Finding a limit with absolute values

**Chokfull** · September 24th 2009, 11:46 AM

My problem is

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

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

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

then
$<br /> \frac {t-|2\pm t|} {(1\pm t)/2}$

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

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

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

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

But the book says the answer is -4!!!!!!!

anyone able to help?

**Plato** · September 24th 2009, 12:19 PM

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

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