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

Question

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.

Answer 1

Yes, what you have done is correct.

Answer 2

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 $$e^{z\epsilon}\ge (z\epsilon)^2 \implies \frac{z\epsilon}{e^{z\epsilon}}\le \frac{z\epsilon}{(z\epsilon)^2}=\frac1{z\epsilon}\to 0$$