$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) + \lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) =$

The limiting value of the function $f(x) = \frac{2\sqrt{2} - (\cos x + \sin x)^3}{1 - \sin 2x}$ as $x \to \frac{\pi}{4}$ is

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 1} \frac{4x+3}{x-2}$

The right hand and left hand limit of the function $f(x)$ are respectively:
$f(x)=\begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

$\lim _{x \rightarrow 0} \frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}} = $

If $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$,then the value of $6 \beta$ is