Question

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

$f(x) = \begin{cases} \frac {|x|} {x}, & \quad \text{if } x { \neq 0}\\\ 0, & \quad \text{if } x { =0} \end{cases}$

Answer 1

$f(x) = \begin{cases} \frac {|x|} {x}, & \quad \text{if } x { \neq 0}\\\ 0, & \quad \text{if } x { =0} \end{cases}$

It is known that, x<0 $\implies$ |x| = -x and x>0 $\implies$ |x| = 0
Therefore,the given function can be rewritten as
$f(x) = \begin{cases} \frac {|x|} {x}=\frac {-x}{x}=-1, & \quad \text{if } x { <0}\\\ 0, & \quad \text{if } x { =0} \\\ \frac {|x|} {x}=\frac {x}{x}=1, & \quad \text{if } n \text{ >0} \end{cases}$
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
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (-1) = -1
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x<0

Case (ii):
If c = 0,then the left hand limit of f at x = 0 is,
$\lim\limits_{x \to 0^-}$ f(x) = $\lim\limits_{x \to 0^-}$ (-1)=-1
The right hand limit of f at x = 0 is,
$\lim\limits_{x \to 0^+}$ f(x) = $\lim\limits_{x \to 0^+}$ (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):
Ifc>0, then f(c) = 1
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (1) =1
∴$\lim\limits_{x \to c}$ 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.