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Sequence $x_n$ such that $x_0 \in [0, 2]$ is defined by $x_{n+1} = \frac{1}{4}x_n(5-x_n)$. For what values of $x_0 \in [0, 2]$ does $x_n$ converges and to what limit?

For $x_0 = 0$, $\lim_{n\to\infty}x_n = 0$. I guess that in other cases $\lim_{n\to\infty}x_n = 1$. The sequence is bounded by $\max(2, \frac{25}{16})$ from AM-GM. How can I prove my assertion that $x_n$ converges to $1$?

I've tried to prove that for $x_0 \in (0, 1)$ sequence increases and for $x_0 \in (1, 2)$ decreases, but when proving first part I got stuck, because I don't know how to prove inductive step that $x_n < 1$ implies $\frac{x_n(5-x_n)}{4} < 1$

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    $\begingroup$ Assume the sequence converges to some $l$, and use the fact that $\{x_{n+1}\}$ and $\{x_n\}$ converge to the same number. You can then calculate precisely the possible limits of the sequence. Note that you still need to proof that the sequence coverges (usually done with arguments about monotonicity of the sequence). $\endgroup$ – Donjim Nov 13 '18 at 10:15
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The function $f(x)=\frac {x(5-x)} 4$ maps $[0,2]$ into itself and it is increasing there. From this it follows by induction that $x_n \in [0,1]$ for all $n$ and that $x_{n+1} \geq x_n$. Hence $a\equiv \lim x_n$ exists and since $a=\frac {a(5-a)} 4$ we get $a=0$ or $a=1$. Since $(a_n)$ is increasing it follows that $a=0$ if $x_0=0$ and $a=1$ otherwise.

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  • $\begingroup$ I understood my mistake. +1 by the brief and objective answer. $\endgroup$ – MathOverview Nov 13 '18 at 10:29

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