1. Let $f(x)=\frac{1}{1-x}, f^{r}(x)=f^{n^{-1}}(f(x))$, then $f^{2002}(2002)=(\quad)$.
A. 2002
B. 0
C. $-\frac{1}{2001}$
D. Does not exist
1. $\mathrm{C}$

Since $f(f(x))=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}, f(f(f(x)))=x$,
thus $f^{2002}(2002)=f^{1999}(2002)=f^{1996}(2002)=\cdots$ $=f(2002)=-\frac{1}{2001}$.