Answer
$$\lim\limits_{x\to1^+}\frac{x^2-9}{x^2+2x-3}=-\infty$$
Work Step by Step
$$A=\lim\limits_{x\to1^+}\frac{x^2-9}{x^2+2x-3}$$
- Consider the numerator:
$x^2-9=(x-3)(x+3)$
- Consider the denominator:
$x^2+2x-3=(x-1)(x+3)$
Therefore,
$$A=\lim\limits_{x\to1^+}\frac{(x-3)(x+3)}{(x-1)(x+3)}$$$$A=\lim\limits_{x\to1^+}\frac{x-3}{x-1}$$
As $x\to1^+$, $\frac{x-3}{x-1}$ approaches $\frac{1-3}{1-1}=\frac{-2}{0}=-\infty$
So, $$\lim\limits_{x\to1^+}\frac{x^2-9}{x^2+2x-3}=-\infty$$
