# Evaluating $\lim_{n \to \infty} n \left( \frac{\log\log n}{\log n} \right)^{\frac{C \log n}{\log\log n}}$

I am trying to evaluate the following limit containing double logarithm terms:

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

It seems that if $$C > 1$$, then the limit tends to zero. On the other hand, if $$C \leq 1$$, then the limit tends to infinity.

I have tried to write it as the ratio

$$\frac{n}{\left( \frac{\log n}{\log \log n} \right)^{\frac{C \log n}{\log \log n}}}$$

to apply L'Hopital's rule. It seems that the ratio of the derivatives does tend to the correct limits, but the form of the derivative of the denominator is not exactly very clear either (e.g. using Wolfram). It is also not very obvious to me why there is a "transition point" at $$C = 1$$ either.

Does anyone know of an easy way to evaluate these limits?

• Are you familiar with logarithmic differentiation? This would make the limit in the denominator from l'Hôpital's rule a bit easier to deal with. Commented Oct 20, 2021 at 18:18

Hint (in order to avoid Hopital and derivatives).

Recall that $$a^b=\exp(b\log(a))$$ for $$a>0$$, and therefore, as $$n\to+\infty$$, \begin{align} n\left( \frac{\log\log n}{\log n} \right)^{\frac{C \log n}{\log\log n}}&= n\exp\left( \frac{C \log n}{\log\log n}\left(\log\log\log n-\log\log n\right) \right)=n^{1-C+\frac{\log\log\log n}{\log\log n}}. \end{align} Since $$\displaystyle \lim_{n\to +\infty} \frac{\log\log\log n}{\log\log n}=0$$, now it should be evident the "transition point" at $$C=1$$.

• Great, it becomes almost obvious with a bit of careful setting out. Thanks!
– JKL
Commented Oct 21, 2021 at 3:21

Set Like that... $$L=\lim_{n \to \infty} n \left( \frac{\log\log n}{\log n} \right)^{\frac{C \log n}{\log\log n}}$$ Then...

$$\log L = \log\lim_{n \to \infty} n \left( \frac{\log\log n}{\log n} \right)^{\frac{C \log n}{\log\log n}} = \lim_{n \to \infty} \log \Bigg(n \left( \frac{\log\log n}{\log n} \right)^{\frac{C \log n}{\log\log n}}\Bigg)$$ So... $$\log L = \lim_{n \to \infty} \bigg( \log n + \frac{C \log n}{\log\log n}\log\left( \frac{\log\log n}{\log n} \right)\bigg)$$ Then substitute $$\log n$$ as $$x$$ $$\log L = \lim_{x \to \infty} \bigg( x + \frac{Cx}{\log x}\log\left( \frac{\log x}{x} \right)\bigg) = \lim_{x \to \infty} x\bigg(1-C + \frac{C\log\log x}{\log x}\bigg)$$ And we know that $$\lim_{t \to \infty} \frac{\log t}t=0$$. So, that turns to... $$\log L \sim x(1-C)=(1-C)\log n \\ L \sim n^{1-C}$$

• What happens if $C=1$? Commented Oct 20, 2021 at 19:19