Let $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \left( \frac{n}{\sqrt{n^4+r^4}} - \frac{2 n r^2}{(n^2+r^2) \sqrt{n^4+r^4}} \right) = \frac{\pi}{k}.$ Using only the principal values of the inverse trigonometric functions,then $k^2$ is equal to:

$\lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2n+1}}+\frac{n}{(n+2) \sqrt{2(2n+2)}}+\frac{n}{(n+3) \sqrt{3(2n+3)}}+\ldots n \text{ terms}\right]=\int_0^1 f(x) d x$,then $f(x)=$

$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\left( {1 + \frac{{{1^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{3^2}}}{{{n^2}}}} \right) \dots \left( {1 + \frac{{{{(n - 1)}^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}$ equals to:

If $[x]$ denotes the greatest integer $\le x$,then $\mathop {\text{Limit}}\limits_{n \to \infty } \frac{1}{n^4} \left( [1^3 x] + [2^3 x] + \dots + [n^3 x] \right)$ equals

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

$\lim _{n}$ ${\rightarrow \infty} \frac{1}{n}\left[\sin \frac{\pi}{4}+\sin \frac{\pi}{12}\left(3+\frac{1}{n}\right)+\sin \frac{\pi}{12}\left(3+\frac{2}{n}\right)+\ldots+\sin \frac{\pi}{3}\right]=$