Prove that $ (lg\; lg\; n)^k=o(lg^\epsilon n)$ for all $0<k,\epsilon$

Question

I am stuck at this problem for a long time:

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Prove that $ (lg\; lg\; n)^k=o(lg^\epsilon n)$ for all $0<k,\epsilon$

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I tried to show that $\lim_{x\to\infty}\frac{ (lg\; lg\; x)^k}{lg^\epsilon x}=0$ and so $\lim_{n\to\infty}\frac{ (lg\; lg\; n)^k}{lg^\epsilon n}=0$ which implies that $ (lg\; lg\; n)^k=o(lg^\epsilon n)$

But I got stuck. I tried L'Hospital, the definition of little $o$ notation and some other stuff but I wasn't able to calculate the limit.

Thanks for any hint or help.

(Note: $lg$ means $log_2$)

Answer 1

At first substitute $\lg(n)$ by $x$ and show that:

$$\lg(x)^k \in o(x^\epsilon)$$ for all $k, \epsilon >0$ when $x \to \infty$. This last inclusion can be proved by the L'Hôpital's rule:

$$\lim_{x \to \infty}{\frac{(\lg(x)^k)'}{(x^\epsilon)'}} = \lim_{x \to \infty}{\frac{k \cdot \lg(x)^{k-1} \cdot \frac{1}{x}}{\epsilon \cdot x^{\epsilon -1}}} = \frac{k}{\epsilon}\cdot \lim_{x \to \infty}\frac{\lg(x)^{k-1}}{x^\epsilon} = 0$$