Let a function $f$ be defined by $f(x)=\begin{cases} \frac{x}{1+e^{1/x}}& \text{if}{\hspace .2 cm} x \neq 0\\ 0 & \text{if}{\hspace .2 cm} x=0 \end{cases}$. Then the value of left hand derivative and right hand derivative are

Left hand derivative 1 right hand derivative 0

$\lim_{x\to 0-}\frac{x}{1+e^{1/x}}/x-0=\lim_{x\to 0-}\frac{x}{1+e^{1/x}}/x=lim_{x\to 0-}1/1+e^{1/x}=1$

Since $e^{1/x}\rightarrow 0$ as $x\rightarrow 0-$.

Since $e^{1/x} \rightarrow \infty$ as $x \rightarrow 0+$

we get $\lim_{x\to 0+}\frac{x}{1+e^{1/x}}/x-0=\lim_{x\to 0+}\frac{x}{1+e^{1/x}}/x=lim_{x\to 0+}1/1+e^{1/x}=0$