The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ is:

For each positive integer $n$,let $y_n = \frac{1}{n} ((n+1)(n+2) \dots (n+n))^{\frac{1}{n}}$. For $x \in \mathbb{R}$,let $[x]$ be the greatest integer less than or equal to $x$. If $\lim_{n \rightarrow \infty} y_n = L$,then the value of $[L]$ is:

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{\sqrt {{n^2} + n} }} + \frac{1}{{\sqrt {{n^2} + 2n} }} + \dots + \frac{1}{{\sqrt {{n^2} + (n - 1)n} }}} \right]$ is equal to

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:

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

$\mathop {Lim}\limits_{n \to \infty } \,\,\sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}{x^2}}}} $,$x > 0$ is equal to