f(x) = p(1/x) * e^(-1/x^2) , x/=0
0 , x=0
p(t) is a nonzero polynomial in t
prove this is differentiable everywhere, i cant prove this limit::
limit( (p(1/h)*e^(-1/h^2))/h, h goes to 0)
Do you mean $\displaystyle \lim_{h\to{0}}p\bigg(\frac{1}{h}\bigg)\cdot\frac{e ^{\frac{-1}{h^2}}}{h}$?
Alternatively you could consider the fact that since e^x is analytic everywhere it is thus differentiable everywhere and work at it from the power series point of view
$\displaystyle \lim_{h\to{0}}p \left ( \frac{1}{h} \right ) \cdot \frac{e^{\frac{-1}{h^2}}}{h}$
You know that p is a non-zero polynomial, so this limit is of the form
$\displaystyle \sum_n a_n \lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n}$
So we need to find
$\displaystyle \lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n}$
for some arbitrary positive integer n.
Let
$\displaystyle x = \frac{1}{h}$
Then
$\displaystyle h^n = \frac{1}{x^n}$
So
$\displaystyle \lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n} = \lim_{x \to \infty} x^n e^{-x^2}$
$\displaystyle = \lim_{x \to \infty} \frac{x^n}{e^{x^2}}$
which is in a slightly more "doable" form.
-Dan
Rewriting this as $\displaystyle \bigg(\frac{1}{h^{n+1}}\bigg)\cdot{e^{\frac{-1}{h^2}}}$
we get $\displaystyle \lim_{h\to{0}}\frac{1-\frac{1}{h^2}+\frac{(\frac{1}{h^2})}{2!}-\frac{(\frac{1}{h^2})^3}{3!}-...\pm\frac{(\frac{1}{h^2})^{n-1}}{(n-1)!}\mp\frac{(\frac{1}{h^2})^n}{n!}}{h^{n+1}}$
can you go from there
huh ?
Ok, and where did the limit of h go ??we get $\displaystyle \lim_{n\to{\infty}}\frac{1-\frac{1}{h^2}+\frac{(\frac{1}{h^2})}{2!}-\frac{(\frac{1}{h^2})^3}{3!}-...}{h^{n+1}}$
can you go from there
Plus, you, can you go from there ?
*After l'Hôpital's rule, power series, here comes the newly trend : MacLaurin series !*
(am kidding)
No I have always liked power series more it was just I was not adept enough at LaTeX yet to use them...now I am...and yes you could go from there...technically the answer would be ∞ due to an infinite amount of ∞ + an infinite amount of 0's...but as you pointed out I read the question wrong so that limit does not apply