$\underset{x\to \mathrm{\infty }}{lim} \frac{x^n}{e^x}=0,(n$ integer), for

If $n$ is a negative integer, then $n=-m,$ where $m\in N.$

$\therefore \underset{x\to \mathrm{\infty }}{lim} \frac{x^n}{e^x}=\underset{x\to \mathrm{\infty }}{lim} \frac{x^{-m}}{e^x}=\underset{x\to \mathrm{\infty }}{lim} \frac{1}{x^m{e}^x}=0$

If $n=0,$ then

If $n\in N,$ then,

$\underset{x\to \mathrm{\infty }}{lim} \frac{x^n}{e^x}=\underset{x\to \mathrm{\infty }}{lim} \frac{n!}{e^x}=0$ [By L' Hospital's Rule]

Hence, $\underset{x\to \mathrm{\infty }}{lim} \frac{x^n}{e^x}=0$ for all values of $n$.