Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \frac{{1 - \cos y}}{{x{y^2}}} \cr
& {\text{Use the product property of limits}} \cr
& = \left( {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \frac{1}{x}} \right)\left( {\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {2,0} \right)} \frac{{1 - \cos y}}{{{y^2}}}} \right) \cr
& = \left( {\mathop {\lim }\limits_{x \to 2} \frac{1}{x}} \right)\left( {\mathop {\lim }\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}}}} \right) \cr
& {\text{Calculating }}\mathop {\lim }\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}}}{\text{ by using the L'Hopitals Rule}} \cr
& \mathop {\lim }\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}}} = \mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{{2y}} = \mathop {\lim }\limits_{y \to 0} \frac{{\cos y}}{2} = \frac{{\cos 0}}{2} = \frac{1}{2} \cr
& {\text{Then}} \cr
& = \left( {\mathop {\lim }\limits_{x \to 2} \frac{1}{x}} \right)\left( {\frac{1}{2}} \right) \cr
& = \left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right) \cr
& = \frac{1}{4} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( { - 1,0} \right)} \frac{{1 - \cos y}}{{x{y^2}}} = \frac{1}{4} \cr} $$
