Show that f (x)= |x-a| |x-a| , & when x 0 2, & when x = 0 is disc… - Vidyadip

Question

Show that $\text{f}\text{ (x)}=\begin{cases}\frac{\text{|x}-\text{a}|}{|\text{x}-\text{a}|}, & \text{when} \text{ x}\neq 0\\2, & \text{when}\text{ x} = 0\end{cases}$ is discontinuous at x = a.

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Answer

The given function can be rewritten as:
$\text{f}\text{(x)}=\begin{cases}\frac{\text{z}-\text{a}}{\text{z}-\text{a}}, & \text{when} \text{ x}> 0\\\frac{\text{a}-\text{x}}{\text{z}-\text{a}}, & \text{when}\text{ x} < 0\\ 1,&\text{when}\text{ x} = \text{a}\end{cases}$
$\text{f}\text{(x)}=\begin{cases}1, & \text{when} \text{ x}> \text{a}\\-1, & \text{when}\text{ x} < \text{a}\\ 1,&\text{when x}= \text{a}\end{cases}$
$\text{f}\text{(x)}=\begin{cases}1, & \text{when} \text{ x}\geq \text{a}\\-1, & \text{when}\text{ x} < \text{a}\end{cases}$
We observe
$(\text{LHL at x}=\text{a})=\lim\limits_{\text{z} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(a}-\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}(-1)=-1$
$(\text{RHL at x}=\text{a})=\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(a}+\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0}(1)=1$
$\therefore\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}$
Thus, f(x) is discontinuous at x = a

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