## Calculus 8th Edition

$\lim\limits_{x \to 4}\frac{x^2+3x}{x^2-x-12}=\frac{(4)^2+3(4)}{(4)^2-(4)-12}=\frac{28}{0}$ so we have to do algebra $\lim\limits_{x \to 4}\frac{x^2+3x}{x^2-x-12}=\lim\limits_{x \to 4}\frac{(x)(x+3)}{(x-4)(x+3)}=\lim\limits_{x \to 4}\frac{x}{x-4}=\frac{(4)}{(4)-4}=\frac{4}{0}$ Therefore, the limit does not exist.