Proof that the solution to cosx = x, is the limit of a recursive sequence. So I've got this question. Exists a sequence $a_n$ such that:
$$a_0 = \frac \pi4, a_n=\cos\left(a_{n-1}\right)$$ 
Prove that $\lim_{n\rightarrow\infty} a_n = \alpha$ Where $\alpha$ is the solution to $\cos x = x$.
There is also a small hint saying I should prove its a Cauchy sequence and then use MVT.
Well, I've proven its Cauchy, and then used the fact that there exists a $L \in \mathbb{R}$ such that:
$$ \lim_{n\rightarrow\infty} \cos(a_{n-1}) = L. $$
Because cos is continuous, it can be written as:
$ \cos(\lim_{n\rightarrow\infty} a_{n-1}) = L $ which is $\cos(L) = L$. I've proven with IVT and the derivative that $\alpha$ from before is unique, therefore $L = \alpha$.
It looks fine to me, but the hint said to use MVT, which I don't really see why, or where to use it. Hopefully you guys will know more.
 A: $g(x):=\cos x-x\Rightarrow g(-{\pi\over 3})>0 ,g({\pi\over 3})<0$. As $g$ is continuous over $\mathbb{R}$, $g$ vanishes somewhere between $(-{\pi\over 3},{\pi\over 3})$. Note that as $\pi\gt 3$, $g$ can't vanish outside $(-{\pi\over 3},{\pi\over 3})$ otherwise it would mean $|\cos (x)|\gt 1$ for some $x$. Also notice $|{d\over dx}\cos x|=|\sin x|< {1\over \sqrt 2}\;\forall\; x\in (-{\pi\over 3},{\pi\over 3})$. As $\cos x$ is positive in $({-\pi\over 3},{\pi\over 3})$, the sole solution of $\cos x=x$, let's call it $\alpha$, must lie in the interval $(0,{\pi\over 3})$ (we can rule out $x=0$ by direct checking).
As cosine is differentiable over $(0,{\pi\over 3})$, using $\mathtt{MVT}$ for $x,y \in (0,{\pi\over 3}) \text{ with } x\neq y$ we see that $\exists c: x< c< y$ and $\cos x-\cos y=(x-y)\sin c\Rightarrow |\cos x-\cos y|\le \left({1\over\sqrt{2}}\right)|x-y|$. As $|\cos x|\le1$, for $n>1,x_n\in(0,{\pi\over 3})$. Therefore for $n>1$ we have
$$|\alpha-x_{n}|=|\cos\alpha-\cos x_{n-1}|\le\left({1\over\sqrt{2}}\right)|\alpha-x_{n-1}|\le\left({1\over\sqrt{2}}\right)^2|\alpha-x_{n-2}|\\
\le...\le\left({1\over\sqrt{2}}\right)^{n}|\alpha-x_0|:=\left({1\over\sqrt{2}}\right)^n\delta
\cdots(1)
$$
So, for $m>n>1$
$$|x_m-x_n|\le|\alpha-x_m|+|\alpha-x_n|\le\left[\left({1\over\sqrt{2}}\right)^m+\left({1\over\sqrt{2}}\right)^n\right]\delta<2\delta\left({1\over\sqrt{2}}\right)^n\cdots(2)
$$
As $2\delta$ is a constant and $\left({1\over\sqrt{2}}\right)< 1$, $(2)$ proves that $\{x_n\}$ is Cauchy and $(1)$ proves that it converges to $\alpha$.
