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

$\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$ is

If ${S_n} = \sum\limits_{k = 1}^n {{a_k}} $ and $\mathop {\lim }\limits_{n \to \infty } {a_n} = a,$ then $\mathop {\lim }\limits_{n \to \infty } \frac{{{S_{n + 1}} - {S_n}}}{{\sqrt {\sum\limits_{k = 1}^n k } }}$ is equal to

Let $f(x)$ be a differentiable function such that $f(0)=0$ and $f^{\prime}(0)=20$. For $x \in \left(0, \frac{\pi}{2}\right]$,if $A(x)=2 f(x) \operatorname{cosec} 4 x+4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x$,then $\lim _{x \rightarrow 0} A(x)=$

$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x} = $