Question
Discuss the continuity of the function $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
✓
Answer
When $\text{x}\neq0,$
$\text{f(x)}=\frac{\text{x}}{|\text{x}|}=\begin{cases}\frac{-\text{x}}{\text{x}}=-1;&\text{x}<0\\\frac{\text{x}}{|\text{x}|}=1;&\text{x}>0\end{cases}$
So, f(x) is a constant function when $\text{x}\neq0,$
Hence, is continuous for all x < 0 and x > 0
Now, Consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{|-\text{h}|}=-1$
$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{|\text{h}|}=1$
So, $\text{LHL}\neq\text{RHL}$
Hence, function is discontinuous at x = 0
$\text{f(x)}=\frac{\text{x}}{|\text{x}|}=\begin{cases}\frac{-\text{x}}{\text{x}}=-1;&\text{x}<0\\\frac{\text{x}}{|\text{x}|}=1;&\text{x}>0\end{cases}$
So, f(x) is a constant function when $\text{x}\neq0,$
Hence, is continuous for all x < 0 and x > 0
Now, Consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{|-\text{h}|}=-1$
$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{|\text{h}|}=1$
So, $\text{LHL}\neq\text{RHL}$
Hence, function is discontinuous at x = 0
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