Evaluating a limit of a function involving absolute values

I have to evaluate the following limit.

$$ \lim \limits_{x \to 0} \frac{ |2x-1| - |2x+1| }{x} $$

I can not open the absolute parts since its zeros, 1/2 and -1/2 are different from zero. Can somebody help ?

Exact, as mentioned in the comments, since $x\to 0$ then we can take $\delta$ as small as we want, i.e. $0<|x|<\delta$ even we can put $0<|x|<\frac{1}{4}$, in this case $|2x-1|=-2x+1$ and $|2x+1|=2x+1$, so $$ \lim_{x\to 0}\frac{|2x-1|-|2x+1|}{x}=\lim_{x\to 0}\frac{-2x+1-2x-1}{x}=\lim_{x\to 0}(-4)=-4. $$ But if that doesn't convince you then you can demonstrate it using $\epsilon-\delta$, as follows: $$\forall\epsilon>0,\;\exists\delta=1/4,\text{ such that }0<|x-0|<\delta=1/4\Rightarrow \color{blue}{|2x-1|=-2x+1\wedge|2x+1|=2x+1}$$ Then we have $\left|\frac{|2x-1|-|2x+1|}{x}+4\right|=|-4+4|=0\color{red}{<\epsilon}$.

What proves that $\displaystyle\lim_{x\to 0}\frac{|2x-1|-|2x+1|}{x}=-4.$

Koro's comment that one can restrict to $|x|<\frac14$ is the best way to do this problem. But here's another fun way if one really wants to deal with the absolute values algebraically: \begin{align*} \lim_{x\to0} \frac{|2x-1|-|2x+1|}{x} &= \lim_{x\to0} \biggl( \frac{|2x-1|-|2x+1|}{x} \cdot \frac{|2x-1|+|2x+1|}{|2x-1|+|2x+1|} \biggr) \\ &= \lim_{x\to0} \frac{(|2x-1|-|2x+1|)(|2x-1|+|2x+1|)}{x(|2x-1|+|2x+1|)} \\ &= \lim_{x\to0} \frac{(2x-1)^2-(2x+1)^2}{x(|2x-1|+|2x+1|)} \\ &= \lim_{x\to0} \frac{-8x}{x(|2x-1|+|2x+1|)} \\ &= -8 \lim_{x\to0} \frac1{|2x-1|+|2x+1|} = -8 \frac1{|{-1}|+|1|} = -4, \end{align*} since $\dfrac1{|2x-1|+|2x+1|}$ is continuous at $x=0$.