If $\lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4$,then $k$ is equal to

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

$\lim _{x \rightarrow 0} \frac{(\operatorname{cosec} x-\cot x)(e^x-e^{-x})}{\sqrt{3}-\sqrt{2+\cos x}} = $

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

$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}} + \frac{3}{{{n^2}}} + \dots + \frac{n}{{{n^2}}}} \right\}$ is

If $\lim_{x \rightarrow 0} [1 + x \ln(1 + b^2)]^{\frac{1}{x}} = 2b \sin^2 \theta$,where $b > 0$ and $\theta \in (-\pi, \pi]$,then the value of $\theta$ is