Discuss the continuity of $f(x)=\sin |x|$.
Let $f(x)=\sin |x|$
This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,
$f=hog$, where $g(x)=|x|$ and $h(x)=\sin x$
$[\because h o g(x)=h(g(x))=h(|x|)=\sin |x|]$
It has to be proved first that $g(x)=|x|$ and $h(x)=\sin x$ are continuous functions.
$g(x)=|x|$ can be written as
$g(x)= \begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}$
Clearly, $g$ is defined for all real numbers.
Let $c$ be a real number.
Case I:
If $c<0$, then $g(c)=-c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c$
$\therefore \lim _{x \rightarrow c} g(x)=g(c)$
Therefore, $g$ is continuous at all points $x<0$
Case III:
If $c=0$, then $g(c)=g(0)=0$
$\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0$
$\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0$
$\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0)$
Therefore, $g$ is continuous at $x=0$
From the above three observations, it can be concluded that $g$ is continuous at all points.
Now, $h(x)=\sin x$
It is evident that $h(x)=\sin x$ is defined for every real number.
Let $c$ be a real number.
Put $x=c+k$
If $x \rightarrow c$, then $k \rightarrow 0$
$h(c)=\sin c$
$h(c)=\sin c$
$\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c} \sin x$
$=\lim _{k \rightarrow 0} \sin (c+k)$
$=\lim _{k \rightarrow 0}[\sin c \cos k+\cos c \sin k]$
$=\lim _{k \rightarrow 0}(\sin c \cos k)+\lim _{h \rightarrow 0}(\cos c \sin k)$
$=\sin c \cos 0+\cos c \sin 0$
$=\sin c+0$
$=\sin c$
$\therefore \lim _{x \rightarrow c} h(x)=g(c)$
So, $h$ is a continuous function.
$\therefore f(x)=h o g(x)=h(g(x))=h(|x|)=\sin |x|$ is a continuous function.