Chapter 12 - Functions of Several Veriables - Review Exercises - Page 960: 35

$$\frac{2}{3}$$

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