Let $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$. The smallest positive integer $k$ such that $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$ is

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$ is equal to

If a function $f$ is defined by $f(x) = \frac{\cot^3 x - \tan x}{\cos(x + \pi/4)}$ for $x \neq \pi/4$,then $\lim_{x \rightarrow \pi/4} f(x) = $

$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4} = $

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