Limiting value of a sequence

Question

Question - Consider the sequence $\langle x_n \rangle , \: n \geq 0$ defined by the recurrence relation $x_{n+1} = c . x^2_n -2$, where $c > 0$.

Suppose there exists a non-empty, open interval $(a, b)$ such that for all $x_0$ satisfying $a < x_0 < b$, the sequence converges to a limit. The sequence converges to the value?

$A. \frac{1+\sqrt{1+8c}}{2c}$

$B. \frac{1-\sqrt{1+8c}}{2c}$

$C. 2$

$D. \frac{2}{2c-1}$

Since it converges, we can write: $x=cx^2 -2$

or

$cx^2 -x -2 = 0$

Solving it gives-

$x=\frac{1\pm \sqrt{1+8c}}{2c}$

So both options (A) and (B) satisfy this.

The answer is (B). I am not able to proceed further. Please help

Answer 1

Let $f(x)=cx^2-2$. Indeed, $f(x)=x\iff x=\frac{1\pm\sqrt{1+8c}}2$. But$$f'\left(\frac{1+\sqrt{1+8c}}2\right)=1+\sqrt{1+8c}>0$$and therefore if $x_0$ is slightly larger than $f(x_0)$, then $f(x_0)>x_0$ and, in fact, the sequence $(x_n)_{n\in\mathbb N}$ is strictly increasing. Therefore, we don't have $\lim_{n\to\infty}x_n=x_0$. So, A. is not an option.

Answer 2

I call $A$ the solution $A$. $ f'(A) >1$. Suppose that $A$ is the limit. There exists a neighborhood $U$ of $A$ such that $n>N$ implies that $x_n$ in $U$ and for every $x$ in $U$, $f'(x) >1$. The condition that $x_0\in(a,b)$ the sequence converges implies that we can suppose $x_0\neq x_1$. We have for $n>N$: $|x_{n+2}-x_{n+1}|=|f(x_{n+1}-f(x_n)|=f'(c_n)|x_{n+1}-x_n |\geq|x_{n+1}-x_n|$. This implies that $lim_n|x_{n+1}-x_n|>0$. Contradiction.