Consider the sequence $\langle x_n\rangle $, $n\ge 0$, defined by the recurrence relation $$x_{n+1} = c \cdot x_n^2 - 2, \quad\text{where $c > 0$.}$$ 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 a value.
A. $(1+\sqrt{1+8c})/2c$
B. $(1-\sqrt{1+8c})/2c$
C. $2$
D. $2/(2c - 1)$
I was trying this problem by putting random values for $c$ and $x_0$ like if $c=1$ and $x_0 = -1$ then the sequence becomes $-1, -1, -1, \dots$ and thus the sequence converges to $-1$. So, I am getting option B as correct.
And again if $c = 1$ and $x_0 = 2$ then the sequence becomes $2,2,2,\dots$ and thus the sequence converges to $2$. So, I am getting option A as correct.
Among the 4 options only one can be correct. They have given option B as correct answer. I don't know how to solve this. Please help.
You did not consider the given condition for $x_0$. There has to be an open interval $(a,b)$ with $x_0\in(a,b)$ such that the sequence converges for all initial values in $(a,b)$. For $c=1$ and $x_0=2$ there is no such interval since any $x_0'>2$ will give a divergent sequence.
Note that the map $x\mapsto cx^2-2$ will always have two fixed points, so those will occur as limits of your sequence. However, at most one of the two fixed points will be stable, so that for initial values near the fixed point your sequence converges to the fixed point.