Calculus: Early Transcendentals 8th Edition

$\lim\limits_{t \to -3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\dfrac{6}{5}$
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: $\lim\limits_{t \to -3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\lim\limits_{t \to -3}\dfrac{(t-3)(t+3)}{(t+3)(2t+1)}=\lim\limits_{t \to -3}\dfrac{t-3}{2t+1}=...$ $...=\dfrac{-3-3}{2(-3)+1}=\dfrac{-6}{-5}=\dfrac{6}{5}$