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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!

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    Minor note: The limit does not **go** to $0$. It **is** $0$. The *function* goes to $0$ as $x$ goes to $0$.2012-11-25

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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, $\lim_{x \to 0} \dfrac{e^{-1/x^2}}{x^k} = 0$

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    Ahh thank you so much! It makes sense now.2012-11-26