Let $$f:[0,1]\to\Bbb R\;\;\mbox{defined by}\;\;\; f(x)=\sqrt{\frac{1+x}{2}}$$
Find:
$$\lim_{x\to0}\left(\lim_{n\to\infty}2^{2n}\left(1-\left(\overbrace {f \circ f \circ f \cdot\cdot\cdot \circ f}^{n}(x)\right)\right)\right)$$
I mainly need help with simplifying the composite function. I'll try to take it on from there. Any (substantial) hints or solutions will be greatly and sincerely appreciated.
Make the substitution $x = \cos \theta$. Then $f(x) = \sqrt{\dfrac{1+\cos \theta}{2}} = \cos \dfrac{\theta}{2}$. Hence, $f^{(n)}(x) = \cos \dfrac{\theta}{2^n}$.
So, $\displaystyle\lim_{n \to \infty}2^{2n}(1-f^{(n)}(x)) = \lim_{n \to \infty}2^{2n}\left(1-\cos \dfrac{\theta}{2^n}\right) = \lim_{n \to \infty}2^{2n}\left(\dfrac{\theta^2}{2 \cdot 2^{2n}}+O\left(\dfrac{1}{2^{4n}}\right)\right) = \dfrac{\theta^2}{2}$ $= \dfrac{1}{2}(\arccos x)^2$.
Taking the limit as $x \to 0$ should be easy.