Examine that

Solution:

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,

$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\sin x$

$[\because(g o h)(x)=g(h(x))=g(\sin x)=|\sin x|=f(x)]$

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 $\lim _{x \rightarrow c} g(x)=\lim _{y \rightarrow c}(-x)=-c$

$\therefore \lim _{x \rightarrow c} g(x)=g(c)$

Therefore, g is continuous at all points x, such that x < 0

 

Case II:

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, such that x > 0

 

Case III:

If $c=0$, then $g(c)=g(0)=0$

$\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0$

$\lim _{x \rightarrow 0^{0}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0$

$\therefore \lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0)$

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

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

$h(c)=\sin c$

$\begin{aligned} \lim _{x \rightarrow c} h(x) &=\lim _{x \rightarrow c} \sin x \\ &=\lim _{k \rightarrow 0} \sin (c+k) \\ &=\lim _{k \rightarrow 0}[\sin c \cos k+\cos c \sin k] \\ &=\lim _{k \rightarrow 0}(\sin c \cos k)+\lim _{h \rightarrow 0}(\cos c \sin k) \\ &=\sin c \cos 0+\cos c \sin 0 \\ &=\sin c+0 \\ &=\sin c \end{aligned}$

$\therefore \lim _{x \rightarrow c} h(x)=g(c)$

Therefore, h is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

Therefore, $f(x)=(g o h)(x)=g(h(x))=g(\sin x)=|\sin x|$ is a continuous function.