Answer
$$\frac{1}{6}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,5} \right)} \frac{{\sqrt {x + y} - 3}}{{x + y - 9}} \cr
& {\text{evaluate using the theorem 12}}{\text{.1}}{\text{, substitute 3 for }}x{\text{ and 1 for }}y{\text{ into }}f\left( {x,y} \right) \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,5} \right)} \frac{{\sqrt {x + y} - 3}}{{x + y - 9}} = \frac{{\sqrt {4 + 5} - 3}}{{4 + 5 - 9}} \cr
& {\text{simplifying}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,5} \right)} \frac{{\sqrt {x + y} - 3}}{{x + y - 9}} = \frac{0}{0}{\text{ indeterminate}} \cr
& {\text{simplify the expression }}\frac{{\sqrt {x + y} - 3}}{{x + y - 9}}{\text{ rationalizing the numerator}} \cr
& = \frac{{\sqrt {x + y} - 3}}{{x + y - 9}}\left( {\frac{{\sqrt {x + y} + 3}}{{\sqrt {x + y} + 3}}} \right) \cr
& = \frac{{\left( {\sqrt {x + y} - 3} \right)\left( {\sqrt {x + y} + 3} \right)}}{{\left( {x + y - 9} \right)\left( {\sqrt {x + y} + 3} \right)}} \cr
& {\text{use }}\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} \cr
& = \frac{{{{\left( {\sqrt {x + y} } \right)}^2} - {{\left( 3 \right)}^2}}}{{\left( {x + y - 9} \right)\left( {\sqrt {x + y} + 3} \right)}} \cr
& = \frac{{x + y - 9}}{{\left( {x + y - 9} \right)\left( {\sqrt {x + y} + 3} \right)}} \cr
& = \frac{1}{{\sqrt {x + y} + 3}} \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,5} \right)} \frac{{\sqrt {x + y} - 3}}{{x + y - 9}} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,5} \right)} \frac{1}{{\sqrt {x + y} + 3}} \cr
& {\text{ evaluate using the theorem 12}}{\text{.1}} \cr
& = \frac{1}{{\sqrt {5 + 4} + 3}} \cr
& = \frac{1}{{3 + 3}} \cr
& = \frac{1}{6} \cr} $$
