$f(x)=\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\cdots}}}$ Let $$f(x)=\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\cdots}}}$$
and $$\phi(x)=\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+\cdots}}}$$
Then find $\lim_{x\to 0}\frac{1-f(x)}{1-\phi(x)}.$
I tried to solve but got stuck.
I simplified $f(x)=\frac{1\pm\sqrt{5-4\cos x}}{2}$ and $\phi(x)=\frac{1\pm\sqrt{1+4x^2}}{2}$ but when i calculated
$\lim_{x\to 0}\frac{1-f(x)}{1-\phi(x)}=\frac{1\mp\sqrt6}{1\mp1}$ ,but the answer given in the book is $\frac{1}{2}$
Where have i gone wrong?Please help me.
 A: Let $$\displaystyle y = \sqrt{(1-\cos x)+\sqrt{(1-\cos x)+\sqrt{(1-\cos x)+..........}}}$$
Then $$\displaystyle y = \sqrt{1-\cos x+y}\Rightarrow y^2 = 1-\cos x+y\Rightarrow y^2-y-2\sin^2 \frac{x}{2} =0$$
So $$\displaystyle y = \frac{1\pm \sqrt{1+8\sin^2 \frac{x}{2}}}{2}\Rightarrow 2y-1 = +\sqrt{1+8\sin^2 \frac{x}{2}}$$
and $$\displaystyle z = \sqrt{x^2+\sqrt{x^2+\sqrt{x^2+.......}}}$$
Then $$\displaystyle z= \sqrt{x^2+z}\Rightarrow z^2=x^2+z\Rightarrow z^2-z-x^2=0$$
So $$\displaystyle z = \frac{1\pm \sqrt{1+4x^2}}{2}\Rightarrow 2z-1 = \sqrt{1+4x^2}$$
So given $$\displaystyle \lim_{x\rightarrow 0}\frac{1-y}{1-z} = \lim_{x\rightarrow 0}\frac{2y-1-1}{2z-1-1} = \lim_{x\rightarrow 0}\frac{\sqrt{1+8\sin^2 \frac{x}{2}}-1}{\sqrt{1+4x^2}-1}$$
So we get $$\displaystyle \lim_{x\rightarrow 0}\frac{\sqrt{1+8\sin^2 \frac{x}{2}}-1}{\sqrt{1+4x^2}-1}\times \frac{\sqrt{1+8\sin^2 \frac{x}{2}}+1}{\sqrt{1+8\sin^2 \frac{x}{2}}+1}\times \frac{\sqrt{1+4x^2}+1}{\sqrt{1+4x^2}+1}$$
So we get $$\displaystyle \lim_{x\rightarrow 0}\frac{8\sin^2 \frac{x}{2}}{4x^2}\times \frac{2}{2} = \frac{1}{2}$$
