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I am having trouble proving the following limit evaluates to $0$:

$$\lim_{n\to\infty}\left(\frac{n^{\ln n}}{(\ln n)^n}\right)=0$$

I tried to use L'Hospital's rule, but couldn't do anything useful with it.

Thanks in advance!!

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What did you get when you were working with l'Hopital's rule? That is what I would suggest using; it might be useful for you to learn if you made any mistakes in your computations. – Clayton Feb 17 '13 at 18:54
Consider that $ln(n) < n \forall n \in \mathbb N$ – Charalampos Filippatos May 15 at 16:31
up vote 11 down vote accepted


$$\ln \left(\frac{n^{\ln{n}}}{(\ln{n})^{n}}\right)=\ln(n)^2-n \ln(\ln(n))= n\left[\frac{\ln(n)^2}{n}- \ln(\ln(n))\right]$$

It is easy to prove that this limit is $- \infty$.

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Ahh okay, that helped a lot, thank you! – Shaktal Feb 17 '13 at 18:59

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