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

How do I explore for which values of the parameter $a$

$$ \lim_{x\to 0} \frac{1}{x^a} e^{-1/x^2} = 0? $$ For $a=0$ it is true, but I don't know what other values.

share|cite|improve this question

Hint: Put $y=1/x$ and note that $ y \to \infty $.

$$ \lim_{y\to \infty} y^a e^{-y^2} = \dots. $$

share|cite|improve this answer

It's true for all $a's$ (note that for $a\leqslant 0$ it's trivial) and you can prove it easily using L'Hopital's rule or Taylor's series for $\exp(-x^2)$.

share|cite|improve this answer

Hint: You may rewrite the equation

$$ \frac{1}{x^a} e^{\frac{-1}{x^2}} = \frac{1}{x^a e^{\frac{1}{x^2}}} = \frac{1}{e^{a \ln(x)+ \frac{1}{x^2}}}$$

share|cite|improve this answer
but now we need to calculate $\lim _{x \to 0}a\ln x +\frac{1}{x^2}$. I don't think it's a simpler limit to figure out. – Amihai Zivan Jan 11 '13 at 12:16

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.