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

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 + x} - \sqrt {3 - x} }}{x} = $

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

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

If $a > 0$,$[\cdot]$ denotes the greatest integer function,$\lim _{x \rightarrow a^{-}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=k$,and $\lim _{x \rightarrow a^{+}}\left(\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right)=l$,then:

Assertion $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
Reason $(R)$: As the value of $x$ decreases,the value of $\frac{1}{x}$ increases.