$$\lim_{x\to 0}\frac{\sin^3x-x^3\operatorname{sgn}\left(1-\left[\frac{x}{\sin^{-1}x}\right]\right)}{x\tan^2 x\sin(\pi\cos x)}$$
(Note that here $\operatorname{sgn}$ denotes the Signum function and $[x]$ denotes the greatest integer less than or equal to $x$.)
The above simplifies to:
$$\lim_{x→0} \frac{\sin^{3}(x)-x^{3}}{x\tan^{2}(x)\sin(\pi(1-\cos x))}$$
then after further simplification i get:
$$=\lim_{x→0} \frac{6}{π}\frac{\sin(x)-x}{x^{3}}$$
$$=\lim_{x→0} \frac{6}{\pi}\frac{-1}{6}=\frac{-1}{π},$$
but in this method I have to calculate the limit of
$$\lim_{x→0} \frac{\sin(x)-x}{x^{3}}=\frac{-1}{6},$$
which is lengthy. Is there any other way to do this without calculating this limit.
We have that
$$\frac{\sin^{3}x-x^{3}}{x\tan^{2}x\sin(π(1-\cos x))} =\frac{x^3-\frac12x^5+o(x^5)-x^3}{x^3(x^2+o(x^2))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}} =\frac{-\frac12x^5+o(x^5)}{(x^5+o(x^5))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}} =\frac{-\frac12+o(1)}{(1+o(1))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}\to \frac{-\frac12}{1\cdot\frac \pi 2\cdot 1}=-\frac 1 \pi$$
Without Taylor's expansion note that we have
$$\frac{\sin^{3}x-x^{3}}{x\tan^{2}x\sin(π(1-\cos x))} =\frac{\sin^{3}x-x^{3}}{x^5}\frac{1}{\frac{\tan^2x}{x^2} \frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}$$
then since by standard limits the second part $\to \frac 2 \pi $ we can analize
$$\frac{\sin^{3}x-x^{3}}{x^5}=\frac{(\sin x-x)(\sin^2 x+\sin x \cdot x+x^2)}{x^5}=\frac{\sin x-x}{x^3}\frac{\sin^2 x+\sin x \cdot x+x^2}{x^2}$$
and since by standard limits $\frac{\sin^2 x+\sin x \cdot x+x^2}{x^2}\to 3$ we need to consider
$$\frac{\sin x-x}{x^3}\to -\frac 16$$
which can be solved as indicated here Are all limits solvable without L'Hôpital Rule or Series Expansion.