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Consider the sequence defined by $$x_1=3 \quad\text{and}\quad x_{n+1}=\dfrac{1}{4-x_n}$$

I can calculate limit by assuming limit exist and solving quadratic equation, but I first wanted to give existence of limit.

I tried to show given sequence is decreasing and bounded below by 0.

I used derivative test as $$f^\prime(x)=\frac{1}{(4-x)^2}$$ but form this I am not able to show

Also, I tried to shoe $x_{n+1}-x_n<0$ but that way also I am not succeed.

Please tell me how to approach such problem

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You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - \dfrac{1}{4-x}.$

Since $g'(x) = 1-\dfrac{1}{(4-x)^2} =\dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_n\leq 3$ for $n>1$, which can be easily established by induction.

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  • $\begingroup$ 1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...) $\endgroup$ – Did Dec 2 '18 at 9:21
  • $\begingroup$ Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully. $\endgroup$ – Did Dec 2 '18 at 10:59
  • $\begingroup$ "This comment is rude or condescending. Learn more in our Code of Conduct." $\endgroup$ – Did Dec 2 '18 at 21:22
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It can be approached in a graphical manner:

  • Draw the graph of $y = \frac{1}{4-x}$ to scale while marking the essentials.
  • Asymptote at $x=4$; Value at $x = 3$ is $1$.
  • Comparing it to the previous value of the sequence would require the plot of $y=x$ on the same axes.
  • Mark the intersection as $x=2-\sqrt3$ whereas $x=2+\sqrt3$ is near $x=4$.

If through, notice that starting the sequence from $x=3$ means that the next value is $1$ from the hyperbola which is well below the straight line. Now to get the next value put $x=1$ and get the next value from the hyperbola, which is again less than $1$ as the straight line depicts.

If you follow the pattern, you would tend to reach the intersection $x=2-\sqrt3$ as the gap between both the curves decreases to zero which gives the limit of the sequence as $x=2-\sqrt3$ (the limit only, not one of the terms of the sequence, since these are all rational numbers).

Also, one can thus say that if $x_1 \in (2-\sqrt3,2+\sqrt3)$ then the sequence would be decreasing and would converge at $x=2-\sqrt3$ and that all the terms $x \in (2-\sqrt3,2+\sqrt3)$.

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    $\begingroup$ Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing? $\endgroup$ – Did Dec 2 '18 at 10:37
  • $\begingroup$ The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy! $\endgroup$ – Sameer Baheti Dec 2 '18 at 10:38
  • $\begingroup$ You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous. $\endgroup$ – Did Dec 2 '18 at 10:42
  • $\begingroup$ It got edited before people saw the flaw I pointed. That would be the only reason. $\endgroup$ – Sameer Baheti Dec 2 '18 at 10:44
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Suppose $p(y)=y^2-4y+1=0$ and let $x_n=y+z_n$ then $$y+z_{n+1}=\frac 1{4-y-z_n}$$ so that $$-yz_n+(4-y)z_{n+1}=y^2-4y+1=0$$ and $$z_{n+1}=\frac y{4-y}z_n$$

So if $0\lt y\lt 2$ the error term $z_n$ has the same sign and is decreasing in magnitude in constant proportion so tends to zero. It is easy to test that there is a root of the quadratic in the required range by noting $p(0)=1, p(2)=-3$


Note: the equations hold for the other possible value of $y$ too, but the inequality does not then apply to prove that the error term tends to zero. Also this was done without computing $y$.

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To show that $x_n$ is decreasing, use induction to prove that $x_n> x_{n+1}$. To show that $x_n$ is bounded below by zero show a stronger assertion that $0<x_n<2+\sqrt{3}$ by using induction.

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  • $\begingroup$ You do not need to invoke $2+\sqrt{3}$. Use $x_1=3$ as the upper bound and accept $\le$ for the comparison with this bound. $\endgroup$ – Oscar Lanzi Dec 2 '18 at 10:49
  • $\begingroup$ Yes, i realized it later. Thanks :). $\endgroup$ – user9077 Dec 2 '18 at 15:11
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Hint: after finding the limit $l$ from $l={1\over 4-l}$ to make sure that the sequence tends to $l$, define $e_n=a_n-l$ and by substituting it in $a_{n+1}={1\over 4-a_n}$ conclude that $e_n \to 0$

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