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 need some help in showing that $\;\displaystyle \lim_{x\to 0}\frac{e^{-1/x^2}}{x^k}=0$.

I tried to take the log of the limit and then use L'Hospital's rule but got stuck.

How should I approach this problem? I'd appreciate any guidance! Thank you in advance!

share|cite|improve this question
Minor note: The limit does not go to $0$. It is $0$. The function goes to $0$ as $x$ goes to $0$. – André Nicolas Nov 25 '12 at 23:38
up vote 2 down vote accepted

Set $1/x = y$. We then get that $$\lim_{x \to 0^+} \dfrac{e^{-1/x^2}}{x^k} = \lim_{y \to \infty} y^k e^{-y^2} = \lim_{y \to \infty} \dfrac{y^k}{e^{y^2}} = 0$$ The last limit follows immediately if $k<0$. If $k \geq 0$, then $$e^{y^2} = 1 + \dfrac{y^2}{1!} + \dfrac{y^4}{2!} + \dfrac{y^6}{3!} + \cdots + \dfrac{y^{2k}}{k!} + \cdots \geq \dfrac{y^{2k}}{k!}$$ Hence, $$0 \leq \lim_{y \to \infty} \dfrac{y^k}{e^{y^2}} \leq \lim_{y \to \infty} k!\dfrac{y^k}{y^{2k}} = \lim_{y \to \infty} \dfrac{k!}{y^{k}} = 0$$ Hence, we have that $$0 \leq \lim_{y \to \infty} \dfrac{y^k}{e^{y^2}} \leq 0 \implies \lim_{y \to \infty} \dfrac{y^k}{e^{y^2}} = 0$$ The same argument works for $x \to 0^-$. Hence, $$\lim_{x \to 0} \dfrac{e^{-1/x^2}}{x^k} = 0$$

share|cite|improve this answer
If $\lim_{y \to \infty}k! \frac{y^k}{y^{2k}}=0$, and $\lim_{y\to infty}\frac{y^k}{e^{y^2}}<\lim_{y \to \infty}k! \frac{y^k}{y^{2k}}$, why is $\lim_{y\to \infty}\frac{y^k}{e^{y^2}}=0$? – Jess Nov 26 '12 at 0:30
Would I use the squeeze theorem somehow? – Jess Nov 26 '12 at 0:32
@Jess I have updated the post. Hopefully it clarifies your question. – user17762 Nov 26 '12 at 0:32
Ahh thank you so much! It makes sense now. – Jess Nov 26 '12 at 0:35

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.