## Calculus with Applications (10th Edition)

$$8$$
\eqalign{ & \mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{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{{{x^2} - 16}}{{x - 4}} = \frac{{\mathop {\lim }\limits_{x \to 4} \left( {{x^2} - 16} \right)}}{{\mathop {\lim }\limits_{x \to 4} \left( {x - 4} \right)}} \cr & {\text{evaluate the limits}}{\text{, replacing }}4{\text{ for }}x \cr & \mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = \frac{{{4^2} - 16}}{{4 - 4}} = \frac{0}{0}{\text{ indeterminate form}} \cr & {\text{factor the difference of two squares }}{x^2} - 16 \cr & = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x + 4} \right)\left( {x - 4} \right)}}{{x - 4}} \cr & = \mathop {\lim }\limits_{x \to 4} \left( {x + 4} \right) \cr & {\text{evaluating the limit when }}x \to 4 \cr & = 4 + 4 \cr & = 8 \cr}