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