Question
show that $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-|\text{x}|}{2}, & \text{when} \text{ x}\neq 0\\2, & \text{when}\text{ x} = 0\end{cases}$ is discontinuous at x = 0.
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Answer
We want, to check the continuty of the function at x = 0.
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-|-\text{h|}}{2}=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-\text{h}}{2}=0$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}-\text{(|h|)}}{2}=0$
$\text{f}(0)=2$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence,The function is discotinuous at x = 0
This is rem ovable discontinuty.
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-|-\text{h|}}{2}=\lim\limits_{\text{h} \rightarrow 0}\frac{-\text{h}-\text{h}}{2}=0$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{ (x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}-\text{(|h|)}}{2}=0$
$\text{f}(0)=2$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence,The function is discotinuous at x = 0
This is rem ovable discontinuty.
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