# One sided limit

• December 15th 2006, 11:32 PM
One sided limit
Hello i still have a problem about graphing absolute values:
Sketch the graph of the function and find the indicated limit if it exists; if the limit does not exist, state the reason:
$G(x) = |2x - 3| - 4$
find:
$a.) \lim_{x \to {\frac{3}{2}}^+}$
$b.) \lim_{x \to {\frac{3}{2}}^-}$
$c.) \lim_{x \to {\frac{3}{2}}}$<--- does it exist?

Thanks and
Happy holidays!
• December 16th 2006, 01:28 AM
The limit exists
The limit exists because a = -4 and b = -4 are the same
nevmind
my problem is only the graphing the absolute value
• December 16th 2006, 03:47 AM
topsquark
First look at where the absolute value portion of the function is 0:
$2x - 3 = 0$

$x = -\frac{3}{2}$

Now split the function into two pieces:
$G(x) = \left \{ \begin{array}{cc} -(2x-3)-4 & x < -\frac{3}{2} \\ (2x-3) - 4 & x \geq -\frac{3}{2} \end{array} \right .$

-Dan