4. In the infinite sequence $\left\{a_{n}\right\}$, $a_{1}=0, a_{n}=\frac{a_{n-1}+4}{a_{n-1}-2}(n \geq 2)$. If $\lim a_{n}=A$, then $A=$
4. -1

Analysis: It is easy to know that $\frac{a_{n}-4}{a_{n}+1}=-\frac{3}{2} \frac{a_{n-1}-4}{a_{n-1}+1}$, so $\frac{a_{n}-4}{a_{n}+1}=\left(-\frac{3}{2}\right)^{n}(-4)$, which means $a_{n}=\frac{5}{\left(-\frac{3}{2}\right)^{n} 4+1}-1$. Therefore, $\lim _{n \rightarrow+\infty} a_{n}=-1$;