## Calculus (3rd Edition)

$\frac{15}{4}$
We have $$\lim _{x\:\rightarrow \:3}\:\frac{x^3-2x^2-9}{x^2-2x-3}= \lim _{x\:\rightarrow \:3}\:\frac{(x-3)(x^2+x+3)}{(x-3)(x+1)}= \lim _{x\:\rightarrow \:3} \frac{x^2+x+3}{x+1}=\frac{15}{4}.$$ Where we factored the numerator and denominator, canceled out common terms and plugged in "3" for $x$.