A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) 1
D) None of these
Correct Answer: B
Solution :
We have, \[\underset{n\to \infty }{\mathop{\lim }}\,\,\,\left[ \,\frac{n}{1+{{n}^{2}}}+\frac{n}{4+{{n}^{2}}}+......+\frac{1}{2n}\, \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{r=1}^{n}{\,\,\frac{n}{{{r}^{2}}+{{n}^{2}}}}=\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{r=1}^{n}{\,\,\frac{n}{{{n}^{2}}\left( 1+\frac{{{r}^{2}}}{{{n}^{2}}} \right)}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{r=1}^{n}{\,\,\frac{1}{n\,\left( 1+\frac{{{r}^{2}}}{{{n}^{2}}} \right)}=\int_{0}^{1}{\frac{dx}{\text{1}+{{x}^{2}}}}}\], \[\left\{ \text{Applying formula, }\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{r=0}^{n-1}{\,\,\left\{ f\left( \frac{r}{n} \right) \right\}.\frac{1}{n}=\int_{0}^{1}{f(x)dx}} \right\}\] \[=[{{\tan }^{-1}}x]_{0}^{1}={{\tan }^{-1}}1-{{\tan }^{-1}}0=\frac{\pi }{4}.\]
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