Question

If $f\left(x\right)=\{\begin{array}{c}\frac{x}{1+exp\left(1/x\right)},x\ne 0\\ 0,x=0\end{array}$, then $f\left(x\right)$ at $x=0$ is

• A. Continuous
• B. Not continuous
• C. Differentiable
• D. Not differentiable

Answer 1

Answer: (A), (D)

Continuous
Not differentiable

Given $f\left(x\right)=\{\begin{array}{c}\frac{x}{1+exp\left(1/x\right)},x\ne 0\\ 0,x=0\end{array}$,
Continuity at $x=0$
$LHL={lim}_{h\to 0}\frac{0-h}{1+e^{1/(0-h)}}$
$={lim}_{h\to 0}\frac{-h}{1+e^{-1/\left(h\right)}}=\frac{-0}{1}=0$
$={lim}_{h\to 0}\frac{h}{1+e^{1/\left(h\right)}}=\frac{0}{1+\infty }=0$
$\therefore$ $LHL=RHL=f\left(0\right)=0$
Hence it is continuous
Differentiability at $x=0$
$LHD={lim}_{h\to 0}\frac{f(0-h)-f\left(0\right)}{0-h}={lim}_{h\to 0}\frac{\frac{0-h}{1+e^{1/(0-h)}}-0}{-h}=1$
$RHD={lim}_{h\to 0}\frac{f(0+h)-f\left(0\right)}{0+h}={lim}_{h\to 0}\frac{\frac{0+h}{1+e^{1/(0+h)}}-0}{h}=0$
$\therefore$ $LHD\ne RHD$
Hence it is not differentiable at $x=0$