Probably L'Hospital

First we simplify $\displaystyle \psi (a)$ :

$\displaystyle \psi (a) = \sqrt{a+\sqrt{a+\sqrt{a+\cdots}}} = \sqrt{a+\psi (a)}$

$\displaystyle => [\psi (a)]^2 - \psi (a) - a =0$ , this is a quadratic equation can be solved easily : $\displaystyle \psi (a) = \frac{1+\sqrt{4a+1}}{2}$ (we took the positive value)*

Now we write our limit and rearrange it :

$\displaystyle \lim_{x \to 0} \frac{1-\psi (1-\cos x)}{1-\psi (x^2)} = \lim_{x \to 0} \frac{1-\sqrt{5-4\cos x}}{1-\sqrt{4x^2 +1}}$

L'Hopital rule seems messy, so rewrite our limit as:

$\displaystyle \lim_{x \to 0} \frac{1-\sqrt{5-4\cos x}}{1-\sqrt{4x^2 +1}}$

$\displaystyle =\lim_{x \to 0} \frac{1+\sqrt{4x^2 +1}}{1+\sqrt{5-4\cos x}}\frac{1+\sqrt{5-4\cos x}}{1+\sqrt{4x^2 +1}}\frac{1-\sqrt{5-4\cos x}}{1-\sqrt{4x^2 +1}}$

$\displaystyle =\lim_{x \to 0} \frac{1+\sqrt{4x^2 +1}}{1+\sqrt{5-4\cos x}} \times \lim_{x \to 0} \frac{1+\sqrt{5-4\cos x}}{1+\sqrt{4x^2 +1}}\frac{1-\sqrt{5-4\cos x}}{1-\sqrt{4x^2 +1}}$

$\displaystyle = \frac{2}{2} \lim_{x \to 0} \frac{1-(5-4\cos x)}{1-(4x^2+1)} = \lim_{x \to 0} \frac{1-\cos x}{x^2} = \lim_{x \to 0} \frac{\sin^2 x}{x^2(1+\cos x)}$

$\displaystyle = \frac{(\lim_{x \to 0} \frac{\sin x}{x})^2}{\lim_{x \to 0} (1+\cos x)} =\boxed{ \frac{1}{2} }$

• I will justify here why I used the positive sign in the quadratic formula rather than the negative one:

Note that the domain of our $\psi$ is $[0 , \infty [$ . we know from the quadratic formula that $\psi (a) = \frac{1+\sqrt{4a+1}}{2}$ or $\psi (a) = \frac{1-\sqrt{4a+1}}{2}$

Using the fact that $\psi$ is positive for all $a$ , we can't use \(\psi (a) = \frac{1-\sqrt{4a+1}}{2} unless $a=0$ hence:

$\psi (a) = \frac{1+\sqrt{4a+1}}{2}$ If $a \in ]0,\infty [$

and

$\psi (a) = \frac{1 \pm \sqrt{4a+1}}{2}$ If $a = 0$ .

Hold your horses for a moment.

$\boxed{\sqrt{N\sqrt{N\sqrt{N\sqrt{N...}}}}=N}$

Proof:

$\sqrt{N\sqrt{N\sqrt{N\sqrt{N...}}}}=N$

$\Rightarrow N\sqrt{N\sqrt{N\sqrt{N...}}}=N^2$

$\Rightarrow N\cdot N=N^2$

$\Rightarrow \boxed{N^2=N^2}$ ,

$\forall N$ .

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So,

$\psi (a)=a$ .

Thus,

$\displaystyle\lim_{x\rightarrow 0}{\frac{1-\psi (1-\cos{x})}{1-\psi (x^2)}}=\displaystyle \lim_{x\rightarrow 0}{\frac{1- (1-\cos{x})}{1-x^2}=\frac{1- (1-1)}{1-0^2}}=\boxed{1}$

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What went wrong?

L-H rule as well as RATIONALIZE is a real pain .... use the approximation (1+x)^n=1+nx .........done in 2 steps :P

Firstly I worked upon the root expression but later I found no use of it . I blindly use L'hosptal rule and find that is works . Thus answer is 0.5. :)