Proving $\lim_{x \to 0}e^{-1/x^2} \cdot \frac{p(x)}{q(x)} = 0$

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.

Thanks!

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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

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 $0$.