TITLE: Help me understanding logic behind limits of recurence relations
QUESTION [7 upvotes]: I was trying to understand how limits of recurence relations are working. 
I have one.
$$a_0 = \dfrac32 ,\ a_{n+1} = \frac{3}{4-a_n} $$
So, from what i know, if this recurence relation has a limit, it have to be bounded and monotonous. 
To check if it's bounded i have to calculate $$\lim_{x \rightarrow \infty} \frac{3}{4-x}$$ and it goes to 0, so it's bounded. 
Now to check if it's monotonous i have to check if $$a_{n+1} - a_n$$ is monotonous. 
$$a_{n+1} - a_n = \frac{3}{4-a_n} - a_n = - \frac{(a_n-3)(a_n-1)}{(a_n-4)}$$
This expression is monotonous ( decreasing ) but only starting at x = 5. Is it enough to say that this is monotonous? 
If it is. We know that $$\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} a_{n+1} $$
So we calculate it $$L = \frac{3}{4-L}$$ and end up with $$L^2 - 4L + 3 = 0 \rightarrow (L-1)(L-3) = 0$$ We know that the limit of this recurence relation can be 1 or 3. 
On our classes there always were some examples that had only 1 possible limit to choose. In this example we have decreasing sequence and $a_0 = \dfrac32$ then one possible limit is 1.
I was starting to check it... What if $a_0 = \dfrac12$? 
What if $a_0 = 5$ or $a_0 = 2$ it turns out that, no matter what starting value we have, sequence always goes to the same limit ( at least in this example ). Is it true for all recurence relations? You can check using wolframalpha clicking here
and just manipulate with starting value. Please help me with this and explain those weird things. I'd be so thankful!

REPLY [2 votes]: To understand the behaviour of a sequence defined by $u_{n+1}=f(u_n)$ with given $u_0$ sometimes it helps to draw the graphs of $y=x$ and $y=f(x)$ on the same picture. Then it becomes possible to "draw" the sequence $u_n$. For example, when $f(x)=\sqrt{x}$ and $u_0=1/5$ the picture looks like this:

The main trick is that once $u_{n+1}$ is drawn, it is easy to find a point $u_{n+1}$ on the horisontal axis by means of "reflection" using the graph $y=x$. 
There are some more details on this on the page where I have have found this image: http://www.fmaths.com/recursivethinking/lesson.php
By the way, this example also demonstrates that the limit can depend on $u_0$. Namely,
$$
\lim_{n\to \infty}u_n=\begin{cases}0, & u_0=0 \\ 1, & u_0\in(0,1], \\ +\infty, & u_0>1\end{cases}
$$