A) 1
B) 0
C) ?1
D) None of these
Correct Answer: B
Solution :
Here \[f(0)=0\] Since \[-1\le \sin \frac{1}{x}\le 1\,\,\Rightarrow \,\,-|\,\,x\,\,|\,\,\le x\sin \frac{1}{x}\le \,\,|\,\,x\,\,|\] We know that \[\underset{x\to 0}{\mathop{\lim }}\,\,\,|\,\,x\,\,|\,=0\] and \[\underset{x\to 0}{\mathop{\lim }}\,\,\,|\,\,x\,\,|\,=0\] In this way \[\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)=0.\]
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