Q116 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics
$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots(2)\right]^{\frac{1}{n}}=$
- A$2 e^{\pi-4}$
- B$e^{\frac{\pi-4}{2}}$
- Correct answerC$2 e^{\frac{\pi-4}{2}}$
- D$\frac{1}{2} e^{\frac{\pi-4}{2}}$
Solution
$L=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)+\ldots . .\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}$
$\begin{aligned} & \log L=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\log \left(1+\frac{1}{n^2}\right)+\log \left(1+\frac{2^2}{n^2}\right)+\ldots .+\log \left(1+\frac{n^2}{n^2}\right)\right] \\ & \log L=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n} \log \left[1+\left(\frac{r}{n}\right)^2\right] \\ & \therefore \quad \log L=\int_0^1 \log \left(1+x^2\right) d x \\ & =\left[\log \left(1+x^2\right) \cdot \int d x-\int \frac{2 x}{1+x^2} \cdot x d x\right]_0^1 \\ & =\left[x \log \left(1+x^2\right)-2 \int \frac{x^2}{1+x^2} d x\right]_0^1 \\ & =\left[x \log \left(1+x^2\right)-2 \int\left(1-\frac{1}{1+x^2}\right) d x\right]_0^1 \\ & =\left[x \log \left(1+x^2\right)-2\left(x-\tan ^{-1} x\right)\right]_0^1 \\ & =\left[\log 2-2\left(1-\frac{\pi}{4}\right)-0\right]=\log 2-2+\frac{\pi}{2} \\ & \Rightarrow \log =\log 2-\left(2-\frac{\pi}{2}\right)=\log 2+\left(\frac{\pi-}{2}\right) \\ & \Rightarrow \log L=\log 2+\log e^{\frac{\pi-4}{2}} \\ & \Rightarrow \log L=\log \left(2 \cdot e^{\frac{\pi-4}{2}}\right) \\ & \therefore \quad L=2 e^{\frac{\pi-4}{2}} \text {. } \\ & \end{aligned}$
Hence, the correct answer is option: 'C'.