# How do you find the limit lim_(t->-3)(t^2-9)/(2t^2+7t+3) ?

${\lim}_{t \to - 3} \frac{{t}^{2} - 9}{2 {t}^{2} + 7 t + 3}$
$= {\lim}_{t \to - 3} \frac{\left(t + 3\right) \left(t - 3\right)}{\left(t + 3\right) \left(2 t + 1\right)}$
by cancelling out $\left(t - 3\right)$'s,
$= {\lim}_{t \to - 3} \frac{t - 3}{2 t + 1} = \frac{\left(- 3\right) - 3}{2 \left(- 3\right) + 1} = \frac{- 6}{- 5} = \frac{6}{5}$