Question

Examine that sin|x| is a continuous function.

Answer 1

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=goh, where g(x)=|x| and h(x)=sinx[∵(goh)(x)=g(h(x))=g(sinx)=|sinx|=f(x)]
It has to be first proved that g(x)=|x| and h(x)=sinx are continuous functions.
g(x)=|x| can be written as
g(x)=g(x)=\left\\{\begin{matrix} -x &if\,x<0 \\\ x&if\,x\geq0 \end{matrix}\right.
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
∴$\lim_{x\rightarrow c}$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
∴$\lim_{x\rightarrow c}$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
$\lim_{x\rightarrow 0^+}$ g(x)=$\lim_{x\rightarrow 0^+}$(x)=0
∴$\lim_{x\rightarrow 0^-}$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)=cos x It is evident that h(x)=sinx is defined for every real number.
Let c be a real number.
Put x=c+k If x→c, then k→0
h(c)=sinc
$\lim_{x\rightarrow c}$h(x)=$\lim_{x\rightarrow c}$ sinx
=$\lim_{x\rightarrow 0}$sin(c+k)
=$\lim_{x\rightarrow c}$[sinccosk+coscsink]
=$\lim_{x\rightarrow c}$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
∴$\lim_{x\rightarrow c}$h(x)=g(c)

Therefore,h is a continuous function.
It is known that for real-valued functions g and h,such that (goh) is defined at c,if g is continuous at c and if f is continuous at g(c), then (fog) is continuous at c.
Therefore,f(x)=(goh)(x)=g(h(x))=g(sinx)=|sinx|is a continuous function.