Study the character of recursive sequence I have tried yesterday to ask this question but probably it was not well written so I have decided to show you my attempt step by step.
I have to study the character of the following:
$$x_{n}=-x_{n-1}^2,\,\, x_1=x\in\mathbb{R}$$
First of all I have observed that for x=0 and x=-1 the sequence is constant so:

*

*if $x=0$ then $\lim_{n\to\infty}x_n=0$, since $x_n=x_{n-1}$ iff $x_{n}=0,1$.

*if $x=-1$. $x_1=-1,\, x_2=-1\,\,x_3=-1....x_n=-1$ then $\lim_{n\to\infty}x_n=-1$
Then $x_{n}-x_{n-1}=-x_{n-1}^2-x_{n-1}$. So:

*

*if $-x_{n-1}^2-x_{n-1}>0$ then $-1<x_{n-1}<0$.
Thus if $-1<x_1<0$ then $-1<x_n<0$ for each $n$. This means that the sequence is increasing and so it is convergent: $\lim_{n\to\infty}x_{n}=sup\{x_n\}=l$. $\textbf{l=0}$ since $x_n<0$.


*if $-x_{n-1}^2-x_{n-1}<0$ then $x_{n-1}<-1\, \vee x_{n-1}>0$.
Thus if $x_1<-1$ then $x_n<-1$ for each $n$: $\lim_{n\to\infty}x_{n}=inf\{x_n\}=l$. Since the possible finite limits are $0,-1$, they are not acceptable. So $\textbf{l=}$ $-\infty$.
Then?
 A: There is at least two methods.
Once you proved by recursion that $x^n = -x^{2^n}$, you deduce the limit $0$ if $|x|<1$, $-1$ if $|x|=1$, $-\infty$ if $|x|>1$.
Other method, which can work even when you have no close formula for $x_n$. First, make a picture showing the graph of $f : x \mapsto -x^2$ with the line of equation $y=x$, and showing graphically the first iterates. Then you see what happens, depending on the initial value.
Since $f : x \mapsto -x^2$ is continuous, if the sequence $(x_n)$ converges, the limit is a fixed point of $f$, i.e. $-1$ or $0$.
When $x=-1$ or $x=0$, the sequence $(x_n)$ is constant.
When $x<-1$, one has $f(x)<x$. The interval $]-\infty,-1[$ is stable under $f$. Hence a recursion shows that $x_n<-1$ for all $n$ and $(x_n)$ decreases. It has a limit $\ell \in [-\infty,-1[$ which can only be $-\infty$.
When $x>1$, one has $x_1<-1$, and you get the same conclusions but only for $n \ge 1$.
When $-1<x<0$, one has $x<f(x)<0$. The interval $]-1,0[$ is stable under $f$. Hence a recursion shows that $-1<x_n<0$ for all $n$ and $(x_n)$ increases. It has a limit $\ell \in ]-1,0]$ which can only be $0$.
