0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

Log the expression base 2. You will immediately see the leading term and conclude that it diverges.

$\endgroup$
  • $\begingroup$ Ok, I will get $2n$ now what about to log $n^{\log_{2}n}$? what I will get? $\endgroup$ – Ofir Attia Jun 25 '14 at 20:27
  • $\begingroup$ No, if you log the expression base 2 you get $n \log 2 - (\log n)^2$ which is $O(n)$, hence... $\endgroup$ – Alex Jun 25 '14 at 20:29
1
$\begingroup$

Hint:
$$2^{2n}=e^{2n\ln{2}}, \;\; n^{\log_{2}n}=e^{\frac{\ln^2{n}}{\ln{2}}},$$thus $$\frac{2^{2n}}{n^{\log_{2}n}}=e^{2n\ln{2}-\frac{\ln^2{n}}{\ln{2}}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.