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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} $

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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 by sreemathi.v
 

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