A) \[f(0+0)=1\]
B) \[f(0-0)=1\]
C) f is continuous at\[x=0\]
D) None of these
Correct Answer: C
Solution :
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,f(x)={{x}^{2}}\sin \frac{1}{x},\] but \[-1\le \sin \frac{1}{x}\le 1\] and \[x\to 0\] Therefore, \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,f(x)=0=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,f(x)=f(0)\] Hence \[f(x)\] is continuous at \[x=0.\]
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