Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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

share|cite|improve this answer
No, It was enough. Thank you. – user6163 May 20 '11 at 23:06
@FUZ Could you please elaborate on how you propose to solve the problem after changing variables. There are many possibilities so it's hard to guess. – Bill Dubuque May 21 '11 at 0:08
@Bill: Quite similar to what you did. But I arrived at $\lim\limits_{y\to\infty}a\ln y - y^2=-\infty$... Maybe I calculated wrong. – FUZxxl May 21 '11 at 13:36
@FUZ No, that's the same as I hinted. After factoring one gets $\ \log\ y\ (a - y^2/{\log\ y})\ $ so it reduces to calculating $\ y^2/{\log\ y}\ $ by L'Hôpital, as I said. But it's not the change of variable $y = 1/x$ that is essential here. Rather it's the taking of logs, since that allows L'Hôpital to succeed. It trades off an exponential for a log. And generally logs work better for L'Hôpital since they go away upon differentiation, but exp's do not. – Bill Dubuque May 21 '11 at 15:29
@Bill Dubuque: I am not very good at limits (I am still in highschool, they will teach these things next year). To prove that, I just sad, that $a\ln y$ grows slower than $y^2$ and thus $a\ln y - y^2\to-\infty$ as $y\to\infty$. I guess that's not a good solution. – FUZxxl May 21 '11 at 15:33

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.

share|cite|improve this answer

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 $a$.

share|cite|improve this answer
No, the above limit is 0. Why do you think it doesn't exist? – Bill Dubuque May 21 '11 at 15:17
Because $\frac{\sqrt{x}}{e^{1/x^2}}$ is undefined for $x < 0$. – JavaMan May 21 '11 at 15:44
But that's a bit artifical since the limit will still be 0 if one allows complex values. – Bill Dubuque May 21 '11 at 16:23
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?) – FUZxxl May 21 '11 at 17:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.