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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$\begingroup$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 $0$.
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