Answer
$$\frac{2}{9}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{{x^2}y}}{{{x^4} + 2{y^2}}} \cr
& {\text{use the law 5 }}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} \frac{{f\left( {x,y} \right)}}{{g\left( {x,y} \right)}}.\,\,\,{\text{theorem 12}}{\text{.2 }}\left( {{\text{see page 887}}} \right){\text{. then}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} {x^2}y}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \left( {{x^4} + 2{y^2}} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} {x^2}y}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} {x^4} + 2\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} {y^2}}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = \frac{{{{\left( 1 \right)}^2}\left( 2 \right)}}{{{{\left( 1 \right)}^4} + 2{{\left( 2 \right)}^2}}} \cr
& {\text{simplifying}} \cr
& = \frac{2}{{1 + 8}} \cr
& = \frac{2}{9} \cr} $$
