Answer
$$\lim _{t \rightarrow-2} \frac{2 t+4}{12-3 t^{2}}
=\frac{1}{6}$$
Work Step by Step
Given $$\lim _{t \rightarrow-2} \frac{2 t+4}{12-3 t^{2}}$$
let $$ f(t) = \frac{2 t+4}{12-3 t^{2}}$$
Since, we have
$$ f(-2)=\frac{-4+4}{12-12}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{t \rightarrow-2} \frac{2 t+4}{12-3 t^{2}}\\
&=\lim _{t \rightarrow-2} \frac{2 t+4}{-3( t^{2}-4)}\\
&=\lim _{t \rightarrow-2} \frac{2(t+2)}{-3(t-2)(t+2)}\\
&=\lim _{t \rightarrow-2} \frac{2}{-3(t-2)}\\
&=\frac{2}{-3(-2-2)}\\
&=\frac{1}{6}
\end{aligned}
