Q116 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots(2)\right]^{1 / n}=$
- A$16 e^{-1}$
- Correct answerB$2 e^{\left(\frac{\pi-4}{2}\right)}$
- C$2 \log 2-1$
- D$2+e^{\left(\frac{\pi-4}{2}\right)}$
Solution
$V=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right) \ldots . .2\right]^{1 / n}$
$\log V=\frac{1}{n}\left\{\log \left(1+\frac{1}{n^2}\right)+\log \left(1+\left(\frac{2}{n}\right)^2\right) \ldots \ldots\right.$
$\begin{aligned} & \left.\log \left(1+\left(\frac{n}{n}\right)^2\right)\right\} \Rightarrow \log V=\int_0^1 \log \left(1+x^2\right) d x \\ & =\left[x \log \left(1+x^2\right)-\int \frac{2 x^2}{1+x^2} d x\right]_0^1 \\ & =\left[x \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x\right]_0^1 \\ & \Rightarrow \log V=\log 2-2+\frac{\pi}{2} \Rightarrow V=2 e^{\left(\frac{n-4}{2}\right)}\end{aligned}$
Hence, the correct answer is option: 'B'.