Let $\cos:\left[0,\dfrac{\pi}{2}\right]\to \left[0,\dfrac{\pi}{2}\right]$, with $\cos^{m}=\cos\circ\cos\circ\cdots\cos\circ$ (m times). Show that, for all $x\in \left[0,\dfrac{\pi}{2}\right]$, exists $\displaystyle\lim_{m\to\infty}{\cos^{m}(x)}=a$, where $a$ is independent of $x$.
My approach: This problem is about the contraction principle. Let $f:\left[0,\dfrac{\pi}{2}\right]\to\left[0,\dfrac{\pi}{2}\right]$, with $f(x)=\cos^{2}(x)$, is a contraction. Indeed, $f'(x)=\sin(\cos(x))\sin(x)$, this implies $$-1\leq\cos(x)\leq 1\implies-\sin(1)=\sin(-1)\leq\sin(\cos(x))\leq\sin(1)$$ Therefore, $\vert\sin(\cos(x))\vert\leq\sin(1)$. Thus, $$\vert f'(x)\vert=\vert\sin(\cos(x))\sin(x)\vert\leq\vert\sin(\cos(x))\vert\vert\sin(x)\vert\leq\vert\sin(\cos(x))\vert\leq\sin(1)=k<1$$ $\therefore f$ is contraction. Then, for $x_{0}\in\left[0,\dfrac{\pi}{2}\right]$ defined the sequences $x_{1}=f(x_{0})$, $x_{2}=f(x_{1})$, .., $x_{n+1}=f(x_{n})$, It is easy to see that $\{x_{n}\}$ is a Cauchy sequence in $\left[0,\dfrac{\pi}{2}\right]$, but I do not know how to follow. Regards!
As $x_n$ is a Cauchy sequence, it has a limit $a\in\left[0,\frac\pi2\right]$. Now if $y\in\left[0,\frac\pi2\right]$ with $y_n\to b$ then
$$0 \leqslant |a-b| = |f(a)-f(b)|\leqslant k|f(a)-f(b)|, $$ and hence $$0\leqslant (1-k)|f(a)-f(b)|\leqslant 0, $$ so that $|f(a)-f(b)|=0$, i.e. $f(a)=f(b)$. Now, $a=f(a)$ and $b=f(b)$, so we conclude that $a=b$ and thus $a$ is independent of $x$.