Existence of the limit of the recurrent sequence I need to calculate the limit of the following sequence:
$$
x_n=1\\
x_{n+1}=\frac{30+x_n}{x_n}
$$
If it is proven that the limit exists, I know how to do it:
$$
a=\lim_{n\to\infty}\\
a=\frac{30+a}{a}\\
a=6
$$
(we choose positive number because the elements of $x_n$ are positive; it is almost obvious).
But I don't know how to prove the existence of the limit. If you look at the difference $|x_{n+1} - x_n|$, it will be alternating (i.e. the sign is different depending on parity).
So, that is the question: how to prove it?
Thank you in advance
 A: You can get rid of the alternation by pairing the iterations:
$$x_{n+2}=\frac{30+x_{n+1}}{x_{n+1}}=\frac{30+31x_n}{30+x_n}>x_n.$$
Thus the even sequence is growing and bounded above by $31$.
A: Hint:  try to prove it is 


*

*monotone, i.e. the difference $x_{n+1}-x_n$ is always positive or always negative

*bounded, i.e. exits numbers $m,M$ such that $m\leq x_n\leq M$ for all $n$.

A: Here is a presentation of the method of adjacent sequences for a similar equation:
Let $a_{n+1}=\dfrac{10}{a_n}-3 \ \ ; \ \ a_1=10$ the find the limits $\lim\limits_{n \to \infty} a_n$
I would like to show a variant, in fact in this case we can even find a closed form for the sequence, and it doesn't cost much more in term of calculations than the method of adjacent sequences.

First we need to notice that $x_n>0$ for all $n$ since $x_1=1$ and there is no subtraction in the induction formula.
Thus the equation is equivalent to $$x_{n+1}x_n=30+x_n$$
Now remark that $30=5\times 5+5$ so let set $y_n=x_n+5$ to cancel the constant term.
We get $(y_{n+1}-5)(y_n-5)=25+y_n\iff$ $$y_{n+1}y_n=5y_{n+1}+6y_n$$
The trick is now to divide by the product (it is $\neq 0$ since $x_n>0$) to get $\ 1=\dfrac 5{y_n}+\dfrac 6{y_{n+1}}$
So let set $z_n=\dfrac 1{y_n}$ and we have a linear equation $$\begin{cases}z_1=\frac 1{x_1+5}=\frac 16\\6z_{n+1}+5z_n=1\end{cases}$$
This solves classicaly to root $-\frac 56$ with initial condition and we find $z_n=\frac 1{11}\left(1-\left(-\frac 56\right)^n\right)$
Since $z_n\to \frac 1{11}\quad$ we get $\quad x_n\to 11-5=6$.
