Infinite limit terms under root Suggest me a hint to solve:$$\psi=\lim_{x\to0}{\frac{\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}-1}{x^2}}$$

My try, 


Let $\displaystyle y=\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}$
$$\implies y^2=1-\cos x+y\iff y^2-y+\cos x-1=0$$
$$y=\frac{1\pm\sqrt{1-4(\cos x-1)}}2=\frac{1\pm\sqrt{5-4\cos x}}2$$
For $y\ge0$ and $\displaystyle x\to0,x\ne0\cos x<1,\sqrt{5-4\cos x}>1\implies 1-\sqrt{5-4\cos x}<0$  hence can be safely discarded
So, $$\lim_{x\to0}{\frac{\sqrt{1-\cos x+\sqrt{1-\cos x+\sqrt{1-\cos x+\cdots}}}-1}{x^2}}$$
$$=\lim_{x\to0}\frac{\sqrt{5-4\cos x}-1}{2x^2}$$
$$=\lim_{x\to0}\frac{5-4\cos x-1}{2x^2}\cdot\frac1{\lim_{x\to0}\sqrt{5-4\cos x}+1}$$
Now $$\lim_{x\to0}\frac{5-4\cos x-1}{2x^2}=\frac42\lim_{x\to0}\frac{1-\cos x}{x^2}$$
$x=2y\implies$ $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{y\to0}\frac{1-\cos2y}{(2y)^2}=\frac24\left(\lim_{y\to0}\frac{\sin y}y\right)^2=\cdots$$
