A conditional limit in a two-cycle map · source reference

Printed question

A sequence of positive real numbers satisfies

$$x_{n+1}=\frac{3-x_n}{2x_n+1},\quad n\ge1.$$

(i) Given convergence to $l$, find $l$ exactly. [3]

(ii) Using a sketch of $y=\frac{3-x}{2x+1}-x$ and considering $x_{n+1}-x_n$, prove $x_{n+1}>x_n$ if $0<x_n<l$, and $x_{n+1}<x_n$ if $x_n>l$. [3]

Printed working

(i) As $n\to\infty$, $x_n\to l,x_{n+1}\to l$.

$$l=\frac{3-l}{2l+1}\Rightarrow l(2l+1)=3-l\Rightarrow2l^2+2l-3=0,$$ $$l=\frac{-2\pm\sqrt{2^2-4(2)(-3)}}{2(2)}=\frac{-2\pm\sqrt{28}}4=\frac{-1\pm\sqrt7}2.$$

Since $l>0$, $l=\frac{-1+\sqrt7}2$.

(ii) $x_{n+1}-x_n=\frac{3-x_n}{2x_n+1}-x_n$.

From the sketch, if $0<x_n<l$, $\frac{3-x_n}{2x_n+1}-x_n>0$, hence $x_{n+1}-x_n>0$ and $x_{n+1}>x_n$.

If $x_n>l$, $\frac{3-x_n}{2x_n+1}-x_n<0$, hence $x_{n+1}-x_n<0$ and $x_{n+1}<x_n$.

Source diagrams redrawn

Printed answer key

(i) $l=\frac{-1+\sqrt7}2$

Mathematical review and qualifications

(i) $l(2l+1)=3-l$ gives $2l^2+2l-3=0$. Positivity selects $l=(\sqrt7-1)/2$. A zero limit cannot satisfy the equation.

(ii) $F(x)-x=(3-2x-2x^2)/(2x+1)$. On $x>0$ the numerator strictly decreases, is positive before $l$, zero at $l$ and negative after $l$, giving the required inequalities. The increment graph starts at 3, crosses once at $l$ and becomes negative.

Qualification: direct composition gives $F(F(x))=x$ wherever defined. Thus a nonconstant admissible sequence alternates between two values and does not converge; the premise in (i) forces the constant sequence $x_n=l$. This preserves the printed source conditional conditional wording.

The printed question assumes convergence only in (i). Do not infer convergence of every sequence from the one-step inequalities.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2011/CJC/Prelim/I/2] Question PDF pages 3; solution PDF pages 9, 10. The original files remain private QA records outside this native content.