If $a, b, c$ and $k$ are non-zero real numbers and $\lim _{x \rightarrow \infty} x\left(a^{\frac{1}{x}}+b^{\frac{1}{x}}+c^{\frac{1}{x}}-3 k^{\frac{1}{x}}\right)=0$,then $k=$

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

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

The quadratic equation whose roots are $l$ and $m$,where
$\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text{ is}
\end{aligned}$

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to-

$\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ is equal to