Question

Find all points of discontinuity of $f$, where $f$ is defined by $f\left(x\right)=$ $\{\begin{array}{c}\frac{|x|}{x},\text{if}x\ne 0\\ 0,\text{if}x=0\end{array}$

Answer 1

The given function is $f\left(x\right)=\{\begin{array}{c}\frac{|x|}{x},\text{if}x\ne 0\\ 0,\text{if}x=0\end{array}$
It is known that, for $x<0,|x|=-x$ and for $x>0,|x|=x$
Therefore given function can be rewritten as
$f\left(x\right)=\{\begin{array}{c}\frac{|x|}{x}=\frac{-x}{x}=-1,\text{if}x<0\\ 0,\text{if}x=0\\ \frac{|x|}{x}=\frac{x}{x}=1,\text{if}x>0\end{array}$
The given function $f$ is defined at all the points of the real line.
Let $c$ be a point on the real line.
Case I:
If $c<0$, then $f\left(c\right)=-1$
$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(-1)=-1$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x<0$
Case II :
If $c=0$, then the left hand limit of f at $x=0$ is,
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}(-1)=-1$
The right hand limit of $f$ at $x=0$ is,
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\left(1\right)=1$
It is observed that the left and right hand limit of $x=0$ do not coincide.
Therefore, $f$ is not continuous at $x=0$
Case III :
If $c>0$, then $f\left(c\right)=1$
$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}\left(1\right)=1$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x$, such that $x>0$
Hence, $x=0$ is the only point of discontinuity of $f$.