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Suppose that $x_0 \geq 2$ and $x_n = 2 + \sqrt{x_{n-1} - 2}$ for all natural $n$. Use the Monotone Convergence Theorem to prove that either $x_n \rightarrow 2$ or $x_n \rightarrow 3$ as $n$ grows.

Attempt: Suppose that $x_0 \geq 2$ and $x_n = 2 + \sqrt{x_{n-1} - 2}$ for all natural $n$. Then the Monotone Convergence Theorem says if the sequence is increasing and bounded above or decreasing and bounded below, then the sequence converges to a limit.

Then, when $x_0 = 2$ we have $x_n = 2$ for all $n$.

I don't know how to continue. I know if I find the limit of $x_n$ on both sides then we get $x = 2$ or $3$. But I don't know how to prove it. Please any feedback/hint or help will be appreciated. Thank you.

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  • $\begingroup$ Try to use induction to prove the monotonic nature of the sequence when $x_0>3$ and $2<x_0<3$. $\endgroup$ – jdoicj Sep 18 '14 at 5:55
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Let $f(x)=2+\sqrt{x-2}$. $f$ has two fixed points: $f(2)=2$ and $f(3)=3$. Look at the graph of $f$: Graph of $f$

If $x_0\ne2,3$ there are two possibilities:

  1. $2<x_0<3$. Then show that $2<x_n<3$, that $\{x_n\}$ is increasing and that the limit is $3$. (Hint: $f(x)>x$ on $(2,3)$.)
  2. $x_0>3$. Then show that $x_n>3$, that $\{x_n\}$ is decreasing and that the limit is also $3$. (Hint: $f(x)<x$ on $(3,\infty)$.)
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