Let $(u_n)$ be a bounded sequence of real numbers such that $\lim_{n\rightarrow+\infty} u_{n+1}-u_n^2=0$ Let $(u_n)$ be a bounded sequence of real numbers such that $\lim_{n\rightarrow+\infty} u_{n+1}-u_n^2=0$. What can we say about $(u_n)$.
Noticing the fact that $(u_n)$ is a bounded sequence, we can use the Bolzano–Weierstrass theorem to say that $0$ and $1$ are the only points where $(u_n)$ can converges. Can we notice more ?
 A: First, we show that $(u_n)_n$ cannot have a subsequence converging to any $L\not\in \{0,1\}.$ Second, we show this implies $u_n$ converges to $0$ or to $1.$
First, suppose $L$ is a limit of the sequence with $0\ne L\ne 1.$ Then $L^2,L^4,L^8,L^{16}$ (etc.) are limits too, and the sequence is bounded, so $0<|L|<1.$ Then $L'=L^{2^k}$ is also a limit for some $k\in \Bbb N$ such that $0<L'<|L|/8$. Take $m_1\in \Bbb N$ such that $n>m_1\implies |u_n^2-u_{n+1}|<|L|/8.$ Take $m_2\in \Bbb N$ with $m_2>m_1$ and $0<|u_{m_2}-L'|<L'.$
Now (1). $|u_{m_2}|<|L|/4.$ (2). If $n\ge m_2$ and if $|u_n|<|L|/4$ then $$|u_{n+1}|<|u_n^2|+|L|/8<|L|^2/16+|L|/8<|L|/16+|L|/8<|L|/4.$$ So by induction on $n\ge m_2$, we have $n\ge m_2\implies |u_n|<|L|/4.$ This contracts the assumption that $L$ is a limit of the sequence.
Second, by the first part, take $n_1\in \Bbb N$ such that $\forall n\in \Bbb N\,(\,n>n_1\implies |u_{n+1}-u_n^2| <1/4\,).$
Take $n_2\in \Bbb N$  such that $\forall n\in \Bbb N\,(\,n>n_2\implies (\,|u_n-1|<1/5\lor |u_n|<1/5\,).$
Suppose that $\{n\in \Bbb N: |u_n-1|<1/5\}$ and $\{n\in \Bbb N: |u_n|<1/5\}$ are both infinite sets. Take some (any) $n>\max (n_1,n_2)$ such that $|u_n-1|<1/5$. Now let $n'$ be the least $m>n$ such that $|u_m|<1/5.$ Then $n'-1\ge n>n_2$ and by def'n of $n'$ we have $\neg (|u_{n'-1}|<1/5),$ so $|u_{n'-1}-1|<1/5.$ Hence, since also $n'-1\ge n>n_1,$ therefore $$ 16/25=(1-1/5)^2< |u_{n'-1}^2|<|u_{n'}|+1/4< 1/5 + 1/4$$ which is absurd.
Therefore $u_n$ converges to $0$ or $u_n$ converges to $1.$
A: Let $m= \liminf u_n$. Because $(u_n)$ is bounded, $m$ cannot be infinity. 
Let$(u_{n_k})$ be a subsequence of $(u_n)$ such that $$\lim_{k \rightarrow+\infty} u_{n_k}=m$$
So we have
$$0 = \lim_{k}( u_{n_k+1}-u_{n_k}^2) \ge m-m^2 $$
Hence $m \ge 1$ or $m \le 0$


If $m \ge 1$, let $M=\limsup u_n \ge \liminf u_n=m  $. Similarly, we can show that 
$$M-M^2 \ge 0$$
which means $0 \le M \le 1$, hence $M=m=1$. Thus
$$\lim u_n =1$$

If $m \le 0$, wee see that 
$$0 = \lim_{k} ( u_{n_k}- u_{n_{k-1}}^2 ) \le \lim_k u_{n_k} =m $$
Thus $m=0$, or
$$\liminf u_n = 0$$
Choosing any $\epsilon \in (0,1/8)$
Because $\lim u_n-u_n^2=0$,  there is a $N_1$ such that $$ u_n-u_n^2<\epsilon \quad \forall n >N_1 $$
Because $\liminf u_n = 0$, there is a $N_2>N_1$ such that $u_n>-\epsilon$ for all $n \ge N_2$ and
$$|u_{N_2}|<\epsilon/2$$
We'll prove now that $$u_n <\frac{1-\sqrt{1- 4\epsilon}}{2}=: a_{\epsilon}$$ for all $n\ge N_2$.
Indeed, $ u_n < \frac{2\epsilon}{1+\sqrt{1-4\epsilon}}= a_{\epsilon} $ with $n=N_2$
If $u_k<a_{\epsilon}$ for  some $n=k \ge N_2$, we see that
because  $ k \ge N_2 \ge N_1$,
$$u_{k+1}-a_{\epsilon}\le u_k^2+\underbrace{\epsilon-a_{\epsilon}}_{=-a_{\epsilon}^2} = \underbrace{(u_k-a_{\epsilon})}_{<0}\underbrace{(u_k+a_{\epsilon})}_{ > 0 \text{ because } u_k>-\epsilon} <0$$
Hence
$$u_n < a_{\epsilon}$$ for all $n\ge N_2$.
Because $\epsilon$ is arbitrary, this implies that $M=\limsup u_n \le 0=m$, thus
$$\lim u_n= 0$$
Side note: Boundedness is not necessary anyways, we can always show it by using the given convergence.
