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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, enter image description here

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You were going right with the latex, why did you stop? – chubakueno Jul 5 '14 at 5:46
@chubakueno it was just wasting time to write such complicated eqn in latex, rather i wrote it by hand – ADG Jul 5 '14 at 5:52
up vote 2 down vote accepted

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 $$\lim_{x\to0}\frac{5-4\cos x-1}{2x^2}=\frac42\lim_{x\to0}\frac{1-\cos x}{x^2}$$

$x=2y\implies$ $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{y\to0}\frac{1-\cos2y}{(2y)^2}=\frac24\left(\lim_{y\to0}\frac{\sin y}y\right)^2=\cdots$$

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Aditya was already there,,, – chubakueno Jul 5 '14 at 5:49
how can you eliminate - sign even though it gives 0 as satisfied by $y\ge0$ – ADG Jul 5 '14 at 5:51
sorry, but the thing i want to ask is: $$y=\frac{1\pm\sqrt{5-4\cos x}}{2}$$ is correct but if you put $x\to0$ or $\cos x\to1 $ , you get $$y\to\frac{1\pm\sqrt{5-4}}{2}=\frac{1\pm1}{2}=0,1$$, both of which are $y\ge0$ – ADG Jul 5 '14 at 6:05
@Aditya, Thanks for your observation. But, I think the other limit is undefined – lab bhattacharjee Jul 5 '14 at 6:12
@Aditya, For $$x\to0,x\ne0\cos x<1,\sqrt{5-4\cos x}>1\implies 1-\sqrt{5-4\cos x}<0$$ hence can be safely discarded – lab bhattacharjee Jul 5 '14 at 6:22


If you use Taylor series, you can start from lab bhattacharjee's answer

$$y=\frac{1\pm\sqrt{5-4\cos x}}2$$ and use the fact that around $x=0$, $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^5\right)$. So,$$5-4\cos (x)=1+2 x^2-\frac{x^4}{6}+O\left(x^5\right)$$ Now, use the fact that $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ and replace $y$ by $(2 x^2-\frac{x^4}{6})$. So,$$\sqrt{5-4\cos (x)}=1+x^2-\frac{7 x^4}{12}+O\left(x^5\right)$$

I am sure that you can take from here and easily find lab bhattacharjee's results.

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