question_answer

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

Answer:

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 \[x=4\]