Find the limit of $U_n$ that satisfies $x_{n+1}=\frac{n+2}{3n+11}(\sqrt{x_n}+\sqrt[3]{7+x_n})$

Find the limit of $U_n$ that satisfies \begin{cases} x_1=\sqrt[2022] \frac{2022}{2021}\\ x_{n+1}=\frac{n+2}{3n+11}(\sqrt{x_n}+\sqrt[3]{7+x_n}), \forall n \in \mathbb{N*} \end{cases}

I am always confused when I comes to a sequence with no fixed pattern, that is the coefficient ( like $\frac{n+2}{3n+11}$) is changed.

One thing I can assume is that for $n \to +\infty$, then $\frac{n+2}{3n+11} \to 1/3$, then assume $L$ is the existed limit, we can solve for $L=1$.

Now what I tried to do is that I subtract both sides in someway so that $x_{n+1}-1$ or $x_n-1$ in someway, but apparently that yield no results ( and doesn't even work as well).

Should I change my approach? What tools should I use?

Clearly, if the sequence converges, $\lim_{n\to \infty} x_{n+1}=\lim_{n\to \infty} x_n.$ We see that $$\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty} \frac{n+2}{3n+11} \left(\sqrt{x_n}+\sqrt[3]{7+x_n}\right)=\lim_{n\to\infty}\frac{1}{3}\left(\sqrt{x_n}+\sqrt[3]{7+x_n}\right)=\lim_{n\to\infty}x_n.$$ Thus, in the limit, we have that $$\frac{1}{3}\left(\sqrt{x_n}+\sqrt[3]{7+x_n}\right)=x_n.$$ Solving gives $x_n=1,$ so $$\lim_{n\to\infty}x_n=1.$$