The value of $\lim _{n \rightarrow \infty} \left( \frac{1}{\sqrt{4n^2-1}} + \frac{1}{\sqrt{4n^2-4}} + \dots + \frac{1}{\sqrt{4n^2-n^2}} \right)$ is

$\mathop {Limit}\limits_{n \to \infty } \frac{1}{n} \left[ 1 + \sqrt {\frac{n}{n + 1}} + \sqrt {\frac{n}{n + 2}} + \sqrt {\frac{n}{n + 3}} + \dots + \sqrt {\frac{n}{n + 3(n - 1)}} \right]$ has the value equal to

$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$

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

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

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