Answer
$$\lim\limits_{x\to-3}\frac{x^2-9}{x^2+2x-3}=\frac{3}{2}$$
Work Step by Step
$$A=\lim\limits_{x\to-3}\frac{x^2-9}{x^2+2x-3}$$
- Consider the numerator:
$x^2-9=(x-3)(x+3)$
- Consider the denominator:
$x^2+2x-3=(x-1)(x+3)$
Therefore,
$$A=\lim\limits_{x\to-3}\frac{(x-3)(x+3)}{(x-1)(x+3)}$$$$A=\lim\limits_{x\to-3}\frac{x-3}{x-1}$$$$A=\frac{-3-3}{-3-1}$$$$A=\frac{3}{2}$$
