Question

$f\left(x\right)=\{\begin{array}{cc}\frac{x}{1+e^{1/x}}& x\ne 0\\ 0& x=0\end{array}$then the left derivative of $f\left(x\right)$ at $x=0$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $Doesnotexists$

Answer 1

The correct option is A $1$

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