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Prove that:

$$\lim\limits_{n\rightarrow \infty}\left({\frac{10^{\log_2{\log_2{n}}}}{\log_2{n}}}\right)=\infty$$

I've tried applying L'hopital to no avail.

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Let $\log_2n=m$ to get $$\lim_{n\rightarrow \infty}({\frac{10^{\log_2{\log_2{n}}}}{\log_2{n}}})=\lim_{m\to\infty}\dfrac{10^{\log_2m}}m$$

Again, let $\log_2m=r\iff m = 2^r$

$$\implies\lim_{m\to\infty}\dfrac{10^{\log_2m}}m=\lim_{r\to\infty}\dfrac{10^r}{2^r}=\lim_{r\to\infty}5^r=?$$

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One could also use $$ 10^{\log_2(x)}=2^{\log_2(10)·\log_2(x)}=x^{\log_2(10)} $$ to find that the fraction inside the limit reduces to $$ \log_2(n)^{\log_2(10)-1}=\log_2(n)^{\log_2(5)} $$

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Another possibility, which doesn't involve substitutions, is to take $\log_2 L$, where $L$ is the limit. The argument then becomes (given the linearity of the operator $\lim$): $$ \log_2 L=\log_2\left(\frac{10^{\log_2\log_2 n}}{\log_2 n}\right)=\log_2\log_2 n\cdot\log_210-\log_2\log_2 n=\left(\log_210-1\right)\log_2\log_2 n $$ since $1=\log_22$, it follows that $\log_210-1=\log_25$, which is greater than one. Next, taking $2^{\log_2 L}$ gives the equivalent simplified argument: $$ L=2^{\log_2 L}=2^{\log_25\cdot\log_2(\log_2 n)}=2^{\log_2(\log_2 n)^{\log_25}}=(\log_2 n)^{\log_25} $$ Applying the limit as $n\to\infty$ gives the result $L=\infty$, since it's a positive monotone increasing function taken to a positive constant power. $\square$

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