Answer
$$16$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 3} \frac{{3{x^2} - 2x - 21}}{{x - 3}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{x \to 3} \frac{{3{x^2} - 2x - 21}}{{x - 3}} = \frac{{\mathop {\lim }\limits_{x \to 3} \left( {3{x^2} - 2x - 21} \right)}}{{\mathop {\lim }\limits_{x \to 3} \left( {x - 3} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}3{\text{ for }}x \cr
& \mathop {\lim }\limits_{x \to 3} \frac{{3{x^2} - 2x - 21}}{{x - 3}} = \frac{{3{{\left( 3 \right)}^2} - 2\left( 3 \right) - 21}}{{3 - 3}} = \frac{0}{0}{\text{ indeterminate form}} \cr
& {\text{factor the trinomial }}3{x^2} - 2x - 21\,{\text{ as }}\left( {x - 3} \right)\left( {3x + 7} \right) \cr
& = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {3x + 7} \right)}}{{x - 3}} \cr
& = \mathop {\lim }\limits_{x \to 3} \left( {3x + 7} \right) \cr
& {\text{evaluating the limit when }}x \to 3 \cr
& = 3\left( 3 \right) + 7 \cr
& = 16 \cr} $$
