Proof that $\lim_{n\to\infty} \frac{2^{\sqrt{2\lg{n}}}}{\lg^2(n)}$ is infinity 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
 A: 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

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*What is the limit $x\to\infty$ of $x^n \cdot e^{-x}$?
A: Actually the easiest way is always to express them in canonical form so that they can be easily compared. For exponentiation, always express it using the same base. Here, you have $2^{\sqrt{2·\lg n}}$ and $(\lg n)^2 = 2^{2·\lg\lg n}$. Can you tell which is bigger? If not, just do one more step of the canonicalization process: $\sqrt{2·\lg n} = \sqrt{2}·2^{1/2·\lg\lg n}$ and $2·\lg \lg n = 2·2^{\lg\lg\lg n}$. Now there are no more exponents, and obviously $1/2·\lg\lg n ≫ \lg\lg\lg n$ as $n → ∞$.
Note that this process is completely systematic, and computer algebra systems use such processes to evaluate limits. For related systematic limit computations, see here and the posts linked from there.
