Question

Let \(x_1\in(0,4)\) and consider the sequence \(\{x_n\}_{n\geq1}\) defined iteratively by
\[ x_{n+1}=2-(4-x_n)^{\frac{1}{2}},\quad n\geq1. \]
Consider the following statements:
(I) \(\{x_n\}\) converges to \(0\).
(II) \(\left\{\dfrac{x_{n+1}}{x_n}\right\}\) converges to \(\dfrac{1}{4}\).

Which of the following statements is correct?

• A. Only statement (I) is correct
• B. Only statement (II) is correct
• C. Both statements (I) and (II) are correct
• D. Neither statement (I) nor statement (II) is correct

Hint:

Rationalise the recursion to get \(x_{n+1}=\dfrac{x_n}{2+\sqrt{4-x_n}}\); this shows the sequence decreases to \(0\), and the ratio \(x_{n+1}/x_n\) tends to \(1/(2+2)=1/4\).

Answer 1

The Correct Option is C

Step 1: Bound and monotonicity directly from an inequality.
For $0<x<4$ we compare $x$ with $x_{n+1}=2-\sqrt{4-x}$. We claim $\sqrt{4-x}>2-x$ for every $x$ in $(0,4)$. If $x\leq2$, then $2-x\geq0$, and squaring both sides gives $4-x>4-4x+x^2$, that is $x(3-x)>0$, true because $0<x\leq2<3$. If $2<x<4$, then $2-x<0\leq\sqrt{4-x}$, so the inequality holds automatically. So $x>2-\sqrt{4-x}=x_{n+1}$.

Step 2: The sequence is trapped and shrinking.
Step 1 shows $x_{n+1}<x_n$ for every term, so the sequence is strictly decreasing. Also $\sqrt{4-x_n}\in(0,2)$ whenever $x_n\in(0,4)$, so $x_{n+1}=2-\sqrt{4-x_n}\in(0,2)$, keeping every later term positive. A decreasing sequence bounded below by $0$ must converge to some limit $L\geq0$.

Step 3: Pin down $L$ using the fixed point equation.
Passing to the limit gives $L=2-\sqrt{4-L}$, so $\sqrt{4-L}=2-L$. Squaring gives $L^2-3L=0$, giving $L=0$ or $L=3$. Because the sequence stays in $(0,2)$, the value $L=3$ cannot be the limit, so $L=0$. Statement (I) holds.

Step 4: Get the ratio from a first order expansion near $0$.
Near $x=0$, write $\sqrt{4-x}=2\sqrt{1-x/4}$. Using $\sqrt{1-u}\approx1-\dfrac{u}{2}$ for small $u=x/4$, gives $\sqrt{4-x}\approx2-\dfrac{x}{4}$. So for small $x_n$,
\[ x_{n+1}=2-\sqrt{4-x_n}\approx2-\left(2-\frac{x_n}{4}\right)=\frac{x_n}{4}. \]

Step 5: Take the limit of the ratio.
So $\dfrac{x_{n+1}}{x_n}\approx\dfrac{1}{4}$ once $x_n$ is small, and since $x_n\to0$ this becomes exact in the limit, giving $\dfrac{x_{n+1}}{x_n}\to\dfrac{1}{4}$. Statement (II) also holds.

Both statements are correct. \[ \boxed{\text{Both statements (I) and (II) are correct}} \]