$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sum_{k=1}^{n} {k^2}}}{{{n^3}}}} \right] = $

$\mathop {\lim }\limits_{n \to \infty } \cos \left( {\pi \sqrt {{n^2} + n} } \right)$,where $n \in I$,is

$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to

Let $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots \left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}$ for $n \geq 2$. Then,the value of $\lim _{n \rightarrow \infty} x_{n}$ is

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|$,$x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)(x^2-5x+6)}{(x^2-6x+8)}$ is equal to

$\mathop {Lim}\limits_{x \to {0^ - }} \sin^{-1}([\tan x])$ $= l$,then $\{l\}$ is equal to,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer and fractional part functions,respectively.