Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{4x - y}}{{\sin y - 1}} \cr
& {\text{Using the limit laws }} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left( {4x - y} \right)}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left( {\sin y - 1} \right)}} \cr
& {\text{then}}{\text{,}} \cr
& = \frac{{4\left( 0 \right) - \left( 0 \right)}}{{\sin \left( 0 \right) - 1}} \cr
& {\text{simplifying}} \cr
& = \frac{0}{{ - 1}} \cr
& = 0 \cr} $$
