Comparing asymptotic complexity of functions $\log{n}$, $(\log{n})^c$ and $\sqrt{n}$
for any $c&gt;1,\epsilon&gt;0$,
$$\lim_{n\to\infty} \frac{(\log n)^c}{n^\epsilon} = 0$$

This means that $\log n=o( (\log n)^c)$ and $(\log n)^c = o(n^\epsilon)$. In particular, for $\epsilon =0.5$, $(\log n)^c = o(\sqrt{n})$.

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