If $a_{n+1}=\cos(a_n)$ for $n\ge0$ and $a_0 \in [-\pi/2,\pi/2]$, find $\lim_{n \to \infty}a_n$ if it exists

Let $a_{n+1}=\cos(a_n)$ for $n\ge0$ and $a_0 \in [0,\pi/2]$

Find $\lim_{n \to \infty}a_n$ if it exists.

I drew some sketches and it does seem like the limit exists, it's probably $x$ such that $\cos(x)=x$

I have no idea how to go about solving this, hints would really be appreciated.

Thank you for your time!

Solution 1:

Let $f=\cos$. Note that $a_1$ is in $I=[0,1]$ and that $f(I)\subset I$. Furthermore, for every $x$ in $I$, $-c\lt f'(x)\leqslant0$ with $c\lt1$ hence $(a_{2n})_{n\geqslant1}$ and $(a_{2n-1})_{n\geqslant1}$ are monotonous sequences with values in $I$ and they both converge to the unique solution $\ell$ of $\cos\ell=\ell$ in $[0,1]$.

(Numerically, $c=\sin(1)\lt.85$, $|a_{n+1}-\ell|\leqslant c\cdot|a_n-\ell|$ for every $n\geqslant1$, which implies that $|a_n-\ell|\leqslant c^{n-1}$ for every $n\geqslant1$, and $\ell\approx0.739085$.)