Answer
$\dfrac{3}{2}$
Work Step by Step
We want to find $\lim\limits_{t \to 1}\dfrac{t^2+t-2}{t^2-1}$, 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 $\dfrac{t^2+t-2}{t^2-1}=\dfrac{(t+2)(t-1)}{(t-1)(t+1)}=\dfrac{t+2}{t+1}$ for all $t \neq 1.$
Thus $\lim\limits_{t \to 1}\dfrac{t^2+t-2}{t^2-1}=\lim\limits_{t \to 1}\dfrac{t+2}{t+1}= \dfrac{1+2}{1+1}=\dfrac{3}{2}.$
