I would like someone to review my solution to this limit, the result (at least to me) is quite surprising.
Assume there is a limit and it is $L$, then
$$\lim_{n \to \infty}\frac{2^{\sqrt{\log (\log(n))}}}{(\log(\log(n)))^2\sqrt n}=L$$
and if we square everything we get
$$\lim_{n \to \infty}\frac{2^{{\log (\log(n))}}}{(\log(\log(n)))^4 n}=L^2$$
From the definition of $\log$: $2^{\log(n)}=n$, then $2^{{\log (\log(n))}}=\log(n)$ and so our limit is
$$\lim_{n \to \infty}\frac{\log(n)}{(\log(\log(n)))^4 n}=\lim_{n\to \infty} \frac{1}{(\log(\log(n)))^4} * \lim_{n \to \infty}\frac{\log(n)}{n}=0*0=L^2$$ which gives us $L=0$
This to me is a very surprising result which indicates perhaps I was wrong. exponential functions are usually very very large, we have an exponent in the numerator, and in the denominator $\sqrt{n}$ is largest.
We have that $2^{\log(\log(n))}>>[\log(\log(n))]^\alpha \text{ } \forall \alpha>1$
So we have to deal with the following limit:
$$\lim_{n\to \infty}\frac{2^{\log(\log(n))}}{n}=\lim_{n\to \infty}\frac{2^{\log(\log(n))}}{2^{\log_2(n)}}=\lim_{n\to \infty}\frac{2^{\log(\log(n))}}{2^{\frac{\log(n)}{\log(2)}}}=0$$
Because it is easy to see that: $\log(n)>>\log(2)\cdot\log(\log(n))$
so the denominator is much larger than the numerator.