Prove that $x_{n+1}=\frac{1}{4-x_n}$ converges Let $x_1 =3.$ Prove that $$x_{n+1}=\frac{1}{4-x_n}$$ converges.
The sequence seems to be converging to $1/4$ as the first few terms are $$3,1,\frac{1}{3},\frac{3}{11},\frac{11}{41},\frac{41}{153}, \cdots$$
In order to prove convergence I know I must make use of the definition that a sequence $(x_n)$ converges to $x$ if there exists an $N$ for all $n \ge N \Rightarrow |x_n-x|\lt \epsilon$. But how do I choose $N$ as a function of $\epsilon$? What can I assume? Could someone show me what a proof for this looks like? Thanks.
 A: First show that $x_{n+1} < x_n$ and note that $x_n \geq 0$.  Since the sequence is bounded below and decreasing, it must converge to some number $x_\infty$.  We then note $$\lim_{n \to\infty} x_{n+1} = \lim_{n\to\infty}\frac{1}{4 - x_n} \iff x_{\infty} = \frac{1}{4 - x_\infty} \iff x_\infty(4-x_\infty) = 1 \iff x_\infty = 2 \pm \sqrt{3}.$$
Since the sequence is decreasing and $x_2 = 1$, we must have that the sequence converges to $2 - \sqrt{3}$.
A: Step 1

Show that $$0<x_n \leq 3\tag1$$ for $n=1,2,\cdots.$

Proof
Obviously, $(1)$ holds for $n=1$. Assume that $(1)$ holds for $n=k$, namely, $0<x_k \leq 3$, then $$0<\frac{1}{4}<x_{k+1}=\frac{1}{4-x_k} \leq 1 \leq 3,$$which implies $(1)$ also holds for $n=k+1$. Thus, by mathematical induction, $(1)$ holds for all $n=1,2,\cdots.$
Step 2

Show that $$x_{n+1}<x_n\tag2$$ for $n=1,2,\cdots.$

Proof
Notice that $x_1=3, x_2=1.$ Hence $x_2<x_1,$ which implies that $(2)$ holds for $n=1$. Assume that $(2)$ holds for $n=k$, namely, $x_{k+1}<x_k$, then $$x_{k+2}=\frac{1}{4-x_{k+1}}<\frac{1}{4-x_{k}}=x_{k+1},$$which implies $(2)$ also holds for $n=k+1$. Thus, by mathematical induction, $(2)$ holds for all $n=1,2,\cdots.$
Combining the two steps, by monotone convergence theorem, we may claim $x_n$ has a limit, which could be denoted as $x.$ Thus, taking the simultaneous limits of both sides of the recursion formula, we have$$x=\frac{1}{4-x}.$$Hence, $x=2 \pm \sqrt{3}.$ But $x_n \leq 3$, hence $x \leq 3,$ which implies that $2+\sqrt{3}>3$ does not satisfy the requirement. As a result, $$x=2-\sqrt{3},$$ which is desired limit.
A: If such a limit say $l$ exists we must have$$l=\dfrac{1}{4-l}$$or $$4l-l^2=1$$which gives us $$l=2\pm \sqrt 3$$since all the terms except the first 2 ones are below $1$ the limit $2+\sqrt3$ is invalid, then we need to show that  the sequence converges to $2-\sqrt 3$. Define $$e_n=x_n-(2-\sqrt 3)$$therefore$$e_{n+1}=x_{n+1}-2+\sqrt 3=\dfrac{1}{4-x_n}-2+\sqrt 3=\dfrac{1-4(2-\sqrt 3)+x_n(2-\sqrt 3)}{4-x_n}=\dfrac{-7+4\sqrt 3+x_n(2-\sqrt 3)}{4-x_n}=(2-\sqrt 3)\dfrac{e_n}{4-x_n}$$since for $n>2$ we have $x_n<1$ by substitution we obtain$$|e_{n+1}|<\dfrac{2-\sqrt 3}{3}|e_n|$$ which means that $e_n\to 0$ or $x_n\to 2-\sqrt 3$
