Answer
$$ - 13$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - 4} \frac{{2{x^2} + 3x - 20}}{{x + 4}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{x \to - 4} \frac{{2{x^2} + 3x - 20}}{{x + 4}} = \frac{{\mathop {\lim }\limits_{x \to - 4} \left( {2{x^2} + 3x - 20} \right)}}{{\mathop {\lim }\limits_{x \to - 4} \left( {x + 4} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}2{\text{ for }}x \cr
& \mathop {\lim }\limits_{x \to - 4} \frac{{2{x^2} + 3x - 20}}{{x + 4}} = \frac{{2{{\left( { - 4} \right)}^2} + 3\left( { - 4} \right) - 20}}{{ - 4 + 4}} = \frac{0}{0}{\text{ indeterminate form}} \cr
& {\text{factor the trinomial }}2{x^2} + 3x - 20{\text{ as }}\left( {2x - 5} \right)\left( {x + 4} \right) \cr
& = \mathop {\lim }\limits_{x \to - 4} \frac{{\left( {2x - 5} \right)\left( {x + 4} \right)}}{{x + 4}} \cr
& = \mathop {\lim }\limits_{x \to - 4} \left( {2x - 5} \right) \cr
& {\text{evaluating the limit when }}x \to - 4 \cr
& = 2\left( { - 4} \right) - 5 \cr
& = - 13 \cr} $$
