Answer
$\frac{5}{7}$
Work Step by Step
$\lim\limits_{r \to -1}\frac{r^2-3r-4}{4r^2+r-3}$ (Use the factorization method)
$=\lim\limits_{r \to -1}\frac{(r+1)(r-4)}{(r+1)(4r-3)}$ (Cancel the common factor)
$=\lim\limits_{r \to -1}\frac{r-4}{4r-3}$
$=\frac{-1-4}{4\cdot (-1)-3}$
$=\frac{-5}{-7}$
$=\frac{5}{7}$
