Answer
$\frac{5}{7}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - x - 6}}{{3{x^2} + 5x - 2}} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - x - 6}}{{3{x^2} + 5x - 2}} = \frac{{{{\left( { - 2} \right)}^2} - \left( { - 2} \right) - 6}}{{3{{\left( { - 2} \right)}^2} + 5\left( { - 2} \right) - 2}} \cr
& \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - x - 6}}{{3{x^2} + 5x - 2}} = \frac{0}{0}{\text{ Indeterminate}} \cr
& \cr
& {\text{Factoring the numerator and denominator}} \cr
& \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - x - 6}}{{3{x^2} + 5x - 2}} = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x - 3} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {3x - 1} \right)}} \cr
& {\text{Cancel the common factor }}x + 2 \cr
& = \mathop {\lim }\limits_{x \to - 2} \frac{{x - 3}}{{3x - 1}} \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to - 2} \frac{{x - 3}}{{3x - 1}} = \frac{{ - 2 - 3}}{{3\left( { - 2} \right) - 1}} = \frac{5}{7} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - x - 6}}{{3{x^2} + 5x - 2}} = \frac{5}{7} \cr} $$
