Chapter 13 - Review - Exercises - Page 942: 12

$\frac{4}{3}$

Let's simplify this rational equation. $\frac{x^2-4}{x^2+x-2}=\frac{(x-2)(x+2)}{(x+2)(x-1)}$. To find the limit lets cancel $x+2$ from both numerator and denominator. Then we get $\frac{x-2}{x-1}$. Plugging in $x=-2$ we get $-4/-3 $ or 4/3.