Exam question on fixed point iteration I am solving the following exam problem.
Problem: An iterative scheme is given by $$ x_{n+1}= \frac{1}{5}\left(16-\frac{12}{x_n} \right).$$ Such a scheme with suitable initial approximation $x_0$ will
(a) not converge
(b) converge to $1.6$
(c) converge to $1.8$
(d) converge to $2$
My attempt. By defining $g(x) = 1/5(16-12/x )$, I found that fixed points of $g(x)$ are $2$ and $6/5$. Thus, given iterative scheme will converge either to $2$ or $6/5$.
Using fixed point theorem I need to find the interval [a, b] show that $g(x)\in [a, b]$ and is continuous and derivative of g(x) should exit on $(a, b)$. Further I have to check that $|g^{'}(x)| \leq r$, $r <1$.
First confusion: I want to know that how to find such interval $[a, b]$ satisfying these conditions?
Second confusion: Here we see that at fixed point $2$, $|g^{'}(2)| <1$. My confusion is then can we conclude that sequence $x_n$ will converge to $2$?
Thank you for your help.
 A: The function
$$g(x):={1\over5}\left(16-{12\over x}\right)$$
has the fixed points $2$ and ${5\over6}$. As
$$g'(x)={12\over 5x^2}\tag{1}$$ we have
$$\bigl|g'(2)\bigr|={3\over5}<1,\qquad \bigl|g'(5/6)\bigr|={5\over3}>1\ .$$
Therefore we try to prove that $2$ is an attractive fixed point. To this end we have to produce an open $x$-interval $J$ containing $2$, and an $r<1$ such that $\bigl|g'(x)\bigr|\leq r$ for all  $x\in J$. Now from $(1)$ we obtain
$$0<g'(x)\leq {12\over 5\cdot 1.6^2}={15\over16}<1\qquad(x\geq1.6
)\ .\tag{2}$$
This shows that we can take $J:=[1.6, \>\infty[\ $ and $r={15\over16}$.
Claim: When $x_n\in J$ then $$|x_{n+1}-2|\leq{15\over16}\>|x_n-2|\ .$$
This implies that we can take any point $x_0\in J$ as starting point, and obtain $\lim_{n\to\infty}x_n=2$.
Proof: By the MVT there  is a point $\xi$ between $2$ and $x_n$ such that
$$|x_{n+1}-2|=\bigl|g(x_n)-g(2)\bigr|=\bigl|g'(\xi)\bigr|\>|x_n-2|\>\leq{15\over16}|x_n-2|\ ,$$
where we have made use of $(2)$, since $\xi\in J$.
Note that $g$ is a Moebius transformation with the two fixed points $2$ and ${5\over6}$. Introducing a new complex coordinate $z$ such that the two fixed points correpond to $z=0$ and $z=\infty$ "conjugates" $g$ into
$$\hat g(z)={3\over5} z\qquad(z\in\bar{\mathbb C})\ .$$
From this we can conclude that any starting point $z_0\ne\infty$, resp. $x_0\ne{5\over6}$, will lead to the desired convergence.
A: Think $\{x_n\}$ as a sequence.
$$x_{n+1}-x_n=-\frac{(x_n-2)(5x_n-6)}{5x_n}$$
When $x_n>2$ then clearly $x_{n+1}-x_n<0$ (check!), and when $x_n<6/5$ then also $x_{n+1}-x_n<0$ (check!). So the sequence $\{x_n\}$ is monotone decreasing , also it is bounded (check!). So it must be convergent. So your option (A) is incorrect.
Let, $\{x_n\}$ converges to $l$. Then,
$$l=\frac{1}{5}\left(16-\frac{12}{l}\right)$$From here, you will get either $l=2$ or $l=1.2$.
From your options , clearly (D) is correct.
NOTE:
If 'converges to $2$' and 'converges to $1.2$' both options are given , then  it is hard to detect that at which point it converges..In that case , it depends on the initial approximation $x_0$.
A: If you calculate $g'(x)=\frac{12}{5x^2}$ so 
$$
g'(x)<1 \qquad \iff x>\sqrt{\frac{12}{5}}
$$
In particular if you take $[a,b]=[1,6 ; 3]$
A: Let $x_{n} =\dfrac 6 5 +\dfrac 4 {5y_{n}}$ 
Then the equation turns into $5y_{n+1}-3y_{n}-2=0 \Rightarrow y_{n}=c\cdot(\dfrac 3 5)^n +1 \Rightarrow \\$
$x_{n}=\dfrac 6 5 +\dfrac 4 {5c\cdot (\dfrac 3 5 )^n+5}$
Now obviously
$\lim_{n\rightarrow \infty }x_{n}=\lim_{n\rightarrow \infty }\dfrac 6 5 +\dfrac 4 {5c\cdot (\dfrac 3 5 )^n+5}=\dfrac 6 5+\dfrac 4 5=2$
