# limit involving absolute values

• Sep 7th 2009, 06:15 PM
darcyrenee08
limit involving absolute values
okay so im in calculus 2 right now, and im going over all my notes from cal 1 and i cant seem to figure this out.

lim x->0- {abs(7x-2)-abs(7x+2)} / x

i hate summer for this reason.(Worried)
• Sep 7th 2009, 07:33 PM
Chris L T521
Quote:

Originally Posted by darcyrenee08
okay so im in calculus 2 right now, and im going over all my notes from cal 1 and i cant seem to figure this out.

lim x->0- {abs(7x-2)-abs(7x+2)} / x

i hate summer for this reason.(Worried)

Note that $\displaystyle \left|7x-2\right|=\left\{\begin{matrix}7x-2 & \text{when }x\geq\frac{2}{7}\\-7x+2 & \text{when }x<\frac{2}{7}\end{matrix}\right.$ and $\displaystyle \left|7x+2\right|=\left\{\begin{matrix}7x+2 & \text{when }x\geq-\frac{2}{7}\\-7x-2 & \text{when }x<-\frac{2}{7}\end{matrix}\right.$

So when we approach 0 from the left, we take $\displaystyle \left|7x-2\right|=-7x+2$ and $\displaystyle \left|7x+2\right|=7x+2$.

Thus, we have

$\displaystyle \lim_{x\to0^-}\frac{\left|7x-2\right|-\left|7x+2\right|}{x}=\lim_{x\to0^-}\frac{-7x+2-(7x+2)}{x}=\lim_{x\to0^-}\frac{-7x+2-7x-2}{x}$ $\displaystyle =\lim_{x\to0^-}\frac{-14x}{x}=-14$
• Sep 8th 2009, 05:36 AM
darcyrenee08
(Rofl) THANKS!!