Show that $\log n = o(n^\epsilon)$
Here is a simple proof which avoids L'Hôpital's rule.

We start with the observation
$$
\log n = \int_1^n \frac{dx}{x} \leq \int_1^n dx \leq n,
$$
and so
$$
\frac{\log n}{n^2} \leq \frac{1}{n} \xrightarrow{n\to\infty} 0.
$$

Given $\epsilon &gt; 0$, using the fact that $\lim_{n\to\infty} n^{\epsilon/2} = \infty$, we see that
$$
0 = \lim_{n\to\infty} \frac{\log (n^{\epsilon/2})}{(n^{\epsilon/2})^2} =
\lim_{n\to\infty} \frac{\frac{\epsilon}{2} \log n}{n^\epsilon}.
$$
It follows that
$$
\lim_{n\to\infty} \frac{\log n}{n^\epsilon} = 0.
$$