Rate of growth of functions

Question

I was comparing the growth rates of $2^{n}$ and $n^{\ln n}$. By L'Hôpital's rule, we have $$\lim_{n\rightarrow\infty} \frac{2^{n}}{n^{\log n}} = \lim_{n \rightarrow \infty } \frac{2^{n}\ln 2}{n^{\ln n}{\frac{2\ln n}{n}}}.$$ Now, I am stuck. What can I do to continue?

Answer 1

Intuitively, exponents matter much more than bases,so you should think $2^n \gt n^{\ln n}$

To prove this, note that $$2^n=(e^{\ln 2})^n=e^{n \ln 2}\\ n^{\ln n}=(e^{\ln n})^{\ln n}=e^{(\ln n)^2}$$ and you should know that $n \ln 2 \gt (\ln n)^2$ or you can prove it by taking a derivative.