Answer
$\frac{1}{4}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{2x - y}}{{4{x^2} - {y^2}}} \cr
& {\text{Evaluating the limit by direct substitution}}{\text{.}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{2x - y}}{{4{x^2} - {y^2}}} = \frac{{2\left( 1 \right) - \left( 2 \right)}}{{4{{\left( 1 \right)}^2} - {{\left( 2 \right)}^2}}} = \frac{0}{0} \cr
& {\text{Factoring the difference of two squares}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{2x - y}}{{4{x^2} - {y^2}}} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{2x - y}}{{\left( {2x + y} \right)\left( {2x - y} \right)}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{1}{{2x + y}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{1}{{2x + y}} = \frac{1}{{2\left( 1 \right) + 2}} = \frac{1}{4} \cr
& {\text{Therefore}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {1,2} \right)} \frac{{2x - y}}{{4{x^2} - {y^2}}} = \frac{1}{4} \cr} $$
