How do you prove the statement lim as x approaches -1.5 for # ((9-4x^2)/(3+2x))=6# using the epsilon and delta definition?

You can't prove it since it isn't 6.
#lim_(x to -1.5) (9-4x)/(3+2)# does not exist. The limit from the left is #+oo# and the limit from the right is #-oo#.

Preliminary analysis
We need to show that for every positive #epsilon#, there is a positive #delta# such that if #0 < abs(x-(-3/2)) < delta# we get #abs((9-4x^2)/(3+2x) - 6) < epsilon#.

Note first that #abs(x-(-3/2)) = abs(x+3/2)#

For every #x# other than #-3/2#, we have:

#(9-4x^2)/(3+2x) = 3-2x#.

So, for every #x# other than #-3/2#, we have:

#abs((9-4x^2)/(3+2x) - 6)= abs((3-2x)-6)#

# = abs(-2x-3) = abs((-4)(x+3/2)) = abs(-4)abs(x+3/2)#.
# = 4abs(x+3/2)#

We want this to be less than #epsilon# and we control, through #delta#, the size of #abs(x+3/2)#

If we make #abs(x+3/2) < epsilon/4#, then we will have

#4abs(x+3/2) < 4(epsilon/4) = epsilon# as desired.

Now we are ready to write the proof:

Proof

Given #epsilon > 0#, let #delta = epsilon/4#. Observe that this #delta# is also positive, as required.

Now if #x# is chosen so that #0 < abs(x-(-3/2)) < delta#, the we have:

#abs((9-4x^2)/(3+2x) - 6)= abs((3-2x)-6)#

# = abs(-2x-3) = abs(-4)abs(x+3/2)#.

# = 4abs(x+3/2) < 4delta = 4(epsilon/4) = epsilon#

That is: if #0 < abs(x-(-3/2)) < delta#, then #abs((9-4x^2)/(3+2x) - 6) < epsilon#.

So, by the definition of limit, #lim_(xrarr-1.5)((9-4x^2)/(3+2x)) = 6#