We have the function $f(x)=x^2-x-12=0$ with roots $r_1=-3$ and $r_2=4$.
We consider the sequence $x_{n+1}=g(x_n), \ n=0,1,2,\ldots $ where $g(x)=\sqrt{x+12}$.
We want to show that $x_n\rightarrow r_2, \ \forall x_0\in [-3,12]$.
Let $x_n\rightarrow c$ then $x_{n+1}\rightarrow c$. We have that $c=g(c)$ and $c$ must be a root of $f$, which is either $-3$ or $4$.
Is my idea correct? Or do we show in an other way that $x_n\rightarrow r_2$ ?
On
The OP might need to know the following:
Proposition 1: Let $(u_n)_{n\ge 0}$ be a convergent sequence and set $v_k = u_{k+1}$. Then
$$ \lim_{n\to \infty} v_n = \lim_{n\to \infty} u_n$$
Proposition 2: Let $(u_n)_{n\ge 0}$ and $(v_n)_{n\ge 0}$ be two sequences that converge to the same limit $L$. Then
$$ \lim_{n\to \infty} u_n^2 = \lim_{n\to \infty} v_n^2 = L^2$$
Your idea is correct, but incomplete.
First you have to show that the sequence converges at all. Hint: Show that the sequence is monotonic and bounded. Distinguish the cases $x_0 < 4$ and $x_0 > 4$.
Then you are correct: If the sequence is convergent then the limit is either $-3$ or $4$. But you have to exclude the possibility that $c=-3$. Hint: $x_n \ge 0$ for $n \ge 1$.