Answer
$\frac{7}{4}$
Work Step by Step
Factor each polynomial:
$$\lim_{x\to -4^-}\frac{x^2+x-12}{x^2+4x}\\=\lim_{x\to -4^-}\frac{(x+4)(x-3)}{(x)(x+4)}.$$
Cancel the common factors: $$\lim_{x\to -4^-}\frac{(x-3)}{(x)}\\=\frac{-4-3}{-4}=\frac{7}{4}$$