Q115 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics
$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots . .\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=$
- A$e$
- B$2 e$
- C$2 e^{\frac{\pi-2}{2}}$
- Correct answerD$2 e^{\frac{\pi-4}{2}}$
Solution
Let $L=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}$
$\begin{aligned} \therefore \quad \log L & =\lim _{n \rightarrow \infty} \frac{1}{n} \\ & {\left[\log \left(1+\frac{1}{n^2}\right)+\log \left(1+\frac{2^2}{n^2}\right)+\ldots . \log \left(1+\frac{n^2}{n^2}\right)\right] }\end{aligned}$
$\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\left(\frac{r}{n}\right)^2\right) \\ & =\int_0^1 \log \left(1+x^2\right) d x \\ & =\left[x \log \left(1+x^2\right)\right]_0^1-\int_0^1 x \cdot \frac{1}{1+x^2} \cdot 2 x d x \\ & =\log 2-2 \int_0^1 \frac{x^2}{1+x^2} d x=\log 2-2 \int_0^1\left(1-\frac{1}{1+x^2}\right) d x \\ & =\log 2-2\left[x-\tan ^{-1} x\right]_0^1=\log 2-2\left[1-\frac{\pi}{4}\right] \\ & =\log 2-2+\frac{\pi}{2}=\log 2-2+\frac{\pi}{2} \\ \therefore \quad L & =e^{e \log 2} \cdot e^{\frac{\pi}{2}-2}=2 e^{\frac{\pi-4}{2}}\end{aligned}$
Hence, the correct answer is option: 'D'.