Giá trị của \(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sin 3x}}{{1 - \cos x}}\) bằng 

Giải chi tiết:

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

\( = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\sin \frac{x}{2}\left( {6\cos \frac{x}{2} - 32{{\sin }^2}\frac{x}{2}{{\cos }^3}\frac{x}{2}} \right)}}{{2{{\sin }^2}\frac{x}{2}}}\)

\( = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{3\cos \frac{x}{2} - 16{{\sin }^2}\frac{x}{2}{{\cos }^3}\frac{x}{2}}}{{\sin \frac{x}{2}}}\)

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

\( = \mathop {\lim }\limits_{x \to {0^ - }} \cot \frac{x}{2}(3 - 4{\sin ^2}x) = - \infty \)

Đáp án cần chọn là: C