If $f(n) = \frac{1}{n} [(n+1)(n+2)(n+3) \ldots (2n)]^{\frac{1}{n}}$, then $\lim_{n \rightarrow \infty} f(n) =$

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^{2}}+\frac{n}{(n+2)^{2}}+\ldots+\frac{n}{(2 n-1)^{2}}\right]$ is equal to ...... .

$\lim _{n \rightarrow \infty}\left(\frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\ldots+\frac{n}{n^2+n^2}\right)=$

Evaluate the following definite integral as the limit of a sum: $\int_{-1}^{1} e^{x} dx$

$\lim _{n}$ ${\rightarrow \infty} \frac{1}{n}\left(\frac{1}{e^{1 / n}}+\frac{1}{e^{2 / n}}+\frac{1}{e^{3 / n}}+\ldots+\frac{1}{e^{2n/n}}\right)=$

Given that $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. If $f: R \rightarrow R$ is defined by $f(x)=x^2+2$, then $\lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]=$