calculus - Evaluate $lim_{nto infty} frac{2^{ln(ln(n))}}{nln(n)}$

I'm having trouble evaluating the limit: $$\lim_{n\to \infty} \frac{2^{\ln(\ln(n))}}{n\ln(n)}$$ (as it looks like, the limit tends to $0$)

This is what I got until now:
$$0\le \lim_{n\to \infty} \frac{2^{\ln(\ln(n))}}{n\ln(n)}\le \lim_{n\to \infty} \frac{2^{\ln(n)}}{n\ln(n)}=\lim_{n\to \infty} \frac{e^{\ln{2^{\ln(n)}}}}{n\ln(n)} = \lim_{n\to \infty} \frac{e^{\ln(n)\ln{2}}}{n\ln(n)}$$

Tried L'Hopital from here, but it seems usefulness.

Would appreciate your advice.

Answer

You're almost there. $$\frac{e^{\ln(n)\ln(2)}}{n\ln(n)}= \frac{(e^{\ln(n)})^{\ln(2)}}{n\ln(n)}=\frac{n^{\ln2}}{n\ln(n)} = \frac{n^{\ln2 -1}}{\ln(n)}$$ which should tend to zero as $\ln2<1$.