Proving that $(a_n) $ defined by induction to be: > $a_1=2.2$, $a_{n+1}=5-\frac6{a_n}$ is converging and finding the limit 
Let there be the sequence $(a_n) $ defined by induction to be:
  $a_1=2.2$, $a_{n+1}=5-\frac6{a_n}$ $\forall n\ge 1$
Prove that the sequence is converging and calculate it's limit.

So what needs to be done is to show that it's increasing and bounded and to find the limit. 
Finding the limit we know that $a_n\to \alpha $ which leads to $ \alpha=5-\frac6{\alpha}$ after solving the quadratic equation we get: $ \alpha_{1,2}=-1,6$ so we know that the limit is 6 because it can't be negative. 
Increasing, by induction, we need to show that $a_n\le a_{n+1}$ so we get to: $a_{k+2}= 5-\dfrac6{a_{k+1}} \ge 5-\dfrac6{a_k}=a_{k+1}\to...\to a_{k}\le a_{k+1}$.
Is my way correct ? how do I show that it's bounded ? anything missing ?
NOTE: I need to know how to solve this rigoursly as I study for a test so please let me know of anything that needs to be mentioned in the test. 
 A: Ok. I don't know how you resolved this in class but here's one way.
Let $f$ be defined by $$f(x) = 5-\frac{6}{x}$$
So we have, $a_1 = \frac{11}{5}$ and $$\forall n \geqslant 1, \qquad a_{n+1} = f(a_n)$$
Study of $f$
$f$ is defined for $x\neq 0$. We are looking for stable intervals by $f$ (interval $I$ such that $f(I) \subset I$).
If we plot the function we can have some idea of what are some stable intervals.
For example, if $x \geqslant 2$, then $\frac{6}{x} \leqslant 3$ and finally $f(x) \geqslant 2$. Idem, if $x \leqslant 3$, then we can show that $f(x) \leqslant 3$. So $I= [2, 3]$ is stable by $f$.
Let's also study the sign of $$g(x) = f(x) - x = 5-x-\frac{6}{x}$$ on $I$.
For $x\in I$, we know that we can differentiate $g$ and got $$g'(x) = \frac{6}{x^2}-1$$
So, if $x^2 < 6$ ($\Leftrightarrow |x| < \sqrt{6}$ and $\sqrt{6} > 2.2$) $g$ is increasing and decreasing otherwise.
We can also (by solving a simple equation) that the roots of $g$ are $2$ and $3$.
So. On $I$, $f(x) \geqslant x$ if $x\in [2,3]$ and $f(x) \leqslant x$ if $x > 3$
We've got everything to end this.
$a_n$ is bounded
We're going to show that $\forall n\in \mathbb{N}, a_n \in I=[2,3]$.
This is clear for $a_1 = 2.2$. Then if we assume that $a_n \in I$ then
$$a_{n+1} = f(a_n) \in I$$ because $a_n \in I$ and $I$ is stable by $f$.
So the property is proved.
$a_n$ is increasing
We now know that $a_n \in [2,3]$ for all $n$.
But, we also know that $$\forall x\in [2,3], \qquad f(x) \geqslant x$$ If we say $x = a_n$ (which is allowed because we're in the right interval) we have $$f(a_n) \geqslant a_n$$
So $$a_{n+1} \geqslant a_n$$ and $a_n$ is increasing.
The end
So we now know that $a_n$ has a limit, say $\ell$. $\ell$ has to have the property $$f(\ell) = \ell$$ We have prove that there's only $\ell =2$ or $\ell =3$. And since for $n > 1$, $a_n >a_1 >2$, we know we have $$a_n \to 3$$
A: Hint: 
$2.2\leq a_{n}\leq3$ implies $2.2\leq a_{n+1}\leq3$. 
You can show that under these conditions $a_{n+1}>a_n$
The equation $$\alpha=5-\dfrac{6}{\alpha}$$ has $\alpha=2$ and $\alpha=3$ as solutions.
