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How to prove $\lim_{x\to\infty}\frac{x^{99}}{e^x}=0$?

I wasn't sure how to do this because both the top and the bottom limits independently turns out to be infinity!


marked as duplicate by Saad, Paramanand Singh, Hans Lundmark, Hanul Jeon, GNUSupporter 8964民主女神 地下教會 Apr 23 '18 at 12:00

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    $\begingroup$ do you know l'Hôpital's rule? $\endgroup$ – qbert Apr 23 '18 at 4:41
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    $\begingroup$ oh also, it's not. $\endgroup$ – qbert Apr 23 '18 at 4:44
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    $\begingroup$ @AndrewLi because 0 is suppose to be the right answer. Exponentials grow much faster than polynomials. $\endgroup$ – welshman500 Apr 23 '18 at 4:45
  • $\begingroup$ my apoligies it seems to be that i mis wrote infinity as 0 ill fix that $\endgroup$ – John Rawls Apr 23 '18 at 5:36

After successively applying L'Hopital's rule $100$ times, you get:

$$\lim_{x\to\infty} {x^{99} \over e^x} = \lim_{x\to\infty} {99! \over e^x} = 0$$

This is due to the fact that exponentials always grow faster than polynomials. $e^x$ will eventually overcome $x^{99}$.


For $x>0$ we have $e^x > \frac{x^{100}}{100!}$, hence $0 < {x^{99} \over e^x}< \frac{100 !}{x}$. This gives ${x^{99} \over e^x} \to 0$ as $x \to \infty$.


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