• # question_answer $f(x)=\left\{ \begin{matrix} \frac{|x-4|}{2(x-4)}, & \text{if}\,\,x\ne 4 \\ 0, & \text{if}\,\,x=4 \\ \end{matrix}at\,\,x=4. \right.$

At $x=4,$ $LHL=\underset{h\to 0}{\mathop{\lim }}\,\,f(A-h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|A-h-4|}{2(4-h-4)}$ $=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\frac{|-h|}{2\,(-h)}=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\frac{h}{-\,2h}=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\left( -\frac{1}{2} \right)=-\frac{1}{2}$ $RHL=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\,f\,(A+h)=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\frac{|4+h-4|}{2\,(4+h-4)}$ $=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\frac{|h|}{2h}=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\frac{h}{2h}=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,\left( \frac{1}{2} \right)=\frac{1}{2}$ and $f(4)=0$ $LHL\ne RHL\ne f(4)$ Hence, the given function is discontinuous at $x=4$