Answer
$$\frac{1}{{2\pi }}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} \frac{{\cos xy + \sin xy}}{{2y}} \cr
& {\text{use the law 5 }}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} \frac{{f\left( {x,y} \right)}}{{g\left( {x,y} \right)}}.\,\,\,{\text{theorem 12}}{\text{.2 }}\left( {{\text{see page 887}}} \right){\text{. then}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} \left( {\cos xy + \sin xy} \right)}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} 2y}} \cr
& {\text{use law 1}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} \left( {\cos xy} \right) + \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} \left( {\sin xy} \right)}}{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} 2y}} \cr
& {\text{law constant multiply}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} \left( {\cos xy} \right) + \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} \left( {\sin xy} \right)}}{{2\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,\pi } \right)} y}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = \frac{{\cos \left( 0 \right) + \sin 0}}{{2\left( \pi \right)}} \cr
& {\text{simplifying}} \cr
& = \frac{1}{{2\pi }} \cr} $$
