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I need to prove that $\lim_{x\rightarrow \infty}\dfrac{x^2}{e^x}=0$.

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    $\begingroup$ Are you allowed to use l'Hôpital's theorem? $\endgroup$ – egreg Nov 13 '14 at 14:14
  • $\begingroup$ Do you know about l'Hopital's rule? $\endgroup$ – Ritz Nov 13 '14 at 14:14
  • $\begingroup$ Do you know that $\lim_{x\to\infty}\frac{x}{e^x}=0$? $\endgroup$ – Hagen von Eitzen Nov 13 '14 at 14:15
  • $\begingroup$ @Ritz Not yet... $\endgroup$ – qexi Nov 13 '14 at 14:24
  • $\begingroup$ @Hagen von Eitzen Yes $\endgroup$ – qexi Nov 13 '14 at 14:24
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We have the well-known inequality $e^t\ge 1+t$ for all $t\in\mathbb R$. Therefore $$e^x=(e^{x/3})^3\ge \left(1+\frac x3\right)^3=1+x+\frac13x^2+\frac1{27}x^3$$ for $x\ge 0$, whence $$0\le\frac{x^2}{e^x}\le \frac{x^2}{1+x+\frac13x^2+\frac1{27}x^3}\to 0.$$

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$$0 \leq \frac{x^2}{e^x}= \frac{x^2}{(e^{x/4})^4}\leq \left(\frac{x}{(1+x/4)^2}\right)^2.$$

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For $x>0$, we have $e^x \geq \dfrac{x^3}{3!}$ (Taylor expansion of $e^x$). Hence, we have $$\dfrac{x^2}{e^x} \leq \dfrac{6}x \to 0 \text{ as }x \to \infty$$

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  • $\begingroup$ Sorry, but I don't know much about Taylor expansion... yet $\endgroup$ – qexi Nov 13 '14 at 14:27

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