A) \[m>0\]
B) \[m<0\]
C) \[m=0\]
D) any value of\[m\]
Correct Answer: A
Solution :
Since \[f(x)\] is continuous at\[x=0\], then, \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f(0)=0\] Now, \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,{{(-h)}^{m}}\sin \left( -\frac{1}{h} \right)\] \[=-\underset{h\to 0}{\mathop{\lim }}\,{{(-h)}^{m}}\sin \left( \frac{1}{h} \right)\] \[=0\], only when\[m>0\] and \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,{{h}^{m}}\sin \left( \frac{1}{h} \right)=0,\]only when\[m>0\] Hence, \[f(x)\] is continuous at\[x=0\], if\[m>0\].
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