To prove that sine is continuous at\(a\)as shown below:
\(\begin{aligned}\mathop {\lim }\limits_{h \to 0} \sin \left( {a + h} \right) &= \mathop {\lim }\limits_{h \to 0} \left( {\sin a\cosh + \cos a\sinh } \right)\,\,\,\,\left( {\sin \left( {{\mathop{\rm A}\nolimits} + B} \right) = \sin {\mathop{\rm A}\nolimits} \cos {\mathop{\rm B}\nolimits} + \cos {\mathop{\rm A}\nolimits} \sin {\mathop{\rm B}\nolimits} } \right)\\ &= \mathop {\lim }\limits_{h \to 0} \left( {\sin a\cosh } \right) + \mathop {\lim }\limits_{h \to 0} \left( {\cos a\sinh } \right)\\ &= \left( {\mathop {\lim }\limits_{h \to 0} \sin a} \right)\left( {\mathop {\lim }\limits_{h \to 0} \cosh } \right) + \left( {\mathop {\lim }\limits_{h \to 0} \cos a} \right)\left( {\mathop {\lim }\limits_{h \to 0} \sinh } \right)\\ &= \left( {\sin a} \right)\left( 1 \right) + \left( {\cos a} \right)\left( 0 \right)\left( {{\mathop{\rm since}\nolimits} \,\,\mathop {\lim }\limits_{\theta \to 0} \sin \theta = 0,\mathop {\lim }\limits_{\theta \to 0} \cos \theta = 1} \right)\\ &= \sin a\end{aligned}\)
Thus, it is proved that\(\mathop {\lim }\limits_{h \to 0} \sin \left( {a + h} \right) = \sin a\).