Answer
$$\ln \left( 2 \right) + 1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \ln \sqrt {xy} \cr
& {\text{write }}\sqrt {xy} {\text{ as }}{\left( {xy} \right)^{1/2}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \ln {\left( {xy} \right)^{1/2}} \cr
& {\text{logarithmic property}} \cr
& = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \frac{1}{2}\ln \left( {xy} \right) \cr
& {\text{law constant multiply}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {{e^2},4} \right)} \ln \left( {xy} \right) \cr
& {\text{evaluate using the theorem 12}}{\text{.1}} \cr
& = \frac{1}{2}\ln \left( {\left( {{e^2}} \right)4} \right) \cr
& = \frac{1}{2}\ln \left( {4{e^2}} \right) \cr
& = \frac{1}{2}\ln \left( 4 \right) + \frac{1}{2}\ln \left( {{e^2}} \right) \cr
& = \ln \left( 2 \right) + 1 \cr} $$
