$$\begin{gathered} Letf\left(x\right)=\frac{x+1}{x-1},fof\left(x\right)=f^2\left(x\right);fofof\left(x\right)=f^3\left(x\right),fofofof\left(x\right)=f^4\left(x\right)\mathrm{....} \\ then{f}^{2008}\left(x\right)= \end{gathered}$$

$\begin{array}{l}f\left(x\right)=\frac{x+1}{x-1}\\ fof\left(x\right)=f\left[f\left(x\right)\right]=f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}\Rightarrow \frac{x+1+x-1}{x+1-x+1}=\frac{2x}{2}=x\\ fofof\left(x\right)=f\left[fof\left(x\right)\right]=f\left(x\right)\\ f^n\left(x\right)isxifniseven,f^{ }\left(x\right)ifnisodd\\ \therefore f^{2008}\left(x\right)isx\end{array}$