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This question already has an answer here:

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!

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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
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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 $x^{99}$.

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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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