#### Answer

$\dfrac{7}{4}$

#### Work Step by Step

We find the left-hand limit as follows:
$\lim\limits_{x \to -4^{-}} \dfrac{x^2+x-2}{x^2+4x} \\=\lim\limits_{x \to -4^{-}} \dfrac{(x+4) (x-3)}{x(x+4)} \\=\lim\limits_{x \to -4^{-}} \dfrac{x-3}{x} \\=\dfrac{-4-3}{-4} \\=\dfrac{7}{4}$