# Find all points of discontinuity of $$f$$, where $$f$$ is defined by $f(x) = \left\{ \begin{array} {1 1} \large \frac{ |\;x\;| }x,& \quad\text{ if x $$\neq$$ 0 }\\ 0 ,& \quad \text{if$x$=0}\\ \end{array} \right.$

$\begin{array}{1 1} \text{The point of discontinuity is x=1} \\ \text{The point of discontinuity is x=-1} \\ \text{The point of discontinuity is x=0} \\ \text{ The point of discontinuity is x=2} \end{array}$

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
LHL :$\lim\limits_{\large x\to 0^-}\big(\large\frac{\mid x\mid}{x}\big)$$=-1. RHL:\lim\limits_{\large x\to 0^+}\big(\large\frac{\mid x\mid}{x}\big)$$=1.$
LHL$\neq$RHL $\neq$ f(0)
$f$ is discontinuous at $x=0$
Step 2:
At $x=c<0,\lim\limits_{\large x\to c}\big(\large\frac{-x}{x}\big)$$=-1 f(c)=-1 Therefore \lim\limits_{\large x\to c}f(x)=f(c)\Rightarrow f is continuous at x=c<0 Step 3: At x=c>0\lim\limits_{\large x\to c}\big(\large\frac{\mid x\mid}{x}\big)=$$\lim\limits_{\large x\to c}\large\frac{x}{x}$$=1$
$\qquad\qquad\qquad\qquad\qquad\qquad=f(c)$
Therefore $\lim\limits_{\large x\to c}f(x)=f(c)\Rightarrow f$ is continuous at x=c>0
Therefore the point of discontinuity is $x=0$
answered May 27, 2013