Let f be defined on D = ℝ - { - 1 , 1 } by f x = | x | 1 - | x | , then

Given:

$f\left(x\right)=\frac{\left|x\right|}{1-\left|x\right|}$

Now,

$\text{LHL}=\underset{x\to 0^{-}}{\lim }f\left(x\right)$

        $=\underset{h\to 0}{\lim }f\left(0-h\right)$

        $=\underset{h\to 0}{\lim }\frac{\left|0-h\right|}{1-\left|0-h\right|}=0$

$\text{RHL}=\underset{x\to 0^{+}}{\lim }f\left(x\right)$

          $=\underset{h\to 0}{\lim }f\left(0+h\right)$

           $=\underset{h\to 0}{\lim }\frac{\left|0+h\right|}{1-\left|0+h\right|}=0$

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

Hence, $f\left(x\right)$ is continuous at $x=0$.

Using differentiation:

$f\left(x\right)=\left\{\begin{array}{l}\frac{-x}{1+x}, x<0\\ \frac{x}{1-x}, x>0\end{array}\right.$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{l}\frac{-1}{1+x}+\frac{x}{{\left(1+x\right)}^2}; x<0\\ \frac{1}{1+x}+\frac{x}{{\left(1-x\right)}^2}; x>0\end{array}\right.$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{l}\frac{-1}{{\left(1+x\right)}^2}; x<0\\ \frac{1}{{\left(1-x\right)}^2}; x>0\end{array}\right.$

$\text{LHD}\ne \text{RHD}$ at $x=0$.

Hence, $f\left(x\right)$ is differentiable in $D$ except at $x=0$.