The sequence $(x_n)$ satisfies $x_0 = 3$ and
\[x_n = \frac{1 + x_{n - 1}}{1 - x_{n - 1}}\]for all $\ge 1.$  Find $x_{12345}.$
A common strategy for such problems is to try to find a pattern or a formula for the sequence. I will start by computing the first few terms of the sequence and see if I notice anything. Using the given formula and the initial value, I get $x_1 = \frac{1 + x_0}{1 - x_0} = \frac{1 + 3}{1 - 3} = -2.$ Then, $x_2 = \frac{1 + x_1}{1 - x_1} = \frac{1 + (-2)}{1 - (-2)} = \frac{-1}{3}.$ Next, $x_3 = \frac{1 + x_2}{1 - x_2} = \frac{1 + \frac{-1}{3}}{1 - \frac{-1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}.$ Finally, $x_4 = \frac{1 + x_3}{1 - x_3} = \frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3.$ Wait a minute, this looks familiar. I started with $x_0 = 3$ and I ended up with $x_4 = 3.$ Let me check by computing $x_5.$ I get $x_5 = \frac{1 + x_4}{1 - x_4} = \frac{1 + 3}{1 - 3} = -2.$ Yes, this is the same as $x_1.$ So the sequence does repeat with a period of four. This means that I can find any term of the sequence by looking at its remainder when divided by four and using the corresponding term from the first four. For example, $x_8 = x_0 = 3$ because $8$ has a remainder of zero when divided by four. Similarly, $x_9 = x_1 = -2$ because $9$ has a remainder of one when divided by four. So, to find $x_{12345}$, I need to find the remainder of $12345$ when divided by four. Therefore, the remainder is one, and $x_{12345} = x_1 = -2.$ 