TITLE: How to prove that $\lim\limits_{x \to 0 }\;x^{-a}e^{\frac{-1}{x^{2}}} =0$ for all a?
QUESTION [5 upvotes]: I'd like your help with this:
I tried using L'Hôpital's Rule and all kinds of arithmetic to prove that
$$\lim_{x \to 0 }\left(x^{-a}e^{\left(\frac{-1}{x^{2}}\right)}\right) = 0$$ for every $a$, and it didn't work.
($a=0$ is trivial)
Any hints?
Thank you.

REPLY [2 votes]: Actually, this result you hope for is false if $a$ is allowed to be an arbitrary real number (but it only fails for stupid reasons).  If $a = -\frac{1}{2}$, for example, then 
$$\lim_{x \to 0^{-}} \frac{\sqrt{x}}{e^{\frac{1}{x^2}}} $$
does not exist.
If one is restricted to integer values of $a$, then the result does as was previously discussed.  Also, the limit $\displaystyle \lim_{x \to 0} |x|^{-a} e^{- \frac{1}{x^2}}$ always exists for all $a$.