question_answer

For the function   \[x=0\], which of the following is correct?

A) \[\underset{x\to 0}{\mathop{\lim }}\,\,f(x)\]does not exist

B)   \[\underset{x\to 0}{\mathop{\lim }}\,\,f(x)=1\]

C)   \[\underset{x\to 0}{\mathop{\lim }}\,\,f(x)\] exist but \[f(x)\]is not continuous at \[x=0\]

D)   \[f(x)\]is continuous at \[x=0\]

Correct Answer: B

Solution :

We have the function Taking limit on both side. \[\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{e}^{1/x}}-1}{{{e}^{1/x}}+1}=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{1-{{e}^{-1/x}}}{1+{{e}^{-1/x}}} \right]\] \[=\frac{1}{1}=1\]                \[\left[ \because \,\,{{e}^{1-/x}}={{e}^{-\infty }}=0 \right]\] \[\Rightarrow \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)=1\] and \[f(0)=0\] \[\therefore \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)\]exists but \[f(x)\] is not continuous at \[x=0\]