Computing $\lim_{x \to \infty} \frac{|x^n|}{e^x}$ How to prove that $e^x$ goes faster to infinity than any polynomial of $x$ without using the Taylor expansion of $e^x$ or L'hopital rule? in other words, the proof that:
$$\lim_{x \to \infty}  |x^n|e^{-x}=0$$
I tried to bound the expression from above by a function greater than $|x^n|$ for $x$ greater than some $\delta$ to apply squeeze theorem. I tried proving the limit directly, but both times I could find no excuse for the existence of such $\delta$ without using the known Taylor expansion.

You can use the root test for series: $$ \sum_{k=1}^{\infty} \frac{k^n}{e^{k}} $$ and check that: $$\sqrt[k]{\frac{k^n}{e^k}} = \frac{\sqrt[k]{k}^n}{e} \stackrel{k\to\infty}{\longrightarrow} \frac{1}{e} < 1$$
so the series is convergent and (using necessary condition):
$$\lim_{k\to\infty} k^n e^{-k} = 0 $$
