A) 1
B) ?1
C) Does not exist
D) None of these
Correct Answer: C
Solution :
\[\underset{x\to 2-}{\mathop{\lim }}\,\,\,\frac{|\,\,x-2\,\,|}{x-2}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{|\,\,2-h-2\,\,|}{2-h-2}=-1\] and \[\underset{x\to 2+}{\mathop{\lim }}\,\,\,\frac{|\,\,x-2\,\,|}{x-2}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{|\,\,2+h-2\,\,|}{2+h-2}=1\] Hence limit does not exist.
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