17 (10 points) Given the quadratic function $f(x)=x^{2}-t x+t$ where $t \in \mathbf{R}$, and it satisfies: (1) The solution set of the inequality $f(x) \leqslant 0$ has exactly one element; (2) If $0<x_{2}<x_{1}$, then the inequality $f\left(x_{2}\right)<f\left(x_{1}\right)$ always holds. Let the sequence $\left\{a_{n}\right\}$ have the sum of its first $n$ terms $S_{n}=f(n)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Find $\lim _{n \rightarrow \infty}\left(\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{n} a_{n+2}}\right)$.
17 (1) Because (1), so $\Delta=0$, i.e., $t^{2}-4 t=0$, so $t=0$, or $t=4$. Also, because (2), $f(x)$ is an increasing function on $(0,+\infty)$, so $t=0$, $f(x)=x^{2}, s_{n}=f(n)=n^{2}$, so
$$
a_{n}=s_{n}-s_{n-1}=2 n-1(n \geqslant 2) .
$$

Also, because $a_{1}=1$ fits $-a_{n}=2 n-1$, so $a_{n}=2 n-1$.
(2) Because $\frac{1}{a_{n} a_{n+2}}=\frac{1}{4}\left(\frac{1}{a_{n}}-\frac{1}{a_{n+2}}\right)$, so
$$
\begin{aligned}
\text { the original expression } & =\lim _{n \rightarrow \infty} \frac{1}{4}\left(\frac{1}{1}-\frac{1}{5}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n+3}\right) \\
& =\lim _{n \rightarrow \infty} \frac{1}{4}\left(1+\frac{1}{3}+\cdots-\frac{1}{2 n+1}-\frac{1}{2 n+3}\right) \\
& =\frac{1}{3} .
\end{aligned}
$$