A) continuous
B) discontinuous
C) increasing
D) differentiable
Correct Answer: B
Solution :
Given, \[f(x)=\frac{{{e}^{1/x}}}{1+{{e}^{1/x}}},\,\,f(0)=0\] \[LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{-1/h}}}{1+{{e}^{-1/h}}}=0\] \[RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{1/h}}}{1+{{e}^{1/h}}}=1\] Since, \[LHL\ne RHL\] So, \[f(x)\] is discontinuous at \[x=0.\]
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