Example 6 (9th China High School Mathematics Olympiad Winter Camp Problem) Find all functions $f$: $[1,+\infty) \rightarrow[1,+\infty)$ that satisfy the following conditions:
(1) $f(x) \leqslant 2(x+1)$;
(2) $f(x+1)=\frac{1}{x}\left[(f(x))^{2}-1\right]$.
Prove that the function $f(x)=x+1$ is one of the solutions.
Below, we prove that $f(x)=x+1$ is the unique solution, i.e., we need to prove that $g(x)=f(x)-(x+1)=0$ always holds. $x+1 \leqslant 2(x+1), g(x+1)+x+2=\frac{[g(x)+(x+1)]^{2}-1}{x}$, i.e., $-x \leqslant g(x) \leqslant x+1$,
$g(x+1)=\frac{[g(x)]^{2}+2(x+1) g(x)+(x+1)^{2}-1-x(x+2)}{x}$
$=\frac{[g(x)]^{2}+2(x+1) g(x)}{x}=\frac{g(x)[g(x)+x+x+2]}{x}$.
Notice that $g(x)+x \geqslant 0$, so we have $g(x)=\frac{x g(x+1)}{g(x)+x+x+2}$.
From $-x-1 \leqslant g(x) \leqslant x+1$, we have $|g(x)| \leqslant x+1$.
Thus, $|g(x)|=\frac{x|g(x+1)|}{[g(x)+x]+x+2} \leqslant \frac{x|g(x+1)|}{x+2} \leqslant \frac{x(x+2)}{x+2}=x$.
Therefore, from $|g(x)| \leqslant x$, we deduce $|g(x+1)| \leqslant x+1$.
Thus, $|g(x)| \leqslant \frac{x(x+1)}{x+2}$, which implies $|g(x+1)| \leqslant \frac{(x+1)(x+2)}{x+3}$, and $|g(x)| \leqslant$ $\frac{\frac{x(x+1)(x+2)}{x+3}}{x+2}=\frac{x(x+1)}{x+3}$.

Let $n$ be a positive integer, $|g(x)| \leqslant \frac{x(x+1)}{x+n}$. Then $|g(x+1)| \leqslant \frac{(x+1)(x+2)}{x+n+1}$, thus $|g(x)| \leqslant$ $\frac{x|g(x+1)|}{x+2} \leqslant \frac{x(x+1)(x+2)}{(x+2)(x+n+1)}=\frac{x(x+1)}{x+n+1}$.
By the principle of mathematical induction, for all positive integers, $|g(x)| \leqslant \frac{x(x+1)}{x+n}$.
Taking $n \rightarrow \infty$, we deduce $|g(x)| \leqslant 0$, i.e., $g(x)=0$ always holds.
This proves that $f(x)=x+1$ is the unique solution.
The problem is given by inequalities and identities. In the proof, inequalities are repeatedly used for estimation. Of course, the key to solving this problem is to see that $f(x)=x+1$, and then use inequalities to estimate, obtaining $f(x)-(x+1) \equiv 0$, and the technique of taking limits as $n \rightarrow+\infty$.