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`