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

Question

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?

Answer 1

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 $C=1$.

Answer 2

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... $$ \log L \sim x(1-C)=(1-C)\log n \\ L \sim n^{1-C} $$