Answer
$$ - 5$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,1} \right)} \frac{{2{x^2} - xy - 3{y^2}}}{{x + y}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}}{\text{, substitute }} - {\text{1 for }}x{\text{ and 1 for }}y{\text{ into }}f\left( {x,y} \right) \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,1} \right)} \frac{{2{x^2} - xy - 3{y^2}}}{{x + y}} = \frac{{2{{\left( 1 \right)}^2} - \left( { - 1} \right)\left( 1 \right) - 3{{\left( 1 \right)}^2}}}{{ - 1 + 1}} \cr
& {\text{simplifying}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,1} \right)} \frac{{2{x^2} - xy - 3{y^2}}}{{x + y}} = \frac{{2 + 1 - 3}}{{ - 1 + 1}}{\text{ }} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,1} \right)} \frac{{2{x^2} - xy - 3{y^2}}}{{x + y}} = \frac{0}{0}{\text{ indeterminate}} \cr
& {\text{simplify the expression }}\frac{{2{x^2} - xy - 3{y^2}}}{{x + y}}{\text{ factoring the numerator}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,1} \right)} \frac{{\left( {x + y} \right)\left( {2x - 3y} \right)}}{{x + y}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,1} \right)} \left( {2x - 3y} \right) \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = 2\left( { - 1} \right) - 3\left( 1 \right) \cr
& = - 2 - 3 \cr
& = - 5 \cr} $$
