# L'Hôpital's rule in proofs

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?

• You could take $n$-th derivatives. Or substitute $u = a^x$. – Daniel Fischer Dec 8 '14 at 16:53

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?
• Not quite. What happens if you differentiate $x^4$ 4 times, for instance? – Cameron Williams Dec 8 '14 at 16:55
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 $0$.