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: Show the sequence stays inside \((0,2)\) after the first step.
For any \(x\in(0,4)\), the quantity \(4-x\) lies in \((0,4)\), so \((4-x)^{1/2}\in(0,2)\). Hence
\[ x_{n+1}=2-(4-x_n)^{1/2}\in(0,2) \]
for every \(n\). So the sequence is bounded, staying between \(0\) and \(2\) from the second term onward.

Step 2: Rewrite the recursion by rationalising.
Multiply and divide by the conjugate:
\[ x_{n+1}=2-(4-x_n)^{1/2}=\frac{\left(2-(4-x_n)^{1/2}\right)\left(2+(4-x_n)^{1/2}\right)}{2+(4-x_n)^{1/2}}=\frac{4-(4-x_n)}{2+(4-x_n)^{1/2}} \]
\[ x_{n+1}=\frac{x_n}{2+(4-x_n)^{1/2}} \]
This clean form is the key to both statements.

Step 3: Show the sequence is decreasing.
For \(x_n\in(0,4)\), we have \((4-x_n)^{1/2}>0\), so the denominator \(2+(4-x_n)^{1/2}>2\). Therefore
\[ x_{n+1}=\frac{x_n}{2+(4-x_n)^{1/2}}<\frac{x_n}{2}<x_n. \]
So \(\{x_n\}\) is strictly decreasing. Together with Step 1 it is also bounded below by \(0\), so by the monotone convergence theorem the sequence converges to some limit \(L\in[0,2)\).

Step 4: Identify the limit \(L\).
Taking \(n\to\infty\) on both sides of \(x_{n+1}=2-(4-x_n)^{1/2}\) and using continuity of the square root, \(L\) must satisfy
\[ L=2-(4-L)^{1/2}\ \Rightarrow\ (4-L)^{1/2}=2-L. \]
Squaring (valid since \(L\leq2\) makes \(2-L\geq0\)):
\[ 4-L=4-4L+L^2\ \Rightarrow\ 3L-L^2=0\ \Rightarrow\ L(3-L)=0. \]
So \(L=0\) or \(L=3\). Since \(L\in[0,2)\), the root \(L=3\) is rejected, leaving \(L=0\). This proves statement (I): \(\{x_n\}\) converges to \(0\).

Step 5: Find the limit of the ratio using the rationalised form.
From Step 2,
\[ \frac{x_{n+1}}{x_n}=\frac{1}{2+(4-x_n)^{1/2}}. \]
Since \(x_n\to0\) as \(n\to\infty\), by continuity \((4-x_n)^{1/2}\to(4-0)^{1/2}=2\). So
\[ \frac{x_{n+1}}{x_n}\to\frac{1}{2+2}=\frac{1}{4}. \]
This proves statement (II).

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