A) Equal to 0
B) Equal to 1
C) Equal to ?1
D) Indeterminate
Correct Answer: A
Solution :
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)=0\] and \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\,-(0+h)=0\] \[\therefore \,\,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)=0\], \[\left( \because \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x) \right)\].
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