Answer
$-\dfrac{1}{2}$
Work Step by Step
We want to find $\lim\limits_{t \to -1}\dfrac{t^2+3t+2}{t^2-t-2}$, but we can't use the quotient rule for limits because the limit of the denominator as $t$ approaches -1 is zero, and if we try to substitute $t=-1$ directly, we get zero in both the numerator and the denominator. This means we have a common factor of $t+1$ in both the numerator and the denominator.
Note that $\dfrac{t^2+3t+2}{t^2-t-2}=\dfrac{(t+2)(t+1)}{(t-2)(t+1)}=\dfrac{t+2}{t-1}$ for all $t \neq -1.$
Thus $\lim\limits_{t \to -1}\dfrac{t^2+3t+2}{t^2-t-2}=\lim\limits_{t \to -1}\dfrac{t+2}{t-1}=\dfrac{-1+2}{-1-1}=\dfrac{1}{-2}=-\dfrac{1}{2}.$
