jmc

We have that $$\begin{array}{rl}{x}_1& =\frac{1+3}{1-3}=-2,\\ x_2& =\frac{1+(-2)}{1-(-2)}=-\frac{1}{3},\\ x_3& =\frac{1+(-1/3)}{1-(-1/3)}=\frac{1}{2},\\ x_4& =\frac{1+1/2}{1-1/2}=3.\end{array}$$Since $x_4=x_0=3,$ and each term depends only on the previous term, the sequence is periodic from here on, with a period of length 4. Hence, $x_{12345}=x_1=\boxed{-2}.$