Let \(P=\lim _{n \rightarrow \infty} \frac{1}{n}[(n+1)(n+2) \ldots \ldots(2 n)]^{1 / n}\)
On applying ' \(\log ^{\prime}\) both sides, we get
\(\log P=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\log \left(\frac{n+1}{n}\right)\left(\frac{n+2}{n}\right) \ldots \ldots\left(\frac{n+n}{n}\right)\right]\)
\(\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r}{n}\right)=\int_0^1 \log (1+x) d x \\
& =[x \log (1+x)]_0^1-\int_0^1 \frac{x}{1+x} d x \\
& =\log 2-(x-\log (1+x))_0^1 \\
& =\log 2-1+\log 2 \\
& =\log 4-1=\log \left(\frac{4}{e}\right) \Rightarrow P=\frac{4}{e}
\end{aligned}\)
Hence, option (d) is correct.