Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\theta \to {0^ + }} \left[ {\ln \left( {\sin 2\theta } \right) - \ln \left( {\tan \theta } \right)} \right] \cr
& {\text{Use the property of logarithms}} \cr
& {\text{ = }}\mathop {\lim }\limits_{\theta \to {0^ + }} \ln \left( {\frac{{\sin 2\theta }}{{\tan \theta }}} \right) \cr
& {\text{Use trigonometric identities and simplify}} \cr
& {\text{ = }}\mathop {\lim }\limits_{\theta \to {0^ + }} \ln \left( {\frac{{2\sin \theta \cos \theta }}{{\frac{{\sin \theta }}{{\cos \theta }}}}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{\theta \to {0^ + }} \ln \left( {\frac{{2\sin \theta {{\cos }^2}\theta }}{{\sin \theta }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{\theta \to {0^ + }} \ln \left( {2{{\cos }^2}\theta } \right) \cr
& {\text{Evaluate the limit}} \cr
& = \ln \left( {2{{\cos }^2}0} \right) \cr
& = \ln 2 \cr} $$
