Example 3 Let $f_{1}(x)=\frac{2}{x+1}$, and $f_{n+1}(x)=$ $f_{1}\left[f_{n}(x)\right], n \in \mathbf{N}^{*}$. Let $a_{n}=\frac{f_{n}(2)-1}{f_{n}(2)+2}$, then $a_{99}=$
Solution: Since $f_{1}(2)=\frac{2}{3}$, then $a_{1}=-\frac{1}{8}$, and $f_{n}(2)=\frac{2}{f_{n-1}(2)+1}$,

so $\frac{f_{n}(2)-1}{f_{n}(2)+2}=\frac{\frac{2}{f_{n-1}(2)+1}-1}{\frac{2}{f_{n-1}(2)+1}+2}=-\frac{1}{2} \cdot \frac{f_{n-1}(2)-1}{f_{n-1}(2)+2}$. That is, $a_{n}=-\frac{1}{2} a_{n-1}$,
thus $a_{99}=-\frac{1}{8}\left(-\frac{1}{2}\right)^{98}=-\frac{1}{2^{1011}}$.