Calculus - Sequence and Function - hard exercise EDIT: With your help I have succeed to prove it! Thanks! I also have another question related to this one - written down here.
Let $a,b \in \mathbb{R}$ such that $a<b$ and let $f:[a,b] \to [a,b]$ be a differentiable function, let $t \in [a,b]$, and consider the sequence $(x_n)_{n=1}^{\infty}$ defined by: $$\left\{\begin{matrix}
x_1=t & \\ 
x_{n+1}=f(x_n) & \forall n\geq 1 
\end{matrix}\right.$$
Suppose that there exists a point $\alpha \in [a,b]$ such that $f(\alpha)=\alpha$.
Prove, while using MVT, that if there exists a $0 \le q <1$ such that $|f'(x)|\le q$ for every $x \in [a,b]$, then $\lim_{n \to \infty}x_n=\alpha$
-
I don't know how to use MVT here, since I don't know nothing about $\alpha$ and $x_n$, I can't take the interval $[\alpha,x_n]$ or $[x_n,\alpha]$, ($\alpha$ and $x_n$ can be equal).
Any help will be awesome!
Thanks a lot!
EDIT:
Now I have a $y \in \mathbb{R}$, and the sequence $(x_n)_{n=0}^{\infty}$:
$$x_0=y ~ ~ , ~ ~ x_{n+1}=cos(x_n), \forall n \ge 0$$
Prove thatt the sequence converge to a limit $0<\alpha<1$.
Any tips?
Thanks again!

By using the MVT you can show:
$$
|f(b) - f(a)| < b-a.
$$
Now, have a look at the Banach fixed point theorem for some inspiration.
https://en.wikipedia.org/wiki/Banach_fixed-point_theorem
