Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.3 One-sided Limits; Continuous Functions - 13.3 Assess Your Understanding - Page 909: 44

$\dfrac{7}{4}$

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