real analysis - How to evaluate the following limit

How can I determine the following limit

$$\lim_{n \to \infty} \frac{-\ln(n)}{n^x}$$ where $x \in [2, \infty)$

WolframAlpha tells me the limit is $0$, but I am not sure how to go about calculating it manually, I suspect L'Hospital plays a role?

Answer

From L'Hospital's Rule, we have

$$\lim_{n\to \infty}\frac{-\ln n}{n^x}=\lim_{n\to \infty}\frac{-1/n}{xn^{x-1}}=-\frac1x\lim_{n\to \infty}\frac{1}{n^x}=0$$

when $x>0$