Let $x_0$ be a transcendental number, $x_{n+1}=\frac{3-x_n}{x_n^2+3x_n-2}$. What is the limit of $x_n$? Let $x_0$ be a transcendental number, $$x_{n+1}=\frac{3-x_n}{x_{n}^{2}+3x_{n}-2}$$
What is the limit of $x_{n}$?
Choose $x_0=\pi$, and is seems that the limit of $x_n$ is $-1$. But what is the proof for this $\pi$ and other numbers? Let
$$f(x)=\frac{3-x}{x^{2}+3x-2}$$
The following may be helpful.
$$f'(x)=\frac{(x-7)(x+1)}{(x^{2}+3x-2)^2}$$
$$f(x)-x=\frac{-(x-1)(x+1)(x+3)}{x^{2}+3x-2}$$
$$f(x)+1=\frac{(x+1)^{2}}{x^{2}+3x-2}$$.

Let $f(x) = \frac{3-x}{x^2+3x-2}$. If $\lim x_n$ exists, then $L = \lim x_{n+1}=\lim x_n$, so set $$L=f(L)$$
There's three solutions to this: $L = -3, -1, 1$. In order to find the correct one, note that for a small neighborhood around $-3$, you have $|f(x)+3|>|x+3|$, and around $1$, you have $|f(x)-1|>|x-1|$. For both $-3$ and $1$, the difference will be made even bigger. Around $-1$ on the other hand, you have $|f(x)+1|<|x+1|$, so the difference is becoming smaller (this is not a rigorous proof but more of an intuitive one).
Thus, for "most" $x_0$, it will converge to $-1$. The only way it will converge to $-3$ or $1$ is if it converges exactly in a finite number of iterations. But for that to be true, it has to be a solution to $$f^n(x_0) = -3$$
(or $1$)
for some $n$, meaning that it must be algebraic. Therefore, for all transcendental, the limit will be $-1$.
