How can I show that $\lim_{n\rightarrow \infty} ((-n^x)/e^n)) = 0$

Question

How can I show that $$\lim_{n\rightarrow \infty} ((-n^x)/e^n)) = 0$$ , where n is a natural number and $x>0$?

I wanted to use l'hôspital, but it doesn't work or rather not defined, if n is a natural number.

Answer 1

$$n^x=e^{x\ln (n)},$$

and thus $$\frac{n^x}{e^n}=e^{x(\ln(n)-n)}.$$ Using the fact that $$\ln(n)-n\underset{n\to \infty }{\longrightarrow }-\infty $$ gives the wished result.