Limit of a function with absolute value $(|x |+ |x-1 |- |x+1 |) ∕ x^2$

Question

Consider $$ f(x)=\frac{|x|+|x-1|-|x+1|}{x^2} $$ Calculate the limit of the function $f$ when $x$ tends to $0$.

Answer 1

HINT

By definition of absolute value for $x\to 0$

• $x-1<0 \implies |x-1|=-(x-1)$
• $x+1>0 \implies |x+1|=x+1$

then

$$f(x)=\frac{|x|+|x-1|-|x+1|}{x^2}=\frac{|x|-(x-1)-(x+1)}{x^2}=\frac{|x|-2x}{x^2}=\frac{\frac{|x|}{x}-2}{x}$$

then consider $x\to 0^+$ and $x\to 0^-$.