Cho dãy số $(x_n)$ xác định bởi $$\{\begin{array}{l}{x}_1=\frac{9}{8}\\ 3(n+3{)}^2{x}_n=3n^2{x}_{n+1}-(2n+3)x_{n+1}{x}_n,\forall n\ge 1\end{array}$$
Tìm $\lim x_n$.

Ta có: $9(n+3{)}^2{x}_n=9n^2{x}_{n+1}-(6n+9)x_{n+1}{x}_n=9n^2{x}_{n+1}-\left((n+3{)}^2-n^2\right)x_{n+1}{x}_n,\forall n\in {\mathbb{Z}}^{+}$
$\iff (n+3{)}^2{x}_n.(x_{n+1}+9)=n^2.x_{n+1}(x_n+9),\forall n\in {\mathbb{Z}}^{+}$.

Rõ ràng $x_n\ne 0,\forall n\in {\mathbb{Z}}^{+}$. Do đó: $\frac{x_{n+1}+9}{x_{n+1}}={\left(\frac{n}{n+3}\right)}^2.\frac{x_n+9}{x_n},\forall n\in {\mathbb{Z}}^{+}$

Nên ta có được $\frac{x_{n+1}+9}{x_{n+1}}=\prod _{k=1}^n{\left(\frac{k}{k+3}\right)}^2.\frac{x_1+9}{x_1}=9.{\left(\prod _{k=1}^n\frac{k}{k+3}\right)}^2,\forall n\in {\mathbb{Z}}^{+}$

Ta có bất đẳng thức phụ: $\frac{1}{x-1}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x}$ với mọi $x>0$.

Khi đó: $\ln \left(\prod _{k=1}^n\frac{k+3}{k}\right)=\sum _{k=1}^n\ln \left(1+\frac{3}{k}\right)>\sum _{k=1}^n\frac{3}{k-3},\forall n\in {\mathbb{Z}}^{+}$.

Cho $n\to +\infty$ ta thu được $\underset{n\to +\infty }{\lim }\ln \left(\prod _{k=1}^n\frac{k+3}{k}\right)=+\infty$

Nên $\underset{n\to +\infty }{\lim }\prod _{k=1}^n\frac{k+3}{k}=+\infty$, kéo theo $\underset{n\to +\infty }{\lim }\prod _{k=1}^n\frac{k}{k+3}=0$.

Từ đó ta thu được $\underset{n\to +\infty }{\lim }\left(1+\frac{9}{x_n}\right)=\underset{n\to +\infty }{\lim }9.{\left(\prod _{k=1}^{n-1}\frac{k}{k+3}\right)}^2=0$.

Vì vậy $\underset{n\to +\infty }{\lim }{x}_n=-9$