$\mathop {\lim }\limits_{x \to 0}(\frac{\sqrt[3]{\cos x}-1 }{\sin 2x}+\frac{1-\sqrt{\cos x} }{\sin2 x})$$=\mathop {\lim }\limits_{x \to 0}(\frac{\cos x-1}{\sin 2x(\sqrt[3]{\cos ^{2}x}+\sqrt[3]{\cos x} +1) }+\frac{1-\cos x}{\sin 2x(1+\sqrt{\cos x}) })$
$=\mathop {\lim }\limits_{x \to 0}(\frac{-2\sin ^{2}\frac{x}{2}}{4\sin \frac{x}{2}\cos \frac{x}{2}\cos x(\sqrt[3]{\cos ^{2}x} +\sqrt[3]{\cos x}+1) }+\frac{2\sin^{2} \frac{x}{2}}{4sin\frac{x}{2}cos\frac{x}{2}cosx(\sqrt{cosx}+1) })$
$rut gon sinx/2 o hai phan so di roi thay x=0 dk ket qua bang 0 nha! $