A) 0
B) 1
C) ?1
D) Does not exist
Correct Answer: D
Solution :
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\sin \,\left( \frac{-1}{h} \right)=\underset{h\to 0}{\mathop{\lim }}\,\,\,-\sin \frac{1}{h}\] = ? 1 (finite number lies between ? 1 to 1) \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\sin \left( \frac{1}{h} \right)\]= (finite number lies between 0 to 1) \[\because \,\,\,\,\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\]; \\[\underset{x\to 0}{\mathop{\lim }}\,\sin \left( \frac{1}{x} \right)\] does not exist.
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