Evaluation $\lim_{n\to \infty}\frac{{\log^k n}}{n^{\epsilon}}$
Write $$ \frac{(\log n)^k}{n^\varepsilon}= \left(\frac{\log n}{n^{\varepsilon/k}}\right)^{\!k} $$ For $r>0$, we have $$ \lim_{x\to\infty}\frac{\log x}{x^r}= \lim_{x\to\infty}\frac{1/x}{rx^{r-1}}= \lim_{x\to\infty}\frac{1}{rx^r}=0 $$
Hint:
For any $c > 0$, $\log n = \frac{1}{c} \log n^c < \frac{n^c}{c}$