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I was asked to prove that $\lim\limits_{x\to\infty}\frac{x^n}{a^x}=0$ when $n$ is some natural number and $a>1$. However, taking second and third derivatives according to L'Hôpital's rule didn't bring any fresh insights nor did it clarify anything. How can this be proven?

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  • $\begingroup$ You could take $n$-th derivatives. Or substitute $u = a^x$. $\endgroup$ – Daniel Fischer Dec 8 '14 at 16:53
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Here's a hint: after doing successive applications of L'Hospital's rule, what you get in the numerator is $n(n-1)\cdots(n-m+1)x^{n-m}$. What you get in the denominator is $(\ln a)^m a^m$. If you differentiate a polynomial of degree $n$ $n$ times, what do you get?

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  • $\begingroup$ infinity^0....? $\endgroup$ – Bak1139 Dec 8 '14 at 16:55
  • $\begingroup$ Not quite. What happens if you differentiate $x^4$ 4 times, for instance? $\endgroup$ – Cameron Williams Dec 8 '14 at 16:55
  • $\begingroup$ youll get the power 0, and since x goes to infinity you'll get infinity^0 $\endgroup$ – Bak1139 Dec 8 '14 at 16:56
  • $\begingroup$ You get a power of zero, which means it is a constant. (You do the limit after simplification, not before.) $\endgroup$ – Cameron Williams Dec 8 '14 at 16:57
  • $\begingroup$ you're right my bad.. $\endgroup$ – Bak1139 Dec 8 '14 at 16:58
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Hint: Use $a^x = e^{x\ln a} > \dfrac{x^{n+1}\left(\ln a\right)^{n+1}}{(n+1)!}$. Thus:

$0 < \dfrac{x^n}{a^x} < \dfrac{(n+1)!}{x\left(\ln a\right)^{n+1}}$. Now by Squeeze theorem we get the limit $0$.

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