Prove that there exists $n\in \mathbb{N}$ s.t. $x_n=\frac12$ Let $(x_n)_n$ a sequence given by $2x_{n+1}=2x_n^2-5x_n+3$ with $x_1\in \mathbb{Q}$. I know that the sequence is convergent. I know that the limit of the sequence should be $\dfrac{1}{2}$ or $3$.
I want to prove that there exists $n\in \mathbb{N}$ s.t. $x_n=\dfrac{1}{2}$ if the sequence goes to $\dfrac{1}{2}$. Similar if the sequence goes to $3$.
I tried by definition with $\epsilon$ but I didn’t succeed.

Let $f(x)=x^2-\frac{5}{2}x+\frac{3}{2}$, so that $x_{n+1}=f(x_n)$. Suppose that $x_n$ converges to $l$, where $l\in\lbrace \frac{1}{2},3\rbrace$. Then
$$
y_n=\frac{x_{n+1}-l}{x_n-l}=\frac{f(x_{n})-f(l)}{x_n-l} \to f'(l) \textrm{ when } n \to \infty \tag{1}
$$
Note that $f'(\frac{1}{2})=-\frac{3}{2}$ and $f'(3)=\frac{7}{2}$. So $|f'(l)| \geq \frac{3}{2}$ in both cases, and hence $|f'(l)| \gt \frac{5}{4}$ in both cases. It follows that there is a $n_0$ such that $|y_n|\gt \frac{5}{4}$ for all $n\geq n_0$. Then
$$|x_{n+1}-l| \geq \frac{5}{4} |x_n-l| \textrm{ for all } n\geq n_0 \tag{2}$$. By induction, we deduce
$$|x_n-l| \geq \big(\frac{5}{4}\big)^{n-n_0}|x_{n_0}-l| \textrm{ for all } n\geq n_0 \tag{3}$$
If $x_{n_0}\neq l$, we would deduce $\lim_{n\to\infty}{|x_n-l|}=\infty$, which is impossible. So $x_{n_0}=l$, which finishes the proof.
