The value of $\underset{n→∞}{\lim}\frac{1}{n}\sum\limits_{r=1}^{n}(\frac{r}{n+r})$ is

1 – ln 2

$\underset{n→∞}{\lim}\frac{1}{n}\sum\limits_{r=1}^{n}\frac{r}{n+r}=\underset{n→∞}{\lim}\frac{1}{n}\sum\limits_{r=1}^{n}(\frac{\frac{r}{n}}{1+\frac{r}{n}})$

$=\int\limits_0^1\frac{x\,dx}{1+x}=\int\limits_0^1dx-\int\limits_0^1\frac{dx}{1+x}=1-In\,2$

Hence (C) is the correct answer.