Find all points of discontinuity of f, where f is defined by

Question:

Find all points of discontinuity of f, where f is defined by

$f(x)=\left\{\begin{array}{l}\frac{|x|}{x} \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$

Solution:

The given function $f$ is $f(x)=\left\{\begin{array}{l}\frac{|x|}{x} \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$

It is known that, $x<0 \Rightarrow|x|=-x$ and $x>0 \Rightarrow|x|=x$

Therefore, the given function can be rewritten as

$f(x)=\left\{\begin{array}{l}\frac{|x|}{x}=\frac{-x}{x}=-1 \text { if } x<0 \\ 0, \text { if } x=0 \\ \frac{|x|}{x}=\frac{x}{x}=1, \text { if } x>0\end{array}\right.$

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 $f(c)=-1$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-1)=-1$

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

Case II:

If $c=0$, then the left hand limit of $f$ at $x=0$ is,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1$

The right hand limit of at x = 0 is,

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(1)=1$

It is observed that the left and right hand limit of f at x = 0 do not coincide.

Therefore, f is not continuous at x = 0

Case III:

If $c>0$, then $f(c)=1$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(1)=1$

$\therefore \lim _{x \rightarrow 0} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>0$

Hence, x = 0 is the only point of discontinuity of f.