## Calculus 8th Edition

$\frac{6}{5}$
$\lim\limits_{t \to -3}\frac{t^2-9}{2t^2+7t+3}=\frac{(-3)^2-9}{2(-3)^2+7(-3)+3}=\frac{0}{0}=$indet so we have to do algebra $\lim\limits_{t \to -3}\frac{t^2-9}{2t^2+7t+3}=\lim\limits_{t \to -3}\frac{(t+3)(t-3)}{(2t+1)(t+3)}=\lim\limits_{t \to -3}\frac{t-3}{2t+1}=\frac{(-3)-3}{2(-3)+1}=\frac{-6}{-5}=\frac{6}{5}$