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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.

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  • $\begingroup$ Good point. Will edit. $\endgroup$ Sep 25, 2020 at 7:36
  • $\begingroup$ 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? $\endgroup$
    – Henry
    Sep 25, 2020 at 8:04
  • $\begingroup$ $\frac{\log y}{y^{\epsilon}}$ 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. $\endgroup$ Sep 25, 2020 at 8:11

2 Answers 2

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Yes, what you have done is correct.

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

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