Prove that: $\lim\limits_{n\rightarrow \infty}({\frac{10^{\log_2{\log_2{n}}}}{\log_2{n}}})=\infty$ 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.
 A: 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 $\log_2m=r\iff m = 2^r$
$$\implies\lim_{m\to\infty}\dfrac{10^{\log_2m}}m=\lim_{r\to\infty}\dfrac{10^r}{2^r}=\lim_{r\to\infty}5^r=?$$
A: 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
$$
\log_2(n)^{\log_2(10)-1}=\log_2(n)^{\log_2(5)}
$$
A: 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. $\square$
