question_answer

Is the function \[f(x)=\left\{ \begin{matrix}    \frac{{{e}^{1/x}}-1}{{{e}^{1/x}}+1}, & \text{if}\,\,x\ne 0  \\    0, & \text{if}\,\,x=0  \\ \end{matrix} \right.\] continuous at x = 0?

Answer:

At \[x=0,\]\[LHL=\underset{h\,\to \,0}{\mathop{\lim }}\,f\,(0-h)=\underset{h\,\to \,0}{\mathop{\lim }}\,\frac{{{e}^{-\,1/h}}-1}{{{e}^{-\,/h}}+1}\] \[=\underset{h\,\to \,0}{\mathop{\lim }}\,\frac{\frac{1}{{{e}^{1/h}}}-1}{\frac{1}{{{e}^{1/h}}}+1}=\frac{0-1}{0+1}=-\,1\] \[RHS=\underset{h\,\to \,0}{\mathop{\lim }}\,f\,(0+h)=\underset{h\,\to \,0}{\mathop{\lim }}\,\frac{{{e}^{1/h}}-1}{{{e}^{1/h}}+1}\] \[=\underset{h\,\to \,0}{\mathop{\lim }}\,\frac{1-\frac{1}{{{e}^{1/h}}}}{1+\frac{1}{{{e}^{1/h}}}}=\frac{1-0}{1+0}=1\] \[\therefore \]\[LHL\ne RHL\] Hence, the given function is discontinuous at\[x=0\].