Solution:
(a) Let an arbitrary real number be “a” therefore, $\lim _{x \rightarrow a^{-}} f(x) \Rightarrow \lim _{h \rightarrow 0} f(a+h)$
So now, $\lim _{h \rightarrow 0} f(a+h)=\lim _{h \rightarrow 0} \sin (a+h) \cdot \cos (a+h)$
$=\lim _{h \rightarrow 0}(\sin a \cos h+\cos a \sin h)(\cos a \cos h-\sin a \sin h)$ $=(\sin a \cos 0+\cos a \sin 0)(\cos a \cos 0-\sin a \sin 0)$
$=(\sin a+0)(\cos a-0)$
$=\sin a \cdot \cos a=f(a)$
In the similar way, $\lim _{x \rightarrow a^{-}} f(x)=f(a)$
$\lim _{x \rightarrow a^{-}} f(x)=f(a)=\lim _{x \rightarrow a^{-}} f(x)$
As a result, $f(x)$ is continuous at $x=a$.
As, an arbitrary real number is “a”, as a result, $f(x)=\sin x \cdot \cos x$ is continuous.