## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to3}\frac{x^2-9}{x^2+2x-3}=0$$
$$A=\lim\limits_{x\to3}\frac{x^2-9}{x^2+2x-3}$$ 1) Find the domain of $\frac{x^2-9}{x^2+2x-3}$ $\frac{x^2-9}{x^2+2x-3}$ is defined at any number in $R$ except where $$x^2+2x-3=0$$$$x=1$$ or $$x=-3$$ So $\frac{x^2-9}{x^2+2x-3}$ is defined at any number in $R$ except $x=1$ and $x=-3$. 2) Since $\frac{x^2-9}{x^2+2x-3}$ is a rational function, it is continuous at any number in its domain, including $x=3$. That means $$A=\lim\limits_{x\to3}\frac{x^2-9}{x^2+2x-3}=\frac{3^2-9}{3^2+2\times3-3}$$$$A=0$$