Let be $a,b$ positive real numbers; $a+\frac {b} {a+\frac {b} {a+\frac {b} {\ddots }}}$ how can we prove whether this has a limit or not? Let, given example is like this

$$3+\dfrac {4} {3+\dfrac {4} {3+\dfrac {4} {\ddots }}}=?$$

I wonder if it has a limit or not, but I have a idea, If we show that this sequence monotonous and has a bound, we can say that this has a limit.
Therefore,firstly, we need to write sequence form;

$$3+\dfrac4{x_n}=x_{n+1}$$

If we assume that this sequence has a limit, we can apply convergence of subsequences theorem, with respect to this theorem we can do the following conclusion.Let assume $x$ equals to limit of the sequence>$$3+\dfrac4{x}=x$$
$x_1=4$ and $x_2=-1$, limit should be positive, so $x=4$.That is;
$$3+\dfrac {4} {3+\dfrac {4} {3+\dfrac {4} {\ddots }}}=4$$

My intuition says me that the sequence is increasing and has a upperbound but I couldn't show, so above proof is not valid for now.

How we can that this is a monotonous and has a bound?

One way to prove convergence is to start with the assumed limit and consider the distance to it.
As you've seen the equation for the assumed limits is a quadratic so we could instead start in the other end and write the recurrence formula:
$$x_{n+1} = f(x) = (p+q) - pq/x_n$$
where $p$ and $q$ are the assumed limits. Now as the iteration is a solution method for the fix-value problem it will converge if the secant to the fixpoint has absolute slope of always less than $1$ (since that would mean that the distance to the solution will decrease in every step).
Now the slope to a fixpoint is:
$${f(x) - f(p)\over x-p} = {p+q-pq/x-p\over x-p} = {q(x - p)\over x(x-p)} = {q\over x}$$
So if always $|q/x_n|<1-\epsilon$ the series will converge to $p$. And if always $|q/x_n|>1+\epsilon$ it will diverge (and do so strictly).
This works on your example since you will have with $p=4$ and $q=-1$ that if $|x-p|\le 1$ that $x \ge 3$ which means that $|q/x| = 1/|x| \le 1/3$. As long as $|x-p|\le 1$ it will approach $p$ strictly and therefore it the $|x-p|\le 1$ will remain valid and it will aproach $p$ strictly indefinitely.
However the sequence is not monotone (as you can easily see by testing). What I mean by strict approaching is that $|x_n-L|$ is strictly decreasing, strict convergence to $L$ is convergence to $L$ that strictly approaches $L$.
