> Ok, i need some advice on this one:
> if
> f(x)=
> {0, if x=0
> p(x)/q(x) *(e^{-1/(x^2)} if x neq 0
> where p and q are poly's, defined off the nonzero roots of q. Prove lim
> x->0 f(x)=0 and f(x) is infinetely differentiable.
> My solution starts like this: lim p(x)/q(x) *(e^{-1/(x^2)}=lim p(x)/q(x) *
> lim e^{-1/(x^2)} if both limits exist and are finite. By L'hopitals rule
> lim p(x)/q(x)=lim p'(x)/q'(x)... and so on. If deg(p(x))>deg(q(x)) then
> this chain equals lim p^(n)(x)/1 which is just some constant. and similiary
> if deg(p(x))<deg(q(x)). Thus since lim e^{-1/(x^2)=0 we have it.
> But for showing its C^inf im not sure what to do...
Hint 1: write this as g(x) e^(-1/x^2) x^m where m is an integer (may be Hint 2: express f' in a form similar to that of f.