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

Here's a homework question I'm struggling with:

Let $p(x),q(x)$ two polynomials (such that $q(x) \ne 0$). Prove that

$$\lim_{x \to 0}e^{-1/x^2} \cdot \dfrac{p(x)}{q(x)} = 0$$ Hint: it is enough to prove so for a one sided limit - explain why!

So, let $f(x)=e^{-x^2}$. Its easy to show that $\lim_{x \to \infty}e^{-x^2} = 0$ and I know that if $f(x)$ is positive and the limit exist then $\lim_{x \to \infty}f(x) = \lim_{x \to 0^+}f(1/x)$ and so $\lim_{x \to 0^+}e^{-1/x^2} = 0$. We know that $\dfrac{p(x)}{q(x)}$ is continuous at $x=0$ (since $q(x) \ne 0$) and so $\lim_{x \to 0^+}e^{-1/x^2} \cdot \dfrac{p(x)}{q(x)} = 0$

Assuming I am right so far, I can't answer the hint - why is proving the one sided limit enough?

And if I made a mistake I'd be happy to know where.


share|cite|improve this question
Does $e^{-1/x^2}$ care whether $x$ approaches $0$ from the right or the left? – Brian M. Scott May 28 '12 at 9:27
I need to prove the two sided limit, so no – yotamoo May 28 '12 at 9:34
True, but I’m not sure that you’ve correctly identified the reason: the point is that squaring $x$ means that $e^{-1/x^2}$ behaves the same no matter on which side $x\to 0$. And since $q(x)\ne 0$, the rational function approaches $p(0)/q(0)$ no matter how $x\to 0$. Thus, it suffices to check the limit from one side. – Brian M. Scott May 28 '12 at 9:39
Oh well - that's what I've missed. I thought you were asking if the question says anything about where x is approaching from. Thanks! – yotamoo May 28 '12 at 9:47

As Brian has pointed out in the comments:

Because the $x$ in $f(x) = e^{-1/x^2}$ is squared, $f(x)$ behaves the same regardless of the direction in which it approaches $0$.

share|cite|improve this answer

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.