Q116 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics
$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}=$
- A$3 e^{\frac{\pi-4}{6}}$
- B$2 e^{\frac{\pi-2}{4}}$
- Correct answerC$2 e^{\frac{\pi-4}{2}}$
- D$4 e^{\frac{\pi-4}{4}}$
Solution
Let $A=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}$ taking log on both side
$$
\begin{gathered}
\log _e A=\lim _{n \rightarrow \infty} \frac{1}{n} \\
{\left[\log \left(1+\frac{1}{n^2}\right)+\log \left(1+\frac{2^2}{n^2}\right) \ldots \log \left(1+\frac{n^2}{n^2}\right)\right]} \\
\log A=\lim _{n \rightarrow \infty} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}\right) \frac{1}{n}=\int_0^1 \log \left(1+x^2\right) d x
\end{gathered}
$$
[Applying formula $\lim _{n \rightarrow \infty} \sum_{r=1}^n f\left(\frac{r}{n}\right) \cdot \frac{1}{n}=\int_0^1 f(x) d x$
$$
=\int_0^5 \log \left(1+x^2\right) \cdot 1 d x
$$
using by parts,
$$
\begin{aligned}
& =\log \left(1+x^2\right) \int 1 d x-\int\left[\frac{d}{d x}\left(\log \left(1+x^2\right) \int 1 d x\right] d x\right. \\
& =x \log \left(1+x^2\right)-\int \frac{2 x}{1+x^2} \cdot x d x \\
& =x \log \left(1+x^2\right)-2 \int \frac{x^2}{1+x^2} d x \\
& =x \log \left(1+x^2\right)-2 \int \frac{1+x^2}{1+x^2}+2 \int \frac{1}{1+x^2} d x \\
& =x \log \left(1+x^2\right)-2 \int 1 d x+2 \tan ^{-1}(x) \\
& =\left[x \log \left(1+x^2\right)-2 x+2 \tan ^{-1}(x)\right]_0^1 \\
& \Rightarrow \log { }_e A=\left[1 \cdot \log 2-2+2 \tan ^{-1}(1)-0+0-0\right]
\end{aligned}
$$
$A=e^{\log 2-2+2 \cdot \frac{\pi}{4}}=e^{\log 2} \cdot e^{\frac{\pi}{2}-2}=2 e^{\frac{\pi-4}{2}}$.
Hence, the correct answer is option: 'C'.