I am trying to prove that $2^{\sqrt{2\lg{n}}}$ is asymptotically larger than $\lg^2(n)$ so that $\lim_{n\to\infty} \frac{2^{\sqrt{2\lg{n}}}}{\lg^2(n)}=\infty$.
I have learned that $2^ \sqrt{2\lg n}=n^ \sqrt{\left(\frac{2}{\lg n}\right)}$ but apart from that I was unable to get any closer to the proof. All I am thinking of currently is to find a useful approximation that will make the limit evident, but still no such identity comes to my mind.
So the question is how to prove this
Edit: lg represents logarithm with base 2
Best Answer
Let $x^2 = 2\log_2(n)$ then
$$\lim_{n\to\infty} \frac{2^{\sqrt{2\log_2(n)}}}{(\log_2(n))^2}=\lim_{x\to \infty} \frac{2^x}{\frac{x^4}4}$$
then use that eventually $2^x=e^{x\log 2}\ge x^5$ or as an alternative L'Hospital's rule.
Refer also to the related