Answer
$$14$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \left( {x{y^8} - 3{x^2}{y^3}} \right) \cr
& {\text{use the law 1 }}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} \left( {f\left( {x,y} \right) + g\left( {x,y} \right)} \right).\,\,\,\left( {{\text{see page 887}}} \right){\text{then}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \left( {x{y^8}} \right) - \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \left( {3{x^2}{y^3}} \right) \cr
& {\text{law constant multiply}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \left( {x{y^8}} \right) - 3\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2, - 1} \right)} \left( {{x^2}{y^3}} \right) \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = \left( 2 \right){\left( { - 1} \right)^8} - 3{\left( 2 \right)^2}{\left( { - 1} \right)^3} \cr
& {\text{simplifying}} \cr
& = 2 + 12 \cr
& = 14 \cr} $$
