Evaluate the limit: $\lim _{n \rightarrow \infty} \frac{3}{n}\left\{1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}+\ldots+\sqrt{\frac{n}{n+3(n-1)}}\right\}$

If $f: R \rightarrow R$ is defined by $f(x)=x+1$,then the value of $\lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+\ldots+f\left(\frac{5(n-1)}{n}\right)\right]$ is:

$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ is equal to

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

Evaluate $\int_{0}^{2} e^{x} dx$ as the limit of a sum.

$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots+\frac{n}{n^2} \sec ^2 \frac{n^2}{n^2}\right]=$