A) \[-1\]
B) \[\infty \]
C) Non-existent
D) None of these
Correct Answer: C
Solution :
\[f(1+0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1+h)-f(1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{[1+h]}{|1+h|}-\frac{[1]}{|1|}}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( \frac{1}{1+h} \right)-1}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{-h}{h(1+h)}=-1\] \[f(1-0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1-h)-f(1)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{[1-h]}{|1-h|}-\frac{[1]}{|1|}}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{0-1}{-h} \right)=\infty \] \[\therefore \] \[r(1+0)\ne f(1-0)\] So, \[f(1)\]does not exist.
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