Question

If $f(x)=\frac{e^{1/x}}{1+e^{1/x}}$ for $x\ne 0$ and $f(0)=0,$ then at $x=0$ the function $f(x)$ is

• A. continuous
• B. discontinuous
• C. increasing
• D. differentiable

Answer 1

Answer: (B)

Given, $f(x)=\frac{e^{1/x}}{1+e^{1/x}},f(0)=0$
$LHL=\underset{x\to 0^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }f(0-h)$
$=\underset{h\to 0}{\lim }\frac{e^{-1/h}}{1+e^{-1/h}}=0$
$RHL=\underset{x\to 0^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(0+h)$
$=\underset{h\to 0}{\lim }\frac{e^{1/h}}{1+e^{1/h}}=1$
Since, $LHL\ne RHL$
So, $f(x)$ is discontinuous at $x=0.$