Đáp án: 13/2
Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 3x + 2x.\sin x}}{{\sin {x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 - \cos 3x}}{{{x^2}}} + \frac{{2x.\sin x}}{{{x^2}}}}}{{\frac{{\sin {x^2}}}{{{x^2}}}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{2{{\sin }^2}\left( {\frac{{3x}}{2}} \right)}}{{{x^2}}} + \frac{{2\sin x}}{x}}}{{\frac{{\sin {x^2}}}{{{x^2}}}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{2.\frac{9}{4}\frac{{{{\sin }^2}\left( {3x/2} \right)}}{{{{\left( {3x/2} \right)}^2}}} + 2.\frac{{\sin x}}{x}}}{{\frac{{\sin {x^2}}}{{{x^2}}}}}\\
= \frac{{\frac{9}{2}.1 + 2.1}}{1} = \frac{{13}}{2}
\end{array}$
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin a}}{a} = 1$
