Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to \pi } \frac{{\sin 2x}}{{1 + {{\cos }^3}x}}\\
= \mathop {\lim }\limits_{x \to \pi } \frac{{2\sin x\cos x}}{{\left( {1 + \cos x} \right)\left( {1 - \cos x + {{\cos }^2}x} \right)}}\\
= \mathop {\lim }\limits_{x \to \pi } \frac{{2\sin x\cos x.\left( {1 - \cos x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)\left( {1 - \cos x + {{\cos }^2}x} \right)}}\\
= \mathop {\lim }\limits_{x \to \pi } \frac{{2\sin x\cos x\left( {1 - \cos x} \right)}}{{\left( {1 - {{\cos }^2}x} \right)\left( {1 - \cos x + {{\cos }^2}x} \right)}}\\
= \mathop {\lim }\limits_{x \to \pi } \frac{{2\sin x\cos x\left( {1 - \cos x} \right)}}{{{{\sin }^2}x.\left( {1 - \cos x + {{\cos }^2}x} \right)}}\\
= \mathop {\lim }\limits_{x \to \pi } \frac{{2\cos x\left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x + {{\cos }^2}x} \right)}}\\
\mathop {\lim }\limits_{x \to \pi } 2\cos x\left( {1 - \cos x} \right) = 2\cos \pi .\left( {1 - \cos \pi } \right) = 2.\left( { - 1} \right).\left( {1 - \left( { - 1} \right)} \right) = - 4\\
\mathop {\lim }\limits_{x \to \pi } \sin x\left( {1 - \cos x + {{\cos }^2}x} \right) = \sin \pi .\left( {1 - \cos \pi + {{\cos }^2}\pi } \right) = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to \pi } \frac{{2\cos x\left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x + {{\cos }^2}x} \right)}} = - \infty \\
\Rightarrow \mathop {\lim }\limits_{x \to \pi } \frac{{\sin 2x}}{{1 + {{\cos }^3}x}} = - \infty
\end{array}\)
