limit of absolute value $$ \lim_{x \to 0} \frac{\lvert2x-1\rvert - \lvert2x+1\rvert}{x} $$
Defining the function piecewise reveals the limit is in fact, continuous about 0
However when I go to solve it in a normal algebraic manner, the $2x$ terms are canceling, leaving me with an undefined output - 0 in the denominator. 
Any hints would be fantastic, I've solved this every way I can think of and I keep getting different answers, none of which are the correct answer.
I did graph this, and it does show a continuous function around $0$ where $f(x) = -4$  
My problem here is that I'm being a huge dunce about absolute values. 
 A: I recognize this problem from Stewart (or at least something similar). Use the fact that 
$$|2x-1| = \begin{cases}
2x-1, & 2x-1 > 0 \\
-(2x-1) = 1 - 2x, & 2x-1 < 0
\end{cases}$$
and $$|2x+1| = \begin{cases}
2x+1, & 2x+1 > 0 \\
-(2x+1) = -2x - 1, & 2x+1 < 0\text{.}
\end{cases}$$
Examine the behavior around $x = 0$ using these two equations.
A: For $\;x\;$ pretty close to zero, $\;2x-1<0\;,\;\;2x+1>0\;$ , so we have the limit
$$\frac{-2x+1-2x-1}x=\frac{-4x}x=-4\xrightarrow[x\to 0]{}-4$$
A: $$\lim_{x \to 0} \frac{|2x-1| - |2x+1|}{x}$$
\begin{align}
   \frac{|2x-1| - |2x+1|}{x} 
   &= \frac{|2x-1| - |2x+1|}{x} \cdot \frac{|2x-1| + |2x+1|}{|2x-1| + |2x+1|} \\
   &= \frac{(2x-1)^2 - (2x+1)^2}{x(|2x-1| + |2x+1|)} \\
   &= \frac{-8}{|2x-1| + |2x+1|} \\
\end{align}
and
$\displaystyle \lim_{x \to 0} \frac{-8}{|2x-1| + |2x+1|} = \frac{-8}{2} = -4$
Added later.
I just had to add this...
When $x$ is "close to" $0$,


*

*$2x - 1$ is negative. So $|2x-1|=-2x+1$.   

*$2x + 1$ is positive. So $|2x+1|=2x+1$.


Hence 
$$\frac{|2x-1| - |2x+1|}{x} 
  = \frac{(-2x+1) - (2x+1)}{x} 
  = \frac{-4x}{x}
  \to -4$$
as $x \to 0$.
