We need to prove, using the epsilon-delta definition of a limit, that $\lim_{x \to -\frac{3}{2}} \frac{9 - 4x^2}{3 + 2x} = 6$.

1. Problem Description

We need to prove, using the epsilon-delta definition of a limit, that $\lim_{x \to -\frac{3}{2}} \frac{9 - 4x^2}{3 + 2x} = 6$.

2. Solution Steps

We want to show that for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - (-\frac{3}{2})| < \delta$, then $|\frac{9 - 4x^2}{3 + 2x} - 6| < \epsilon$.

First, let's simplify the expression inside the absolute value:

$|\frac{9 - 4x^2}{3 + 2x} - 6| = |\frac{9 - 4x^2 - 6(3 + 2x)}{3 + 2x}| = |\frac{9 - 4x^2 - 18 - 12x}{3 + 2x}| = |\frac{-4x^2 - 12x - 9}{3 + 2x}|$.

We can factor the numerator:

$-4x^2 - 12x - 9 = -(4x^2 + 12x + 9) = -(2x + 3)^2$.

So, we have:

$|\frac{-(2x + 3)^2}{3 + 2x}| = |\frac{-(2x + 3)(2x + 3)}{2x + 3}|$.

As $x \to -\frac{3}{2}$, $2x + 3 \neq 0$, so we can simplify further to:

$|-(2x + 3)| = |2x + 3| = |2(x + \frac{3}{2})| = 2|x + \frac{3}{2}| = 2|x - (-\frac{3}{2})|$.

Now we want to find a $\delta > 0$ such that if $0 < |x - (-\frac{3}{2})| < \delta$, then $2|x - (-\frac{3}{2})| < \epsilon$.

So, $2|x + \frac{3}{2}| < \epsilon$.

Dividing by 2, we get $|x + \frac{3}{2}| < \frac{\epsilon}{2}$.

Thus, we can choose $\delta = \frac{\epsilon}{2}$.

Let $\epsilon > 0$. Choose $\delta = \frac{\epsilon}{2}$.

If $0 < |x - (-\frac{3}{2})| < \delta$, then

$|\frac{9 - 4x^2}{3 + 2x} - 6| = 2|x + \frac{3}{2}| = 2|x - (-\frac{3}{2})| < 2\delta = 2(\frac{\epsilon}{2}) = \epsilon$.

3. Final Answer

For any $\epsilon > 0$, choose $\delta = \frac{\epsilon}{2}$. Then, if $0 < |x - (-\frac{3}{2})| < \delta$, we have $|\frac{9 - 4x^2}{3 + 2x} - 6| < \epsilon$. Therefore, $\lim_{x \to -\frac{3}{2}} \frac{9 - 4x^2}{3 + 2x} = 6$.