Standard results won't help you in this

Substitute $x=\cos {2\theta}, \ \ \theta \in [0,\pi/2]$

After doing so, we get

$f(x)=f(\cos {2\theta})=\cos {\theta}$

$f^2(x)=f(\cos {\theta})=\cos {\dfrac{\theta}{2}}$

$\vdots$

$f^n(x)=\cos {\dfrac{\theta}{2^{n-1}}}$

So, the required limit

$\begin{aligned} L &=\lim_{n \rightarrow \infty} 2^{2n+1} (1-f^n(x)) \\ &=\lim_{n \rightarrow \infty} 2.2^{2n}\left(1-\cos {\dfrac{\theta}{2^{n-1}}}\right) \\ &=\lim_{n \rightarrow \infty} 4.2^{2n} \sin^2 {\dfrac{\theta}{2^n}} \\ &=4\theta^2\\ &=(\cos^{-1} x)^2 \end{aligned}$

Plugging in $x=-1$ , we get $L=\pi^2$ .

just put x=cos (a) , this is a standard question in almost all textbooks ,,

I just do math for fun, I'm only 15 and im running out of math problems to do. Can you make more math problems.