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I'm working on a computer science problem where I need to solve:

$\lim_{n\to\infty}\frac{({\log n})^{\log n}}{n}$

Since this limit goes to infinity I'm trying to use L'hopitals rule, but I can't figure out how to differentiate $({\log n})^{\log n}$.

In case it helps, it is $\log_2 n$.

Thanks.

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  • $\begingroup$ $\log n^{\log n} = (\log n)^2$, no? $\endgroup$ – peterwhy Sep 26 '15 at 13:22
  • $\begingroup$ Um, that limit has nothing to do with the title about differentiating. $\endgroup$ – Thomas Andrews Sep 26 '15 at 13:29
  • $\begingroup$ Sorry, I wrote it wrong. It's $({\log n})^{\log n}$. $\endgroup$ – user274269 Sep 26 '15 at 13:30
  • $\begingroup$ It still has nothing to do with the derivative of anything. It is just a limit. $\endgroup$ – Thomas Andrews Sep 26 '15 at 13:31
  • $\begingroup$ I need to differentiate $({\log n})^{\log n}$ to solve the limit. I want to use L'hopitals rule, so that even if the limit turns out to be $\infty$ it is in a simpler form. $\endgroup$ – user274269 Sep 26 '15 at 13:33
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let $y= (\log(n))^{\log(n)}$

$\implies \log(y) = \log(n)\log(\log(n))$

$\implies \frac{y'}{y} = \frac{\log(\log(n))}{n} + \log(n)\frac{1}{\log(n)}\frac{1}{n}$

$y' = (\log(n))^{\log(n)}(\frac{\log(\log(n)) + 1}{n})$

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HINT:

$$\log n^{\log n}=\log n \log n=(\log n)^2. $$

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  • $\begingroup$ Sorry, I actually mean $({\log n})^{\log n}$. I'm unfamiliar with Latex. $\endgroup$ – user274269 Sep 26 '15 at 13:31
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For evaluating the limit directly, substitute $m=\log n$. Then $n={\rm e}^m$ and $$\lim_{n\to\infty}{1\over n}(\log n)^{\log n}=\lim_{m\to\infty}\Bigl({m\over {\rm e}}\Bigr)^m$$ which is obviously divergent.

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One more thing to notice: once you've substituted $\log n = t$ you get $$ \lim_{t \to \infty} \bigg( \frac{t}{e} \bigg)^t $$ which, using Stirling's expansion grows roughly like $ \frac{t!}{\sqrt{2 \pi t}}$ and, of course, diverges.

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Can you differentiate $ x^x $ with respect to $x$?

Repeat the same with $ x= \log n $ and finally multiply by $\frac{1}{n}$ using Chain rule.

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