Question

Test the continuity of the function on f(x) at the origin:
$f\left(x\right)=\left\{\begin{array}{ll}\frac{x}{\left|x\right|},& x\ne 0\\ 1,& x=0\end{array}\right.$

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{x}{\left|x\right|},x\ne 0\\ 1,x=0\end{array}\right.$

We observe
(LHL at x = 0) =$\underset{x\mathit{\to }{0}^{\mathit{-}}}{\lim }f\left(x\right)=\underset{h\mathit{\to }0}{\lim }f\left(0-h\right)=\underset{h\mathit{\to }0}{\lim }f\left(-h\right)$
=​$\underset{h\to 0}{\lim }\frac{-h}{\left|-h\right|}=\underset{h\to 0}{\lim }\frac{-h}{h}=\underset{h\to 0}{\lim }-1=-1$

(RHL at x = 0)​ =$\underset{x\mathit{\to }{0}^{+}}{\lim }f\left(x\right)=\underset{h\mathit{\to }0}{\lim }f\left(0+h\right)=\underset{h\mathit{\to }0}{\lim }f\left(h\right)$
=​$\underset{h\to 0}{\lim }\frac{h}{\left|h\right|}=\underset{h\to 0}{\lim }\frac{h}{h}=\underset{h\to 0}{\lim }1=1$


$\therefore \underset{x\mathit{\to }{\mathit{0}}^{+}}{\lim }f\left(x\right)\ne \underset{x\mathit{\to }{\mathit{0}}^{-}}{\lim }f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at the origin.