Question
Prove that $\text{f(x)}=\begin{cases}\frac{\text{x}-|\text{x}|}{\text{x}},&\text{x}\neq0\\2,&\text{x}=0\end{cases}$ is discontinuous at x = 0.

Answer

The given function can be rewritten as
$\text{f(x)}=\begin{cases}\frac{\text{x}-\text{x}}{\text{x}},&\text{when }\text{ x}>0\\\frac{\text{x}+\text{x}}{\text{x}},&\text{when }\text{ x}<0\\2,&\text{when }\text{ x}=0\end{cases}$
$\text{f(x)}=\begin{cases}0,&\text{when }\text{ x}>0\\2,&\text{when }\text{ x}<0\\2,&\text{when }\text{ x}=0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}2=2$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}0=0$
$\therefore\lim_\limits{\text{x}\rightarrow0^-}\text{f}(\text{x})\neq\lim_\limits{\text{x}\rightarrow0^+}\text{f}(\text{x})$
Thus, f(x) is discontinuous at x = 0

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