Calculate $\lim_{n \to \infty} \dfrac{\ln (\log_a (n))-\ln (\log_n (a))}{\ln (n)}$ I have to calculate $\lim_{n \to \infty} \dfrac{\ln (\log_a (n))-\ln (\log_n (a))}{\ln (n)}$
My idea Let $a_n = \dfrac{\ln (\log_a (n))-\ln (\log_n (a))}{\ln (n)}$, then
$$e^{a_n}= (\log_a (n))^{1/\ln(n)} \cdot \left( \dfrac{1}{\log_n (a)} \right)^{1/\ln(n)} $$
And I don't know to continue
 A: You might know that for $ n,a>1$
$$\log_n(a)=\frac{\ln(a)}{\ln(n)}$$
$$=\frac{1}{\log_a(n)}$$
and
$$\ln(\log_n(a))=-\ln(\log_a(n)).$$
So the limit to evaluate is
$$\lim_{n\to+\infty}\frac{2\ln(\log_a(n))}{\ln(n)}=$$
$$2\lim_{n\to+\infty}\frac{\ln(\ln(n))-\ln(\ln(a))}{\ln(n)}=$$
$$2\lim_{N\to+\infty}\frac{\ln(N)}{N}-0=0$$
A: *

*HINT

$$1≠a>0, \log_an=u,n=a^u$$
$$\begin{align}\lim_{n \to \infty} \frac{\ln (\log_a (n))-\ln (\log_n (a))}{\ln (n)}&=\lim_{u\to\infty} \frac{\ln \left( \frac{u}{\frac 1u}\right)}{u\ln a}\\
&=\lim_{u\to\infty} \frac{2\ln u}{u\ln a}\\
&=\frac{2}{\ln a}\times \lim_{u\to\infty}\frac{\ln u}{u}\end{align}$$
Then, note that
$$\lim_{u\to\infty} \frac{\ln u}{u}=0.$$
A: \begin{align*}
    a_n &= \frac{\ln(\log_a(n))-\ln(\log_n(a))}{\ln(n)}\\
&= \frac{1}{\ln(n)}\Big(\ln\Big(\frac{\ln(n)}{\log(a)}\Big)-\ln\Big(\frac{\ln(a)}{\ln(n)}\Big)\Big)\\
&= \frac{2}{\ln(n)}\Big(\ln\Big(\frac{\ln(n)}{\ln(a)}\Big)\Big)\\
&= \frac{2\ln(\ln(n))}{\ln(n)} - \frac{2\ln(\ln(a))}{\ln(n)}
\end{align*}
Now since $a$ is constant $\lim_{n\to\infty} \frac{2\ln(\ln(a))}{\ln(2)} = 0$. Thus
\begin{align*}
\lim_{n\to\infty} a_n &= \lim_{n\to\infty}\frac{2\ln(\ln(n))}{\ln(n)} - \frac{2\ln(\ln(a))}{\ln(n)}\\
&= \lim_{n\to\infty}\frac{2\ln(\ln(n))}{\ln(n)}\\
&= \lim_{\ln(n)\to\infty}\frac{2\ln(\ln(n))}{\ln(n)}=\lim_{x\to\infty}\frac{2\ln(x)}{x} = 0\\
\end{align*}
