th derivative of the function:

$f(x)=e^{-1/x^2}$ if $x\neq0$ , $f(0)=0$ is:

$f^n(x)=r(x)e^{-1/x^2}$

$r(x)$ is a rational function. Now I want to prove $f^n(0)=0$ for all derivatives.

I assumed $f(0)=0, f'(0)=0,.......f^{n-1}(0)=0$

Then calculate using the definition of deriviative:

$ f^n(0) =\lim_{h \to 0} \frac{f^{n-1}(x+h)-f^{n-1}(x)}{h} = \lim_{h \to 0} \frac{f^{n-1}(h)-f^{n-1}(0)}{h} = \lim_{h \to 0}\frac{r(h)e^{-1/h^2}}{h} $

But I don't figure out how to show this equals to $0$

amit
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  • I read that before, but it didn't explain the limit calculation – amit Jan 08 '23 at 13:00
  • When calculating using the definition of derivative there will be some polynomial divided by $h$, how do u know that would not affect the limit? – amit Jan 08 '23 at 13:32
  • Formula of derivative of product of functions. – Mittens Jan 08 '23 at 13:37
  • I still don't understand that, can you write it down? – amit Jan 08 '23 at 14:32
  • once you have that for $x\neq0$, $f^{(n)}(x)=P_n(1/x)f(x)$ where $P_n$ is some polynomial, and $f^{(n)}(0)=0$, you can use the L'Hospital type argument to show that $\lim_{x\rightarrow0}\frac{f^{(n)}(x)-f^{(n)}(0)}{x}=\lim_{x\rightarrow\infty}\frac{P_n(1/x)f(x)}{x}=0$, or to simplify things, you can set $y=1/x$ to get $\lim_{|y|\rightarrow\infty}yP(y)e^{-y^2}$ recall that exponential growth is faster than polynomial growth. – Mittens Jan 08 '23 at 15:00
  • Got that except one thing, what if the rational function does not equal $P(1/x)$? For example, if the rational function has $x$ both in numerator and denominator, what polynomial fits that? – amit Jan 08 '23 at 15:29
  • The rational function that comes with $f^{(n)}$ is in fact of the form $P_n(1/x)$ where $P_n$ is a polynomial. Try by induction to see that is the case. – Mittens Jan 08 '23 at 17:01

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