$\lim _{n}$ ${\rightarrow \infty}\left(\frac{1}{1+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)=$

The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$ is equal to:

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}=$

$\lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{4 n^2-1^2}}+\frac{1}{\sqrt{4 n^2-2^2}}+\frac{1}{\sqrt{4 n^2-3^2}}+\dots+\frac{1}{\sqrt{4 n^2-n^2}}\right\}=$

$\lim _{n \rightarrow \infty}\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}}-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]=$

For a sufficiently large value of $n$,the sum of the square roots of the first $n$ positive integers,i.e.,$\sqrt{1} + \sqrt{2} + \sqrt{3} + \dots + \sqrt{n}$,is approximately equal to: