Answer
$\lim\limits_{x \to -1.5} \frac{9-4x^2}{3+2x} = 6$
Work Step by Step
Let $\epsilon \gt 0$ be given.
Let $\delta = \frac{\epsilon}{2}$
Suppose that $\vert x-(-1.5) \vert \lt \delta$.
Then: $\vert \frac{9-4x^2}{3+2x}-6 \vert = \vert \frac{(3-2x)(3+2x)}{3+2x}-6 \vert = \vert (3-2x)-6 \vert = \vert -2x-3 \vert = \vert 2x+3 \vert = 2~\vert x+1.5 \vert \lt 2\delta = 2(\frac{\epsilon}{2}) = \epsilon$
Therefore, $\lim\limits_{x \to -1.5} \frac{9-4x^2}{3+2x} = 6$
