Question

Let $f\left(x\right)=\frac{1}{1+e^{1/x}}$ for $x\ne 0$ and $f\left(0\right)=0.$ Then

• A. $f\left(0^{+}\right)$ does not exist
• B. $f\left(0^{-}\right)$ is equal to zero
• C. $f^{\prime }\left(0^{-}\right)$ is equal to $1$
• D. $f^{\prime }\left(0^{+}\right)$ is equal to zero

Answer 1

The correct option is D $f^{\prime }\left(0^{+}\right)$ is equal to zero

$f\left(x\right)=\{\begin{array}{cc}\frac{1}{1+e^{1/x}},& x\ne 0\\ 0,& x=0\end{array}$

$LHD=f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{\frac{1}{1+e^{-1/h}}}{-h}\to -\infty$

$RHD=f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{\frac{1}{1+e^{1/h}}}{h}$
$=\underset{h\to 0}{lim}\frac{1/h}{1+e^{1/h}}$
$=\underset{h\to 0}{lim}\frac{\frac{-1}{h^2}}{e^{1/h}\left(\frac{-1}{h^2}\right)}=0$


$f\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{1}{1+e^{1/h}}=0$
$f\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{1}{1+e^{-1/h}}=1$