Answer
$\lim\limits_{t \to -1}\frac{(t^{2}+3t+2)}{(t^{2}-t-2)} = -\frac{1}{3}= 0.33$
Work Step by Step
Simplify:
$\frac{(t^{2}+3t+2)}{(t^{2}-t-2)} =\frac{(t+1)(t+2)}{(t-2)(t+1)} = \frac{(t+2)}{(t-2)}$
Now,
$\lim\limits_{t \to -1}\frac{(t^{2}+3t+2)}{(t^{2}-t-2)} = \lim\limits_{t \to -1}\frac{(t+2)}{(t-2)} = -\frac{1}{3}= 0.33$
