# Thread: Finding a limit with absolute values

1. ## Finding a limit with absolute values

My problem is

$\displaystyle \lim_{x\to0} \frac {|2x-1|-|2x+1|} {x}$

So if i set $\displaystyle t=|2x-1|$ and $\displaystyle c=|2x+1|$ i can get

$\displaystyle 2x-1=\pm t$ and $\displaystyle 2x+1=\pm c$
$\displaystyle x=\frac {1\pm t} {2}$ and $\displaystyle x=\frac {-1\pm c} {2}$
$\displaystyle \frac {1\pm t} {2}=\frac {-1\pm c} {2}$
$\displaystyle |2\pm t|=c$

then
$\displaystyle \frac {t-|2\pm t|} {(1\pm t)/2}$

and as $\displaystyle x$ approaches $\displaystyle 0$, $\displaystyle t$ approaches $\displaystyle 1$ so

$\displaystyle \lim_{t\to1} \frac {2t-2|2\pm t|} {1\pm t}$

since $\displaystyle t$ approaches $\displaystyle 1$ and the denominator cannot be $\displaystyle 0$

$\displaystyle \frac {2-2|2\pm t|} {1\pm t=2}$
$\displaystyle 1-|1\pm 2|=0$ or $\displaystyle 2$

But the book says the answer is -4!!!!!!!
anyone able to help?

2. $\displaystyle - .5 < x < .5\;\& \,\frac{{\left| {2x - 1} \right| - \left| {2x + 1} \right|}}{x} = \frac{{\left( { - 2x + 1} \right) - \left( {2x + 1} \right)}} {x}$