Prove a function is not analytic but it is smooth

[Valhalla]

given

$$f(x)=\left\{\begin{array}{cc}e^{-x^{-2}},& \mbox{ if } x!=0 \\ 0, \mbox{ if } x=0 \end{array}\right$$

show that the function is C-infinity smooth but not analytic

What I have done so far (besides verify that it is continuous) is examine the first few derivatives I found they are in the form of a sum of a rational function of $$\frac{c}{x^n}$$ (where c is some number) times the exponential function. I now need to show that all these derivatives right hand and left hand limit equal 0. So I stated the nth derivative could be composed of the sum of some rational function times the exponential. Then the limit of this function is equal to the product of the limit of the two functions. I get zero times infinity. Is this a legal move? Can I say that zero times infinity is zero?

let $$p(x)=\frac{c}{x^n} \ \ q(x)=e^{-x^{-2}}$$

then $$\lim_{\substack{x\rightarrow0}} p(x)q(x) = \lim_{\substack{x\rightarrow0}} p(x) *\lim_{\substack{x\rightarrow0}} q(x)$$
I want to use that to show that the derivative will always be continuous. Therefore it will be C infinity smooth. Then I am going to do a power series expansion of the exponential and work on figuring out why it isn't analytic.

 

[StatusX]

Try the substitution y=1/x and change the limit to y-> infinity.

 

[Valhalla]

I tried the substitution

I get p(x)=c*y^n and q(x)= e^(-y^2)

which when I take the product of the limits as y goes to infinity I still have infinity times zero. Am I allowed to say that infinity times zero is zero?

 

[StatusX]

Well, obviously you need to take the limit. Exponentials grow faster than any polynomial.