A) continuous at\[x=0\]
B) not continuous at\[x=0\]
C) both continuous and differentiable at\[x=0\]
D) not defined at\[x=0\]
E) continuous but not differentiable at\[x=0\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{1-{{e}^{-1/x}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{1/x}}}{{{e}^{1/x}}-1}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{1/x}}}{{{e}^{1/x}}}\] (by LHospitals rule) \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{1-{{e}^{-1/x}}}=1\] and \[f(0)=0\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{1-{{e}^{-1/x}}}\ne f(0)\] \[\therefore \]Function is not continuous at\[x=0\].
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