Answer
$\lim_{t\to-3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\dfrac{6}{5}$
Work Step by Step
$\lim_{t\to-3}\dfrac{t^{2}-9}{2t^{2}+7t+3}$
Try to evaluate the limit applying direct substitution:
$\lim_{t\to-3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\dfrac{(-3)^{2}-9}{2(-3)^{2}+7(-3)+3}=...$
$...=\dfrac{9-9}{2(9)-21+3}=\dfrac{0}{18-21+3}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Factor the numerator and the denominator of the function and simplify:
$\lim_{t\to-3}\dfrac{t^{2}-9}{2t^{2}+7t+3}=\lim_{t\to-3}\dfrac{(t-3)(t+3)}{(t+3)(2t+1)}=...$
$...=\lim_{t\to-3}\dfrac{t-3}{2t+1}=...$
Try to evaluate using direct substitution again:
$...=\lim_{t\to-3}\dfrac{t-3}{2t+1}=\dfrac{-3-3}{2(-3)+1}=\dfrac{-6}{-6+1}=\dfrac{-6}{-5}=\dfrac{6}{5}$
