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?
 A: 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 $x>0$
A: As $x>0$
$$\lim_{n\rightarrow \infty }\dfrac{-\ln(n)}{n^x} = \lim_{n\rightarrow \infty } -\ln(n) e^{-x ln(n)} = 0$$
A: Let $y$ be such that $0 < y < x$. Then we can see that $n^{y} > 1$ for all integers $n > 1$. Hence we have the inequality $$\log n^{y} < n^{y} - 1 < n^{y}\tag{1}$$ or $$\log n < \frac{n^{y}}{y}$$ It thus follows that $$0 < \frac{\log n}{n^{x}} < \frac{1}{yn^{x - y}}\tag{2}$$ for integers $n > 1$. Taking limits as $n \to \infty$ and noting that $x - y > 0$ so that $n^{x - y} \to \infty$ and using Squeeze Theorem we get $$\lim_{n \to \infty}\frac{\log n}{n^{x}} = 0$$ and hence $$\lim_{n \to \infty}\frac{-\log n}{n^{x}} = 0$$ Note that the result holds for any $x \in (0, \infty)$ and not just for $x \in [2, \infty)$. You can see that there is no need to use L'Hospital's Rule (or any high level theorems) and the result is a direct consequence of the fundamental inequality of $\log$ function namely $$\log t < t - 1\tag{3}$$ for $t > 1$. This has been used in $(1)$ with $t = n^{y}$.
