The sequence $(x_n)$ : $x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-(\sqrt{x_{n+1}^2+1}-1)(\sqrt{x_n^2+1}-1)}$ The sequence $(x_n)$ is defined by the formula:
$$\left\{\begin{array}{cc}
x_1=1, x_2=\sqrt{3}\\
x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-\big(\sqrt{x_{n+1}^2+1}-1\big)\big(\sqrt{x_{n}^2+1}-1\big)}, \quad n=1,2,3\dots
\end{array} \right.$$
Find $\lim\limits_{n\to \infty}x_n$.
I see:
$$\begin{array}{rl}
x_{n+2}&=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-\big(\sqrt{x_{n+1}^2+1}-1\big)\big(\sqrt{x_{n}^2+1}-1\big)}\\
&=\frac{x_{n}\big(\sqrt{x_{n+1}^2+1}-1 \big)+x_{n+1}\big(\sqrt{x_{n}^2+1}-1 \big)}{x_{n+1}x_n-\frac{x_n^2x_{n+1}^2}{\big(\sqrt{x_{n+1}^2+1}+1\big)\big(\sqrt{x_{n}^2+1}+1\big)}}\\
&=\frac{x_{n+1}\big(\sqrt{x_{n}^2+1}+1\big)+x_{n}\big(\sqrt{x_{n+1}^2+1}+1\big)}{\big(\sqrt{x_{n+1}^2+1}+1\big)\big(\sqrt{x_{n}^2+1}+1\big)-x_nx_{n+1}}
\end{array}$$
That's all I can do. So, I hope hints from you. Thank you.

By rationalising the denominator, $x_{n+2}$ simplifies to $$\frac{\sqrt{x_n^2+1}\sqrt{x_{n+1}^2+1}+x_nx_{n+1}-1}{x_n+x_{n+1}}.$$
Letting $$x_n=\frac12\left(p_n-\frac1{p_n}\right),$$ that is (say) $$p_n=x_n+\sqrt{x_n^2+1}$$ and simplifying further, you get $$x_{n+2}=\frac{p_np_{n+1}-1}{p_n+p_{n+1}}$$
So, if $p_n=\cot\theta_n$, then $$x_{n+2}=\cot(\theta_n+\theta_{n+1}).$$
Furthermore, $$x_n=\frac12\left(\cot\theta_n-\frac1{\cot\theta_n}\right)=\cot{2\theta_n}.$$
It follows that $2\theta_{n+2}=\theta_n+\theta_{n+1},$ so that you can solve for $\theta_n$ explicitly from the initial conditions. The solution is
$$\frac{2\theta_2+\theta_1}{3}+\frac{4\theta_2-4\theta_1}{3}\left(-\frac12\right)^n,$$ which has limit $$\frac{2\theta_2+\theta_1}{3}$$.
You know that $\cot2\theta_1=1$ and $\cot2\theta_2=\sqrt{3}$, so that $\theta_1=\frac\pi8$ and $\theta_2=\frac\pi{12}$. Therefore, the limit of $\theta_n$ is $\frac{7\pi}{72}$, and so the limit of $x_n$ is $\cot\frac{7\pi}{36}$.
