Answer
$\lim\limits_{t \to -3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\dfrac{6}{5}$
Work Step by Step
Apply direct substitution first:
$\lim\limits_{t \to -3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\dfrac{(-3)^{2}-9}{2(-3)^{2}+7(-3)+3}=\dfrac{0}{0}$ $(Indeterminate$ $Form)$
Apply factorization to the numerator:
$t^{2}-9=(t-3)(t+3)$
Apply factorization to the denominator:
$2(2t^{2}+7t+3)=4t^2+2(7t)+6=\dfrac{(2t+6)}{2}(2t+1)=(t+3)(2t+1)$
Evaluate the limit:
$\lim\limits_{t \to -3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\lim\limits_{t \to -3}\dfrac{(t-3)(t+3)}{(t+3)(2t+1)}=\lim\limits_{t \to -3}\dfrac{t-3}{2t+1}=...$
$...=\dfrac{-3-3}{2(-3)+1}=\dfrac{-6}{-5}=\dfrac{6}{5}$
