How to find $\lim\limits_{n\rightarrow \infty}\frac{(\log n)^p}{n}$ How to solve $$\lim_{n\rightarrow \infty}\frac{(\log n)^p}{n}$$

If $p\leq 0$,then the result is $0$, which is trivial. We need consider the case when $p>0$ only. Let $ \alpha=\frac1 p>0$, then 
\begin{align*} \lim_{n \to \infty}\frac{\ln n}{n^{\alpha}}&=\lim_{n \to \infty}\frac{\ln(n+1)-\ln n}{(n+1)^{\alpha}-n^{\alpha}}&\text{Stolz}\\ &=\lim_{n \to \infty}\frac{\ln\left(1+\frac{1}{n}\right)}{n^{\alpha}\left[\left(1+\frac{1}{n}\right)^{\alpha}-1\right]}&\ln(1+x)\sim x,(1+x)^{\alpha}\sim \alpha x(x \to 0)\\ &=\lim_{n \to \infty}\frac{\frac{1}{n}}{n^{\alpha}\cdot\frac{\alpha}{n}}\\ &=\lim_{n \to \infty}\frac{1}{\alpha n^{\alpha}}\\ &=0 .\end{align*}
Therefore
$$\lim_{n \to \infty}\frac{\ln^p n}{n}=\lim_{n \to \infty}\left(\frac{\ln n}{n^{\alpha}}\right)^p=\left(\lim_{n \to \infty}\frac{\ln n}{n^{\alpha}}\right)^p=0.$$
