Answer
$$\frac{{19}}{9}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 6} \frac{{2x + 7}}{{x + 3}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{x \to 6} \frac{{2x + 7}}{{x + 3}} = \frac{{\mathop {\lim }\limits_{x \to 6} \left( {2x + 7} \right)}}{{\mathop {\lim }\limits_{x \to 6} \left( {x + 3} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}6{\text{ for }}x \cr
& \mathop {\lim }\limits_{x \to 6} \frac{{2x + 7}}{{x + 3}} = \frac{{2\left( 6 \right) + 7}}{{6 + 3}} \cr
& {\text{simplifying}} \cr
& \mathop {\lim }\limits_{x \to 6} \frac{{2x + 7}}{{x + 3}} = \frac{{19}}{9} \cr} $$
