A) 1
B) \[-1\]
C) 0
D) does not exist
Correct Answer: D
Solution :
\[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|}{h}=1\] \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|-h|}{-h}=-1\] \[\because \] \[f(0+0)\ne f(0-0)\] \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{|x|}{x}\]does not exist.
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