The 38th question, the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=1$, and for any $n \in Z^{+}$, it holds that $x_{n+1}=x_{n}+3 \sqrt{x_{n}}+\frac{n}{\sqrt{x_{n}}}$, try to find the value of $\lim _{n \rightarrow+\infty} \frac{n^{2}}{x_{n}}$.
The 38th question,
Solution: According to the conditions, we have:
$$
\begin{array}{l}
x_{n+1}=x_{n}+3 \sqrt{x_{n}}+\frac{n}{\sqrt{x_{n}}}=x_{n}+2 \sqrt{x_{n}}+\left(\frac{n}{\sqrt{x_{n}}}+\sqrt{x_{n}}\right) \geq x_{n}+2 \sqrt{x_{n}}+2 \sqrt{n}>\left(\sqrt{x_{n}}+1\right)^{2} \\
\Rightarrow \sqrt{x_{n+1}}>\sqrt{x_{n}}+1 \\
\Rightarrow \sqrt{x_{n}}>\sqrt{x_{1}}+(n-1)=n
\end{array}
$$

Furthermore, according to the conditions, we have:
$$
\begin{array}{l}
x_{n+1}=x_{n}+3 \sqrt{x_{n}}+\frac{n}{\sqrt{x_{n}}}x_{n}+3 \sqrt{x_{n}}+\frac{n}{\frac{3 n-1}{2}}>x_{n}+3 \sqrt{x_{n}}+\frac{2}{3}=\left(\sqrt{x_{n}}+\frac{3}{2}\right)^{2}-\frac{19}{12} \\
\Rightarrow \frac{19}{12}>\left(\left(\sqrt{x_{n}}+\frac{3}{2}\right)-\sqrt{x_{n+1}}\right)\left(\left(\sqrt{x_{n}}+\frac{3}{2}\right)+\sqrt{x_{n+1}}\right)>2 \sqrt{x_{n}}\left(\left(\sqrt{x_{n}}+\frac{3}{2}\right)-\sqrt{x_{n+1}}\right) \\
>2 n\left(\left(\sqrt{x_{n}}+\frac{3}{2}\right)-\sqrt{x_{n+1}}\right) \\
\Rightarrow \sqrt{x_{n+1}}>\left(\sqrt{x_{n}}+\frac{3}{2}\right)-\frac{19}{24 n} \\
\Rightarrow \sqrt{x_{n}}=\sqrt{x_{1}}+\sum_{k=1}^{n-1}\left(\sqrt{x_{k+1}}-\sqrt{x_{k}}\right)>1+\frac{3(n-1)}{2}-\frac{19}{24}\left(\sum_{k=1}^{n-1} \frac{1}{k}\right)>1+\frac{3(n-1)}{2}-\frac{19}{24}(1+ \\
\left.\sum_{k=2}^{n-1} \ln \frac{k}{k-1}\right)=\frac{3 n-1}{2}-\frac{19}{24}(1+\ln (n-1))
\end{array}
$$

Thus, we know:
$$
\begin{array}{l}
\frac{3 n-1}{2}-\frac{19}{24}(1+\ln (n-1)) \leq \sqrt{x_{n}} \leq \frac{3 n-1}{2} \\
\Rightarrow \lim _{n \rightarrow+\infty} \frac{n}{\sqrt{x_{n}}}=\frac{2}{3} \\
\Rightarrow \lim _{n \rightarrow+\infty} \frac{n^{2}}{x_{n}}=\frac{4}{9}
\end{array}
$$

In conclusion, the required $\lim _{n \rightarrow+\infty} \frac{n^{2}}{x_{n}}=\frac{4}{9}$.