Here, $\displaystyle \mathrm{A}=\lim_{n\to \infty} \frac{1}{n}\left( \frac{n}{n+1}+\frac{n}{n+2}+\frac{n}{n+3}+\cdots+\frac{n}{2n} \right)=\lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n} \left( \frac{n}{n+r} \right)$
It's a limiting sum of which the $n^{\mathrm{th}}$ term of the sequence $\frac{n}{n+1},\frac{n}{n+2},\frac{n}{n+3},\cdots,\frac{n}{n+n}$ converges to exactly $\frac{1}{2}$. So, this sum can be evaluated by a definite integral.
Now $\frac{n}{n+r}=\frac{1}{1+\frac{r}{n}}$. Let $x=\frac{r}{n}$. $\therefore \frac{n}{n+r}=\frac{1}{1+x}\tag{i}$
Notice that for each term i.e. for $r=1,2,3,...$ we have $x=\frac{1}{n},\frac{2}{n},\frac{3}{n},...,\frac{n}{n}$.
As $n \to \infty$, the first term $x=\frac{1}{n} \to 0$. So the initial integral limit is $x=0$. On the other hand, the last term $x=\frac{n}{n}=1$. So the final integral limit is $x=1$.
Now from the continuous set $[0,1]$, for the sake of integration, we need $\Delta x=\frac{1-0}{n}=\frac{1}{n} \tag{ii}$
Now using no$\mathrm{(i)}$ and no$\mathrm{(ii)}$, we obtain
$\displaystyle \mathrm{A}=\lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n} \left( \frac{n}{n+r} \right)=\lim_{n\to \infty} \sum_{r=1}^{n}\Delta x \left( \frac{1}{1+x} \right)=\int_{0}^{1}\frac{1}{1+x}\mathrm{d}x$
Hence $$\begin{align} \mathrm{A}&=\int_{0}^{1}\frac{1}{1+x}\mathrm{d}x \\ &=[\log_{e}(1+x)]_{0}^{1}\\ &=\log_{e}(2)-\log_{e}(1)\\ &=\log_{e}(2) \end{align}$$
So the correct answer is C.