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 ?
 A: 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 $\displaystyle\lim_{x\to 0}\frac{|2x-1|-|2x+1|}{x}=-4.$
A: 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 $x=0$.
