Answer
$$2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,2} \right)} \frac{{{y^2} - 4}}{{xy - 2x}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}}{\text{, substitute 2 for }}x{\text{ and 2 for }}y{\text{ into }}f\left( {x,y} \right) \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,2} \right)} \frac{{{y^2} - 4}}{{xy - 2x}} = \frac{{{{\left( 2 \right)}^2} - 4}}{{\left( 2 \right)\left( 2 \right) - 2\left( 2 \right)}} \cr
& {\text{simplifying}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,2} \right)} \frac{{{y^2} - 4}}{{xy - 2x}} = \frac{0}{0}{\text{ indeterminate}} \cr
& {\text{simplify the expression }}\frac{{{y^2} - 4}}{{xy - 2x}}{\text{ factoring the diffence of squares and the }} \cr
& {\text{denominator}}{\text{. then}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,2} \right)} \frac{{\left( {y + 2} \right)\left( {y - 2} \right)}}{{x\left( {y - 2} \right)}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,2} \right)} \frac{{y + 2}}{x} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = \frac{{2 + 2}}{2} \cr
& = 2 \cr} $$
