Answer
$\lim\limits_{t \to 1}\frac{(t^{2}+t-2)}{(t^{2}-1)} = \frac{3}{2} = 1.5$
Work Step by Step
Simplify:
$\frac{(t^{2}+t-2)}{(t^{2}-1)} = \frac{(t+2)(t-1)}{(t+1)(t-1)} = \frac{(t+2)}{(t+1)}$
$\lim\limits_{t \to 1}\frac{(t^{2}+t-2)}{(t^{2}-1)} = \lim\limits_{t \to 1}\frac{(t+2)}{(t+1)} = \frac{3}{2} = 1.5$