calculus - How to prove $lim_{xtoinfty}frac{x^{99}}{e^x}=infty$?

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!

Answer

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 $x^{99}$.