2. Given $f(x)=\frac{1+x}{1-x}$, let $f_{1}(x)=f(x)$, $f_{k+1}(x)=f\left(f_{k}(x)\right)(k=1,2, \cdots)$. Then $f_{2007}(x)$ $=(\quad)$.
(A) $\frac{1+x}{1-x}$
(B) $\frac{x-1}{x+1}$
(C) $x$
(D) $-\frac{1}{x}$
2. B.
$$
\begin{array}{l}
f_{1}(x)=\frac{1+x}{1-x}, f_{2}(x)=\frac{1+f_{1}}{1-f_{1}}=-\frac{1}{x}, \\
f_{3}(x)=\frac{1+f_{2}}{1-f_{2}}=\frac{x-1}{x+1}, f_{4}(x)=\frac{1+f_{3}}{1-f_{3}}=x . \\
\text { Therefore, } f_{4 n+1}(x)=\frac{1+x}{1-x}, f_{4 n+2}(x)=-\frac{1}{x}, \\
f_{4 n+3}(x)=\frac{x-1}{x+1}, f_{4 n}(x)=x .
\end{array}
$$

Since 2007 is of the form $4 n+3$, the answer is (B).