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$\left(u_n\right)$ is a sequence defined by recurrence as follows:

$ \begin{cases} u_1=\displaystyle\frac{8}{3}\\ u_{n+1}=\displaystyle\frac{16}{8-u_n}, \forall n\in \mathbb{N} \end{cases} $

The first part of this question is to show that $u_n<4, \forall n\in \mathbb{N}$ which I have done by induction, the second part is to show that the sequence is monotonically increasing and I have done that too.

The third part is to show that $\left(u_n\right)$ converges and that is easy with the previous two parts done but it asks to determine the limit and I'm not sure it's liquid that the limit is 4. I've done it computationally and verified it should be so, but I don't find it immediate just because we have shown $$u_n<4, \forall n\in \mathbb{N}$$ that this value should be considered the limit. Why not $3.9$?

Is there an analytic way of determining the value of this limit?

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    $\begingroup$ The limit must satisfy $L=\frac{16}{8-L}$ $\endgroup$ – kingW3 Mar 30 '17 at 20:18
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The limit of this sequence is $4$. If $l$ is the limit of the sequence we have : $\lim\limits_{n \rightarrow +\infty} u_n = \lim\limits_{n \rightarrow +\infty} u_{n+1} \Leftrightarrow l = \frac{16}{8-l}$ Hence we finally get $l = 4$ so the limit of the sequence is $4$

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At $n \rightarrow \infty$ , $u_n \approx u_{n+1}$

$$\lim_{n \to \infty} u_{n}=\lim_{n \to \infty}\displaystyle\frac{16}{8-u_n} \implies \lim_{n \to \infty}u_n(8-u_n)=16 \implies \lim_{n \to \infty}u_n=4$$

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