Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t} = 1\\
\mathop {\lim }\limits_{x \to a} \frac{{{{\sin }^2}x - {{\sin }^2}a}}{{{x^2} - {a^2}}}\\
= \mathop {\lim }\limits_{x \to a} \frac{{\left( {\sin x - \sin a} \right).\left( {\sin x + \sin a} \right)}}{{\left( {x - a} \right).\left( {x + a} \right)}}\\
= \mathop {\lim }\limits_{x \to a} \frac{{2.cos\frac{{x + a}}{2}.\sin \frac{{x - a}}{2}.2.\sin \frac{{x + a}}{2}.\cos \frac{{x - a}}{2}}}{{\left( {x - a} \right)\left( {x + a} \right)}}\\
= 4.\left( {\mathop {\lim }\limits_{x \to a} \frac{{\sin \frac{{x - a}}{2}}}{{x - a}}.\mathop {\lim }\limits_{x \to a} \left( {\frac{{\sin \frac{{x + a}}{2}}}{{x + a}}.\cos \frac{{x + a}}{2}.cos\frac{{x - a}}{2}} \right)} \right)\\
= 2.\mathop {\lim }\limits_{x \to a} \frac{{\sin \frac{{x - a}}{2}}}{{\frac{{x - a}}{2}}}.\left( {\frac{{\sin a}}{{2a}}.\cos a.1} \right)\\
= 2.1.\frac{{\sin a}}{{2a}}.\cos a\\
= \frac{{\sin 2a}}{{2a}}
\end{array}\)
