Definite Integral as Limit of a Sum

The value of $\underset{n\to \infty }{\lim }\sum _{r=1}^{r=4n}\frac{\sqrt{n}}{\sqrt{r}{\left(3\sqrt{r}+4\sqrt{n}\right)}^2}$ is equal to

The value of $\underset{\mathrm{a}}{\overset{b}{\int }}f\left(x\right)dx=\left(b-a\right)\underset{n\to \infty }{\lim }\frac{1}{n}[f\left(a\right)+f\left(a+h\right)+\mathrm{...}+f\left(a+\left(n-1\right)h\right)],$ where $h=\frac{b-a}{n}$, $f\left(x\right)=x^2+x+2; a=0, b=2$ as limits of sum would be

The value of $\underset{n\to \infty }{\lim }\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{6n}\right)$ is

$\underset{n\to \infty }{\lim }\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{6n}\right)$ is equal to

The limit $\underset{n\to \infty }{\lim }\frac{1}{n^{2020}}\sum _{k=1}^n{k}^{2019}$

If $L=\underset{n\to \infty }{\lim }\left(\frac{1^5+2^5+3^5+....+n^5}{n^6}\right)+\underset{n\to \infty }{\lim }\left(\frac{1^4+2^4+3^4+....+n^4}{n^5}\right)-\underset{n\to \infty }{\lim }\left(\frac{1^3+2^3+....+n^3}{n^4}\right)$, then the value of $L$ is

Suppose the limit $L=\underset{n\to \infty }{\lim }\sqrt{n}{\int }_0^1\frac{1}{{\left(1+x^2\right)}^n}dx$ exists and is larger than $\frac{1}{2}.$ Then,

$\underset{n\to \infty }{\lim }\frac{1+2^4+3^4+...............+n^4}{n^5}=$

$\underset{n\to \infty }{\lim }\left(\frac{\sqrt{1}+2\sqrt{2}+3\sqrt{3}+\dots +n\sqrt{n}}{n^{\frac{5}{2}}}\right)=$

The value of $\underset{n\to \infty }{\lim } \sum _{k=1}^n\frac{k^{\frac{1}{a}} \left\{n^{a-\frac{1}{a}} +k^{a-\frac{1}{a}}\right\}}{n^{a+1}}=$______

The value of $\underset{n\to \infty }{\lim }{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}$ is equal to

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{ k^{{\displaystyle \frac{1}{a}}}\left(n^{a-\frac{1}{a}}+k^{a-\frac{1}{a }}\right)}{n^{a+1}}$ is equal to

The value of $\underset{n\to \infty }{\lim }\underset{ r=1}{\overset{n}{ \sum }}\frac{1}{\sin \left\{\frac{\left(n+r\right)\pi }{4n}\right\}}.\frac{\pi }{n}$ is equal to

$\underset{n\to \infty }{\lim }\left[\frac{1}{n}+\frac{n^2}{{\left(n+1\right)}^3}+\frac{n^2}{{\left(n+2\right)}^3}+...+\frac{1}{8n}\right]$ is equal to

The limit $L=\underset{n\to \mathrm{\infty }}{\lim }{\sum }_{r=1}^n\frac{n}{n^2+r^2}$ satisfies

The value of $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}+.....+\frac{1}{4n}\right)$ is equal to

The value of $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{e^{\frac{1}{n}}}{n^2}+\frac{2e^{\frac{2}{n}}}{n^2}+\frac{3e^{\frac{3}{n}}}{n^2}+\dots +\frac{2e^2}{n}\right)$ is

The value of $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\sum _{r=1}^n\left(\frac{2r}{n^2}\right)e^{\frac{r^2}{n^2}}$ is equal to 

$\underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\left[\frac{1}{n+1}+\frac{1}{n+2}+\dots ..+\frac{1}{6n}\right]$ is equal to

The value of $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{n}\left\{\sqrt[n]{5}+\sqrt[n]{5^2}+...+\sqrt[n]{5^n}\right\}$ is equal to