A square-root iteration crosses its limit · source reference

Printed question

A sequence of positive numbers satisfies

$$x_{n+1}=\sqrt{10-3x_n}\quad(n=1,2,3,\ldots).$$

(i) Given convergence to $l$, find $l$. [2]

(ii) Prove $x_{n+1}^2-l^2=3(l-x_n)$. [2]

(iii) Use (ii) to show that if $x_n<l$, then $x_{n+1}>l$. [2]

Printed working

(i) Since the sequence converges to $l$, as $n\to\infty$, $x_n\to l,x_{n+1}\to l$.

$$l=\sqrt{10-3l}\Rightarrow l^2=10-3l\Rightarrow l^2+3l-10=0\Rightarrow(l-2)(l+5)=0.$$

$l=2$ or $l=-5$ (rejected), so $l=2$.

(ii)

$$x_{n+1}^2-l^2=10-3x_n-2^2=6-3x_n=3(l-x_n).$$

(iii) Since $x_n<l$, $l-x_n>0$. Hence $x_{n+1}^2-l^2>0$ and $(x_{n+1}-l)(x_{n+1}+l)>0$. As $x_{n+1}+l>0$, $x_{n+1}-l>0$, proving $x_{n+1}>l$.

Printed answer key

(i) 2

Mathematical review and qualifications

(i) $l=\sqrt{10-3l}$ implies $(l-2)(l+5)=0$. Nonnegativity rejects $-5$, so $l=2$.

(ii) $x_{n+1}^2-l^2=10-3x_n-4=3(2-x_n)=3(l-x_n)$.

(iii) If $x_n<l$, this expression is positive. Since $x_{n+1}+l>0$, factoring $(x_{n+1}-l)(x_{n+1}+l)$ gives $x_{n+1}>l$.

The source has a square root over $10-3x_n$; plain text extraction loses it. The recurrence is assumed defined and positive at every step.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2012/IJC/Prelim/II/3] Question PDF pages 3; solution PDF pages 7. The original files remain private QA records outside this native content.