Question

Let $f\left(x\right)=\frac{1}{1+e^{\frac{1}{x}}}$ 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

$f\left(x\right)=\frac{1}{1+e^{\frac{1}{x}}}$

as $x\to 0^{+}$

$\frac{1}{x}\to +\infty$

$\Rightarrow {lim}_{x\to 0^{+}}\frac{1}{1+e^{\frac{1}{x}}}=\frac{1}{1+e^{\infty }}=\frac{1}{\infty }=0$

as $x\to 0^{-}$

$\frac{1}{x}\to -\infty$

${lim}_{x\to 0^{-}}\frac{1}{1+e^{\frac{1}{x}}}=\frac{1}{1+e^{-\infty }}=\frac{1}{1+\frac{1}{\infty }}=\frac{1}{1+0}=1$