Answer
$$7$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 3x - 10}}{{x - 2}} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 3x - 10}}{{x - 2}} = \frac{{\mathop {\lim }\limits_{x \to 2} \left( {{x^2} + 3x - 10} \right)}}{{\mathop {\lim }\limits_{x \to 2} \left( {x - 2} \right)}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}2{\text{ for }}x \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 3x - 10}}{{x - 2}} = \frac{{{{\left( 2 \right)}^2} + 3\left( 2 \right) - 10}}{{2 - 2}} = \frac{0}{0}{\text{ indeterminate form}} \cr
& {\text{factor the trinomial }}{x^2} + 3x - 10 \cr
& = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x + 5} \right)\left( {x - 2} \right)}}{{x - 2}} \cr
& = \mathop {\lim }\limits_{x \to 2} \left( {x + 5} \right) \cr
& {\text{evaluating the limit when }}x \to 2 \cr
& = 2 + 5 \cr
& = 7 \cr} $$
