question_answer

Examine the continuity of the function \[f(x)=\left\{ \begin{matrix}    \frac{|\sin \,x|}{x} & x\ne 0  \\    1, & x=0\,\,\text{at}\,\,x=0  \\ \end{matrix} \right.\]

Answer:

We have f(0) = 1 \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|sin(0+h)|}{(0+h)}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|sin\,h|}{-\,h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{h}=1\] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,(0-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|sin(-\,h)|}{-\,h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|-\sin \,h|}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{-\,h}=-\,1\] \[\therefore \]      \[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\ne \underset{x\,\to \,{{0}^{-}}}{\mathop{\lim }}\,f(x).\] So, \[\underset{x\,\to \,0}{\mathop{\lim }}\,f(x)\] does not exist Hence, f(x) is discontinuous at x = 0.