A) \[\underset{x\to 0+}{\mathop{\lim }}\,f(x)=1\]
B) \[\underset{x\to 0-}{\mathop{\lim }}\,f(x)=1\]
C) \[f(x)\]is continuous at\[x=0\]
D) None of these
Correct Answer: C
Solution :
\[f(0)=0\]; \[f(0-)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{-h}{{{e}^{-1/h}}+1}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{-h}{1+\frac{1}{{{e}^{1/h}}}}=0\] \[f(0+)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{h}{{{e}^{1/h}}+1}=0.\]
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