Answer
$$ - 8$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,2} \right)} \frac{{x{y^3}}}{{x + y}} \cr
& {\text{Using the limit laws }} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,2} \right)} x{y^3}}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,2} \right)} \left( {x + y} \right)}} \cr
& {\text{then}}{\text{, substituting }} - 1{\text{ for }}x{\text{ and 2 for }}y \cr
& = \frac{{\left( { - 1} \right){{\left( 2 \right)}^3}}}{{ - 1 + 2}} \cr
& {\text{simplifying}} \cr
& = \frac{{ - 8}}{1} \cr
& = - 8 \cr} $$
