Prove that this function differentiable endles times

[transgalactic]

prove that this function differentiable endles times on x=0 ?? http://img502.imageshack.us/img502/6778/83126617mm0.th.gif

i was told
"once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable"
differentiable around zero part:
$$f'(0)=\lim_{x->0}\frac{e^\frac{-1}{x^2}-0}{x-0} \\ =\lim_{x->0}\frac x{2e^{1/x^2}}$$
i know that exponentials grow faster then polinomials but the power of the exp is 1/x^2
so the denominator goes to infinity but faster then the numenator so the expression
goes to 0.
power series around zero part:
the power series of e^x
$$g(x)=e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^3)\\$$
i put -1/x^2 instead of x
$$g(x)=e^x=1+\frac{-1}{x^2}+\frac{x^{4}}{2}+\frac{-x^6}{6}+O(-x^6)\\$$

what now??

 

[HallsofIvy]

The first derivative of [tex]e^{-1/x^2}[/itex] is, of course, $e^{-1/x^2}(2x^{-3})$. The second derivative, using the product rule is $e^{-1/x^2}(4x^{-6})+ e^{-1/x^2}(-6x^{-4})$.

Can you use "proof by induction" to show that every derivative is $e^{-1/x^2}$ times terms with negative powers of x? It doesn't really matter what those powers are because, as x goes to 0, the exponential will "dominate" and the derivative will go to 0 as x goes to 0 so we can take every derivative to be 0 at x= 0.

That means, by the way, that we could form its Taylor series at x= 0 (its MacLaurin series) as $0+ 0x+ 0x^2+ \cdot\cdot\cdot= 0$. But obviously f(x) is NOT identically 0. This is an example of a function that is infinitely differentiable so has Taylor series at 0 but is NOT equal to that series except at 0.

 

[transgalactic]

my power series is fine
i took the power series for e^t and substitutes t=-1/x^2

so i got a power series and differentiation of the first around 0

is it a solution

how to do it by induction?