# $\lim_{n\rightarrow \infty} \frac {(n!)^{\frac {1}{n}}}{n}$ and $\lim_{n\rightarrow \infty} \frac {1}{n}\ln {2n \choose n}$

I have two questions:

$$\lim_{n\rightarrow \infty} \frac {(n!)^{\frac {1}{n}}}{n}$$ and $$\lim_{n\rightarrow \infty} \frac {1}{n}\ln {2n \choose n}$$

I realised that I had to do this with the help of integration, and in both cases I had to come to the same situation $\int_{0}^{1}\ln \space xdx$. But then I'm having $(x\times ln\space x-x)|_{0}^{1}$ and am stuck in the $0\times ln \space 0$ thing. What should I do with it?

• For the first one, see this. Feb 26 '14 at 19:55
• @DavidMitra, the second answer to that question dectly states $\int_{0}^{1}ln \space xdx=-1$, but I am stuck there. Feb 26 '14 at 20:02
• Feb 26 '14 at 20:19
• – Ben
Feb 27 '14 at 0:23

A simple substitution lets you see whats happening. Substitute $x=1/y$ into the limit so that it becomes $$\frac{-ln(y)}{y}.$$ As x approaches zero y approaches infinity, and it's a little more intuitive that the above limit should go to zero. You can prove it with L'Hopital.