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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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    $\begingroup$ Minor note: The limit does not go to $0$. It is $0$. The function goes to $0$ as $x$ goes to $0$. $\endgroup$ – André Nicolas Nov 25 '12 at 23:38
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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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  • $\begingroup$ If $\lim_{y \to \infty}k! \frac{y^k}{y^{2k}}=0$, and $\lim_{y\to infty}\frac{y^k}{e^{y^2}}<\lim_{y \to \infty}k! \frac{y^k}{y^{2k}}$, why is $\lim_{y\to \infty}\frac{y^k}{e^{y^2}}=0$? $\endgroup$ – Jess Nov 26 '12 at 0:30
  • $\begingroup$ Would I use the squeeze theorem somehow? $\endgroup$ – Jess Nov 26 '12 at 0:32
  • $\begingroup$ @Jess I have updated the post. Hopefully it clarifies your question. $\endgroup$ – user17762 Nov 26 '12 at 0:32
  • $\begingroup$ Ahh thank you so much! It makes sense now. $\endgroup$ – Jess Nov 26 '12 at 0:35

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