Example 11 Find the value of $\sqrt{1989+1985 \sqrt{1990+1986 \sqrt{1991+1987 \sqrt{\cdots}}}}$.
Let $f(x)=$
$$
\sqrt{x+(x-4) \sqrt{(x+1)+(x-3) \sqrt{(x+2)+(x-2) \sqrt{\cdots}}}} \text {, then }
$$
$f(x)$ satisfies the following functional equation:
$$
[f(x)]^{2}=x+(x-4) f(x+1) .
$$

From the identity $(x-2)^{2}=x+(x-4)(x-1)$, we know that $f(x)=x-2$ is one of its solutions.

Since $f(1989)$ is a uniquely determined value, it should have only one solution.
We will prove that when $x \geqslant 6$, the original functional equation has only one solution $f(x)=x-2$.
First, when $x \geqslant 6$, $f(x)>\sqrt{(x-4) \sqrt{(x-3) \sqrt{(x-2) \sqrt{\cdots}}}}$
$$
\begin{aligned}
> & \sqrt{(x-4) \sqrt{(x-4) \sqrt{(x-4) \sqrt{\cdots}}}} \\
& =(x-4)^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=(x-4) \\
& \geqslant \frac{1}{2}(x-2) .
\end{aligned}
$$

Second, when $x \geqslant 6$,
$$
\begin{aligned}
f(x) & <\sqrt{(2 x-4) \sqrt{(2 x-2) \sqrt{2 x \sqrt{(2 x+2) \sqrt{\cdots}}}}} \\
& <\sqrt{2(x-2) \sqrt{4(x-2) \sqrt{8(x-2) \sqrt{\cdots}}}} \\
& =(x-2)^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots} \cdot 2^{\frac{1}{4}} \cdot 4^{\frac{1}{4}} \cdot 8^{\frac{1}{8} \cdots} \\
& =(x-2) 2^{\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\cdots}=4(x-2) .
\end{aligned}
$$

Therefore, $\frac{1}{2}(x-2)<f(x)<4(x-2) \quad(x \geqslant 6)$,
$$
\frac{1}{2}(x-1)<f(x+1)<4(x-1) \text {. }
$$

From $f^{2}(x)=x+(x-4) f(x+1)$, we have
$$
\begin{aligned}
\frac{1}{2}(x-2)^{2} & <\frac{1}{2} x+(x-4) f(x+1)<f^{2}(x) \\
& <4 x+(x-4) f(x+1)<4(x-1)^{2},
\end{aligned}
$$

Thus, $\frac{1}{\sqrt{2}}(x-2)<f(x)<\sqrt{4}(x-2) \quad(x \geqslant 6)$.
Assume $\left(\frac{1}{2}\right)^{2^{-k}}(x-2)<f(x)<4^{2^{-k}}(x-2)$, then
$$
\left(\frac{1}{2}\right)^{2^{-k}}(x-1)<f(x+1)<4^{2^{-k}}(x-1) \text {. }
$$

Therefore, $\left(\frac{1}{2}\right)^{2^{-k}}(x-2)^{2}<\left(\frac{1}{2}\right)^{2^{-k}}+(x-4) f(x+1)<f^{2}(x)$
$$
\begin{array}{l}
<4^{2^{-k}}+(x-4) f(x+1)<4^{2^{-k}}(x-2)^{2}, \\
\left(\frac{1}{2}\right)^{2^{-(k+1)}}(x-2)<f(x)<4^{2^{-(k+1)}}(x-2) \quad(x \geqslant 6) .
\end{array}
$$

This proves that for all natural numbers $n$,
$$
\sqrt[2^{n}]{\frac{1}{2}}(x-2)<f(x)<\sqrt[2 n]{4}(x-2) \quad(x \geqslant 6) .
$$

Let $n \rightarrow+\infty$ to get $f(x)=x-2(x \geqslant 6)$, then $f(1989)=1987$.