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$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sin x - 2\sin \frac{x}{2}} \right)}^2} + {{(1 - \cos x)}^3}}}{{\sin x\sin 2x - 8\cos x{{\sin }^2}\frac{x}{2} - \frac{4}{3}{{\sin }^4}x}} = ?$

 The given limit $ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {2\sin \frac{x}{2}\cos \frac{x}{2} - 2\sin \frac{x}{2}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{x}{2}} \right)}^3}}}{{2{{\sin }^2}x\cos x - 8\cos x{{\sin }^2}\frac{x}{2} - \frac{4}{3}{{\sin }^4}x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}\frac{x}{2}{{\left( {\cos \frac{x}{2} - 1} \right)}^2} + 8{{\sin }^6}\frac{x}{2}}}{{8{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}\cos x - 8\cos x{{\sin }^2}\frac{x}{2} - \frac{4}{3}.16{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}\frac{x}{2}.4{{\sin }^4}\frac{x}{4} + 8{{\sin }^6}\frac{x}{2}}}{{8{{\sin }^2}\frac{x}{2}\cos x\left( {{{\cos }^2}\frac{x}{2} - 1} \right) - \frac{{64}}{3}{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{8{{\sin }^2}\frac{x}{2}\left( {2{{\sin }^4}\frac{x}{4} + {{\sin }^4}\frac{x}{2}} \right)}}{{ - 8{{\sin }^2}\frac{x}{2}\cos x{{\sin }^2}\frac{x}{2} - \frac{{64}}{3}{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{8{{\sin }^2}\frac{x}{2}\left( {2{{\sin }^4}\frac{x}{4} + 16{{\sin }^4}\frac{x}{4}{{\cos }^4}\frac{x}{4}} \right)}}{{ - 8{{\sin }^4}\frac{x}{2}\cos x - \frac{{64}}{3}{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{16{{\sin }^2}\frac{x}{2}{{\sin }^4}\frac{x}{4}\left( {1 + 8{{\cos }^4}\frac{x}{4}} \right)}}{{ - 8{{\sin }^4}\frac{x}{2}\left( {\cos x + \frac{8}{3}{{\cos }^4}\frac{x}{2}} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{{\sin }^4}\frac{x}{4}\left( {1 + 8{{\cos }^4}\frac{x}{4}} \right)}}{{4{{\sin }^2}\frac{x}{4}{{\cos }^2}\frac{x}{4}\left( {\cos x + \frac{8}{3}{{\cos }^4}\frac{x}{2}} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - {{\tan }^2}\frac{x}{4}\left( {1 + 8{{\cos }^4}\frac{x}{4}} \right)}}{{2\left( {\cos x + \frac{8}{3}{{\cos }^4}\frac{x}{2}} \right)}} = 0$