Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 15

Determine the $\lim\limits_{t \rightarrow -3} \quad \displaystyle \frac{t^2-9}{2t^2+7t+3}$, if it exists



$
\begin{equation}
\begin{aligned}
\lim\limits_{t \rightarrow -3} \quad \displaystyle \frac{t^2-9}{2t^2+7t+3} &= \lim\limits_{t \rightarrow -3}
\frac{
(t-3)\cancel{(t+3)}
}
{
(2t+1)\cancel{(t+3)}
}
&& \text{(Get the factor and cancel out like terms)}\\

\lim\limits_{t \rightarrow -3} \quad \displaystyle \frac{t-3}{2t+1} &= \frac{-3-3}{2(-3)+1} && \text{(Substitute value of } t \text{ and simplify)}
\end{aligned}
\end{equation}\\
\boxed{\lim\limits_{t \rightarrow -3} \quad \displaystyle \frac{t^2-9}{2t^2+7t+3} = \frac{6}{5}}
$