Answer
$\frac{{11}}{{10}}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - 5} \frac{{2{x^2} + 9x - 5}}{{{x^2} - 25}} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to - 5} \frac{{2{x^2} + 9x - 5}}{{{x^2} - 25}} = \frac{{2{{\left( { - 5} \right)}^2} + 9\left( { - 5} \right) - 5}}{{{{\left( { - 5} \right)}^2} - 25}} \cr
& \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{x \to - 5} \frac{{2{x^2} + 9x - 5}}{{{x^2} - 25}} = \frac{{50 - 45 - 5}}{{25 - 25}} = \frac{0}{0}{\text{ Indeterminate}} \cr
& \cr
& {\text{Factoring the numerator and denominator}} \cr
& \mathop {\lim }\limits_{x \to - 5} \frac{{2{x^2} + 9x - 5}}{{{x^2} - 25}} = \mathop {\lim }\limits_{x \to - 5} \frac{{\left( {2x - 1} \right)\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} \cr
& {\text{Cancel the common factor }}x + 5 \cr
& = \mathop {\lim }\limits_{x \to - 5} \frac{{2x - 1}}{{x - 5}} \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to - 5} \frac{{2x - 1}}{{x - 5}} = \frac{{2\left( { - 5} \right) - 1}}{{ - 5 - 5}} = \frac{{ - 10 - 1}}{{ - 10}} = \frac{{11}}{{10}} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to - 5} \frac{{2{x^2} + 9x - 5}}{{{x^2} - 25}} = \frac{{11}}{{10}} \cr} $$
