Q81 / 170 · Manipal (MET) 2017 · Mathematics
$\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{4 n^2-1}}+\frac{1}{\sqrt{4 n^2-2^2}}+\ldots+\frac{1}{\sqrt{3 n^2}}\right)$
equal to
- A0
- B1
- C$\frac{\pi}{3}$
- Correct answerD$\frac{\pi}{6}$
Solution
$\begin{aligned} & \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{4 n^2-1}}+\frac{1}{\sqrt{4 n^2-2^2}}+\ldots+\frac{1}{\sqrt{3 n^2}}\right) \\ = & \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{\sqrt{4-\left(\frac{1}{n}\right)^2}}+\frac{1}{\sqrt{4-\left(\frac{2}{n}\right)^2}}+\ldots+\frac{1}{\sqrt{4-\left(\frac{n}{n}\right)^2}}\right] \\ = & \lim _{h \rightarrow 0} h\left[\frac{1}{\sqrt{4-h^2}}+\frac{1}{\sqrt{4-(2 h)^2}}+\ldots+\frac{1}{\sqrt{4-(n h)^2}}\right] \\ = & \int_0^1 \frac{d x}{\sqrt{4-x^2}}=\left[\sin ^{-1} \frac{x}{2}\right]_0^1 \\ = & \sin ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}(0)=\frac{\pi}{6}\end{aligned}$
Hence, the correct answer is option: 'D'.