(3) Find all functions $f:[1,+\infty) \rightarrow[1,+\infty)$ that satisfy the following conditions:
(1) $f(x) \leqslant 2(x+1)$;
(2) $f(x+1)=\frac{1}{x}\left((f(x))^{2}-1\right)$.
For $\forall x \geqslant 11^{1}$, let
$$
g(x)=f(x)-(x+1)
$$

Using the conditions of the problem, $\forall x \geqslant 1$, we have
$$
-x \leqslant g(x) \leqslant x+1
$$

Obviously, $\forall x \geqslant 1$.
$$
\begin{aligned}
g(x+1)= & f(x+1)-(x+2)= \\
& \frac{1}{x}\left((f(x))^{2}-1\right)-(x+2)= \\
& \frac{1}{x}\left((g(x)+x+1)^{2}-1\right)-(x+2)= \\
& \frac{1}{x}\left((g(x)+x+1)^{2}-1-x(x+2)\right)= \\
& \frac{1}{x} g(x)(g(x)+2(x+1))
\end{aligned}
$$

Now, we prove that for $\forall x \geqslant 1$, we have
$$
g(x)=0
$$

By contradiction, if there exists $x_{0} \geqslant 1$, such that $g\left(x_{0}\right)>0$, then, by (3), we have
$$
\begin{aligned}
g\left(x_{0}+1\right)= & \frac{1}{x_{0}} g\left(x_{0}\right)\left(g\left(x_{0}\right)+2\left(x_{0}+1\right)\right)> \\
& \frac{x_{0}+2}{x_{0}} g\left(x_{0}\right)
\end{aligned}
$$

Now, we use mathematical induction to prove: $\forall k \in \mathbf{N}_{+}$ (where $\mathbf{N}_{+}$ is the set of all positive integers), we have
$$
g\left(x_{0}+k\right)>\frac{\left(x_{0}+k\right)\left(x_{0}+k+1\right)}{x_{0}\left(x_{0}+1\right)} g\left(x_{0}\right)
$$

When $k=1$, by (5) we get (6). Assume for some $k \in \mathbf{N}_{+}$, (6) holds. Then $g\left(x_{0}+k\right)>0$. Consider the case $k+1$, in (3), let $x=x_{0}+k$, we have $g\left(x_{0}+k+1\right)=\frac{1}{x_{0}+k} g\left(x_{0}+k\right)\left(g\left(x_{0}+k\right)+2\left(x_{0}+k+1\right)\right)>$
(1) $\forall$ means for all.
$$
\begin{array}{l}
\frac{1}{x_{0}+k} g\left(x_{0}+k\right)\left(x_{0}+k+2\right)> \\
\frac{\left(x_{0}+k+1\right)\left(x_{0}+k+2\right)}{x_{0}\left(x_{0}+1\right) g\left(x_{0}\right)} g\left(x_{0}\right)
\end{array}
$$

The last inequality above is derived using the induction hypothesis (6). Thus, (6) holds for any positive integer $k$.
Since $g\left(x_{0}\right)>0$, take a sufficiently large positive integer $k$, such that
$$
g\left(x_{0}\right) \frac{\left(x_{0}+k\right)}{x_{0}\left(x_{0}+1\right)}>1
$$

Then, by (6) and (8), we have
$$
g\left(x_{0}+k\right)>x_{0}+k+1
$$

This contradicts the right inequality in (2).
If there exists $x_{0} \geqslant 1$, such that $g\left(x_{0}\right)<0$. Similarly, by (3) and (2), we have
$$
\begin{aligned}
g\left(x_{0}+1\right)= & \frac{1}{x_{0}} g\left(x_{0}\right)\left(\left(g\left(x_{0}\right)+x_{0}\right)+\left(x_{0}+2\right)\right) \leqslant \\
& \frac{x_{0}+2}{x_{0}} g\left(x_{0}\right)
\end{aligned}
$$

Similarly to the proof of (6), we can get $\forall k \in \mathbf{N}_{+}$
$$
g\left(x_{0}+k\right) \leqslant \frac{\left(x_{0}+k\right)\left(x_{0}+k+1\right)}{x_{0}\left(x_{0}+1\right)} g\left(x_{0}\right)
$$

Since $g\left(x_{0}\right)<0$, take a very large positive integer $k$, such that
$$
g\left(x_{0}\right) \frac{\left(x_{0}+k\right)}{x_{0}\left(x_{0}+1\right)}<-1
$$

From (1D) and (12), we have
$$
g\left(x_{0}+k\right)<-\left(x_{0}+k+1\right)
$$

This contradicts the left inequality in (2).
Thus, (4) holds. Using (1), $\forall x \geqslant 1$, we have
$$
f(x)=x+1
$$
(44)