A) \[0\]
B) \[1\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{6}\]
E) \[\frac{2\pi }{3}\]
Correct Answer: D
Solution :
\[\underset{h\to \infty }{\mathop{\lim }}\,\left( \frac{1}{\sqrt{4{{n}^{2}}-1}}+\frac{1}{\sqrt{4{{n}^{2}}-{{2}^{2}}}}+....+\frac{1}{\sqrt{3{{n}^{2}}}} \right)\] \[=\underset{h\to \infty }{\mathop{\lim }}\,\frac{1}{n}\left[ \frac{1}{\sqrt{4-{{\left( \frac{1}{n} \right)}^{2}}}}+\frac{1}{\sqrt{4-{{\left( \frac{2}{n} \right)}^{2}}}} \right.+\] \[\left. ....+\frac{1}{\sqrt{4-{{\left( \frac{n}{n} \right)}^{2}}}} \right]\] \[=\underset{h\to 0}{\mathop{\lim }}\,h\left[ \frac{1}{\sqrt{4-{{h}^{2}}}}+\frac{1}{\sqrt{4-{{(2h)}^{2}}}}+..... \right.\] \[\left. +\frac{1}{\sqrt{4-{{(nh)}^{2}}}} \right]\] \[=\int_{0}^{1}{\frac{dx}{\sqrt{4-{{x}^{2}}}}}=\left[ {{\sin }^{-1}}\frac{x}{2} \right]_{0}^{1}\] \[={{\sin }^{-1}}\left( \frac{1}{2} \right)-{{\sin }^{-1}}(0)=\frac{\pi }{6}\]
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