تابع در $x=0$ پیوسته است، پس:

${f}'-\left( 0 \right)= \displaystyle{\lim_{x \to 0^-}}  \frac{f\left( x \right)-f\left( 0 \right)}{x-0}=  \displaystyle{\lim_{x \to 0^-}} \frac{\left| x \right|\left[ x \right]-0}{x-0}= \displaystyle{\lim_{x \to 0^-}} \frac{\left( -x \right)\left( -1 \right)}{x}=1$

${f}'+\left( 0 \right)= \displaystyle{\lim_{x \to 0^-}} \frac{f\left( x \right)-f\left( 0 \right)}{x-0}= \frac{\left| x \right|\left[ x \right]-0}{x-0}= \displaystyle{\lim_{x \to 0^-}} \frac{x\times 0}{x}=0\Rightarrow {f}'-\left( 0 \right)-{f}'+\left( 0 \right)=1-0=1$