$\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$

Let $[.]$ denote the greatest integer function. Assertion $(A) : \lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$. Reason $(R) : f(x) = x - 1, g(x) = [x], h(x) = x$ and $\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{h(x)}{x} = 1$.

If $f(x) = \text{Sgn}(\text{Sgn}(\text{Sgn}(x)))$,then $\mathop {\lim }\limits_{x \to 0} f(x)$ is equal to :-

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(\tan^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}} = $

If $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$,then the value of $ab^3$ is

The value of $\mathop {\lim }\limits_{x \to 2} \frac{{{3^{x/2}} - 3}}{{{3^x} - 9}}$ is