# Evaluate the Limit of : $\lim_{n \to \infty}\frac{2^{2n}}{n^{\log_{2}n}}$

I am trying to evaluate the limit of the following: $$\lim_{n \to \infty}\frac{2^{2n}}{n^{\log_{2}n}}$$ what I did so far is:
$$\dots \lim_{n \to \infty}\frac{n\ln{4}}{\log_{2}n\ln{n}}$$ every step from here like: using L'Hôpital's rule or keep simplify the expression did not success.
any suggestions?

• Ok, I will get $2n$ now what about to log $n^{\log_{2}n}$? what I will get? – Ofir Attia Jun 25 '14 at 20:27
• No, if you log the expression base 2 you get $n \log 2 - (\log n)^2$ which is $O(n)$, hence... – Alex Jun 25 '14 at 20:29
$$2^{2n}=e^{2n\ln{2}}, \;\; n^{\log_{2}n}=e^{\frac{\ln^2{n}}{\ln{2}}},$$thus $$\frac{2^{2n}}{n^{\log_{2}n}}=e^{2n\ln{2}-\frac{\ln^2{n}}{\ln{2}}}.$$