Answer
$$ - 1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3,1} \right)} \frac{{{x^2} - 7xy + 12{y^2}}}{{x - 3y}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}}{\text{, substitute 3 for }}x{\text{ and 1 for }}y{\text{ into }}f\left( {x,y} \right) \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3,1} \right)} \frac{{{x^2} - 7xy + 12{y^2}}}{{x - 3y}} = \frac{{{{\left( 3 \right)}^2} - 7\left( 3 \right)\left( 1 \right) + 12{{\left( 1 \right)}^2}}}{{3 - 3\left( 1 \right)}} \cr
& {\text{simplifying}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3,1} \right)} \frac{{{x^2} - 7xy + 12{y^2}}}{{x - 3y}} = \frac{0}{0}{\text{ indeterminate}} \cr
& {\text{simplify the expression }}\frac{{{x^2} - 7xy + 12{y^2}}}{{x - 3y}}{\text{ factoring the numerator}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3,1} \right)} \frac{{\left( {x - 4y} \right)\left( {x - 3y} \right)}}{{x - 3y}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {3,1} \right)} \left( {x - 4y} \right) \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = 3 - 4\left( 1 \right) \cr
& = - 1 \cr} $$
