$\lim_{n \to \infty} \left [ \frac{1}{(1+n)} + \frac{1}{(2+n)} + - - - - - + \frac{1}{(n+n)} \right ]$
$\lim_{n \to \infty}\sum_{r=1}^n \frac{1}{n+r}$ = $\lim_{n \to \infty}\sum_{r=1}^n \frac{\frac{1}{n}}{1+\frac{r}{n}}$
$\lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^n \frac{1}{1+\frac{r}{n}}$
Which is a Riemann sum for the integral
=$\int\limits_{0}^{1}\frac{1}{1+x}\, dx$
=$[log|x+1| ]_0^1$
=$log2 - log1$
=$log2$
Alternate Method:
We may use Botez-Catalan identity and immediately get that
$\lim_{n \to \infty} \left [ \frac{1}{(1+n)} + \frac{1}{(2+n)} + - - - - - + \frac{1}{(n+n)} \right ]$=$\lim_{n \to \infty} \left [ 1- \frac{1}{2} +\frac{1}{3}-\frac{1}{4}+ ......... +(-1)^{2n+1} \frac{1}{2n} \right ]$
=$\lim_{n \to \infty} \left [ 1- \frac{1}{2} +\frac{1}{3}-\frac{1}{4}+ ......... +(-1)^{n+1} \frac{1}{n} \right ]$
=Log2
Hence,Option(A)$log2$ is the correct choice.