# Proving that $\log\log x = o((\log x)^{\epsilon})$

I am trying to show that $$\log\log x = o((\log x)^{\epsilon})$$ for all $$\epsilon > 0$$.

Attempt:

We wish to show that $$\lim_{x \to \infty}\frac{\log \log x}{(\log x)^{\epsilon}} \rightarrow 0$$

Let $$x = e^{y}$$ then we have $$\frac{\log y}{y^{\epsilon}} \rightarrow 0$$. Hence the result follows. Is this correct?

Thanks.

• Good point. Will edit. Sep 25, 2020 at 7:36
• Your substitution changes the question to proving $\log y =o(y^\epsilon)$. Is asserting $\frac{\log y}{y^{\epsilon}} \rightarrow 0$ enough of a proof for that? Sep 25, 2020 at 8:04
• $\frac{\log y}{y^{\epsilon}}$ is a fairly standard result, so I just assumed the reader would know it. The proof is fairly simple, pretty much using the same method as above. Sep 25, 2020 at 8:11

As noticed your proof is fine, to justify the latter limit we can use again $$y = e^{z}\to \infty$$ then
$$\lim_{y\to \infty}\frac{\log y}{y^{\epsilon}} =\lim_{z\to \infty}\frac1\epsilon\frac{z\epsilon}{e^{z\epsilon}} = 0$$
since eventually $$e^{z\epsilon}\ge (z\epsilon)^2 \implies \frac{z\epsilon}{e^{z\epsilon}}\le \frac{z\epsilon}{(z\epsilon)^2}=\frac1{z\epsilon}\to 0$$