A) \[f(x)\] is continuous at \[x=0\] but not differentiable at \[x=0\]
B) \[f(x)\] is continuous as well as differentiable at \[x=0\]
C) \[f(x)\] is discontinuous at \[x=0\]
D) None of these
Correct Answer: C
Solution :
| Given :\[f\left( x \right)=\left\{ \begin{align} & \frac{x}{\sqrt{{{x}^{2}}}}\,\,\,\,\,,\,\,x\ne 0 \\ & 0\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x=0 \\ \end{align} \right.\] |
| \[\therefore \]\[f\left( x \right)=\left\{ \begin{align} & \frac{x}{\left| x \right|}\,\,\,\,\,,\,\,\,x\ne 0 \\ & 0\,\,\,\,\,\,\,\,,\,\,\,x=0 \\ \end{align} \right.\] |
| \[\therefore \]\[f\left( 0 \right)=0\] |
| R.H.L |
| \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{0+h}{\left| 0+h \right|}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{h}=1\] |
| L.H.L |
| \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{(0-h)}{\left| (0-h) \right|}=\underset{h\to 0}{\mathop{\lim }}\,\frac{-h}{h}=-1\] |
| \[R.H.L\ne L.H.L\] |
| i.e. \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)\] |
| \[\therefore \]\[f\left( x \right)\]is discontinuous at \[x=0\] |
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