Answer
$$ - 6$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \frac{{10uv - 2{v^2}}}{{{u^2} + {v^2}}} \cr
& {\text{use the law 5 }}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {a,b} \right)} \frac{{f\left( {x,y} \right)}}{{g\left( {x,y} \right)}}.\,\,\,{\text{theorem 12}}{\text{.2 }}\left( {{\text{see page 887}}} \right){\text{. then}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {10uv - 2{v^2}} \right)}}{{\mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {{u^2} + {v^2}} \right)}} \cr
& {\text{use law 1}} \cr
& = \frac{{\mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {10uv} \right) - \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {2{v^2}} \right)}}{{\mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {{u^2}} \right) + \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {{v^2}} \right)}} \cr
& {\text{law constant multiply}} \cr
& = \frac{{10\mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {uv} \right) - 2\mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {{v^2}} \right)}}{{\mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {{u^2}} \right) + \mathop {\lim }\limits_{\left( {u,v} \right) \to \left( {1, - 1} \right)} \left( {{v^2}} \right)}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = \frac{{10\left( 1 \right)\left( { - 1} \right) - 2{{\left( { - 1} \right)}^2}}}{{{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& = \frac{{ - 10 - 2}}{{1 + 1}} \cr
& = - 6 \cr} $$