The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ is

The $\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}$ equals

$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = $

$\lim \limits_{x \rightarrow 0} \left(\tan \left(\frac{\pi}{4}+x\right)\right)^{\frac{1}{x}}$ is equal to

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right]$