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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.

3 Answers 3

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Try setting $x = y^{-1}$. Then it should be easy. If you need more help, just ask.

  • 0
    @FUZ Yes, while that int$u$ition on growth is indeed correct, it does not amount to a rigorous proof. One simple way to pro$v$e it is via L'Hôpital's rule - as I mentioned.2011-05-21
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HINT $\ $ Take the $\rm\:log\:,\:$ put $\rm\: z = 1/x\:$ to reduce it to $\rm\: z^2/\:log(z)\to \infty\:$ as $\rm\: z\to\infty\:,\:$ by L'Hôpital.

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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 $a$.

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    Why is it undefined? Then we have $\lim_{x\to0^-}{\sqrt x}/{\exp x^{-2}} = \lim_{x\to0^-}i\sqrt{|x|}/{\exp x^{-2}}$ which is the same except for an additional $i$. (Am I wrong?)2011-05-21