$$
\begin{aligned}
& \text { Let } A=\lim _{n \rightarrow \infty} \frac{1}{n}[(n+1)(n+2)(n+3) \ldots(2 n)]^{\frac{1}{n}} \\
& =\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \ldots\left(1+\frac{n}{n}\right)\right]^{\frac{1}{n}} \\
& \therefore \log A=\lim _{n \rightarrow \infty} \sum_{\Gamma}^n \frac{1}{n} \log \left(1+\frac{r}{n}\right) \\
& =\int_0^1 \log (1+x) d x=[x \log (1+x)]_0^2-\int_0^1 \frac{x}{1+x} d x \\
& =\log 2-\int_0^1\left(1-\frac{1}{1+x}\right) d x \\
& =\log 2-[x-\log (1+x)]_0^2 \\
& =\log 2-[1-\log 2] \\
& =\log 2-1+\log 2=\log 4-1 \\
& \therefore \quad \log A=\log 4-1 \\
& \Rightarrow \log \left(\frac{A}{4}\right)=-1 \Rightarrow \frac{A}{4}=e^1 \\
& \Rightarrow \quad A=\frac{4}{e} \Rightarrow \lim _{n \rightarrow \infty} f(n)=\frac{4}{e} \\
&
\end{aligned}
$$