Evaluation of $\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}$ 
Evaluate
  $$\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}.$$

$\underline{\bf{My\;\;Try}}::$ Let $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdots (n+n)}{(1)\cdot (2)\cdot (3)\cdots (n)}\right)$
$\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+\cdots+\ln \left(\frac{n+n}{n}\right)\right\}$
$\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)$
Now Using Reinman Sum
$\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx =  \ln (2)$
My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test
If yes then please explain here
Thanks

Notice $\displaystyle \sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n}$ and $\displaystyle \binom{2n}{n} \ge \binom{2n}{k}$ for all $0 \le k \le 2n$, we have
$$
\frac{2^{2n}}{2n+1}\le \binom{2n}{n} \le 2^{2n}
\quad\implies\quad \log 2 - \frac{\log{(2n+1)}}{2n} \le \frac{1}{2n} \log \binom{2n}{n} \le \log 2$$
Since $\quad\displaystyle \lim_{n\to\infty} \frac{\log(2n+1)}{2n} = 0\quad$, 
we get $\quad\displaystyle \lim_{n\to\infty} \frac{1}{2n}\log\binom{2n}{n} = \log 2$.
