Calculate $\lim_{n \to \infty} \dfrac{\ln (\log_a (n))-\ln (\log_n (a))}{\ln (n)}$

Question

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

Answer 1

• 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

$$\lim_{u\to\infty} \frac{\ln u}{u}=0.$$

Answer 2

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 $$\lim_{n\to+\infty}\frac{2\ln(\log_a(n))}{\ln(n)}=$$

$$2\lim_{n\to+\infty}\frac{\ln(\ln(n))-\ln(\ln(a))}{\ln(n)}=$$ $$2\lim_{N\to+\infty}\frac{\ln(N)}{N}-0=0$$

Answer 3

\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*}