Giải thích các bước giải:
Áp dụng: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\) ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sin x.\sin 2x.\sin 3x.....\sin nx}}{{n!.{x^n}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sin x.\sin 2x.\sin 3x.....\sin nx}}{{x.2x.3x.4x.....\left( {nx} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}.....\mathop {\lim }\limits_{x \to 0} \frac{{\sin nx}}{{nx}}\\
= 1.1.1.1.1.....1\\
= 1
\end{array}\)
