## University Calculus: Early Transcendentals (3rd Edition)

$\lim\limits_{x \to -5}\frac{(x^{2}+3x-10)}{(x+5)} = \lim\limits_{x \to -5}(x-2) = -7$
Simplify: $\frac{(x^{2}+3x-10)}{(x+5)} = \frac{(x-2)(x+5)}{(x+5)} = (x-2)$ Now, $\lim\limits_{x \to -5}\frac{(x^{2}+3x-10)}{(x+5)} = \lim\limits_{x \to -5}(x-2) = -7$