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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Hanul Jeon
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John Rawls
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6do you know l'Hôpital's rule? – operatorerror Apr 23 '18 at 04:41
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8oh also, it's not. – operatorerror Apr 23 '18 at 04:44
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1@AndrewLi because 0 is suppose to be the right answer. Exponentials grow much faster than polynomials. – welshman500 Apr 23 '18 at 04:45
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1my apoligies it seems to be that i mis wrote infinity as 0 ill fix that – John Rawls Apr 23 '18 at 05:36
2 Answers
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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 $x \to \infty$.
Fred
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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 $x^{99}$.
Andrew Li
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