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Could you help me evaluating this limit?

$$ \lim_{x\to 0}\frac{1}{x}\cdot\left[\arccos\left(\frac{1}{x\sqrt{x^{2}- 2x\cdot \cos(y)+1}}-\frac{1}{x}\right)-y\right] $$

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3  
L'Hospital's rule? –  Doctor Dan Aug 12 '13 at 17:19
    
@ Doctor Dan I'm not sure of it, but does the form $\arccos(+\infty-\infty)$ allow the application of De L'Hopital's Rule? –  Andrea L. Aug 12 '13 at 17:31
1  
@BinaryBurst, could you please confirm the question is not modified by the edit? –  lab bhattacharjee Aug 12 '13 at 17:32
    
I strongly suspect that using Taylor polynomials is easier here. Start from the inside, using $(1+u)^{-1/2}=1-u/2+3u^2/4+O(u^3)$ with $u=x^2+2x\cos y$ and work your way outwards. –  Harald Hanche-Olsen Aug 12 '13 at 17:39
    
It is the same question :) And thank you for the nice editing. –  BinaryBurst Aug 12 '13 at 17:46

1 Answer 1

up vote 4 down vote accepted

Notice: I changed what I think a typo otherwise the limit is undefined.

By the Taylor series we have (and we denote $a=\cos(y)$) $$\frac{1}{\sqrt{x^{2}-2xa+1}}=1+xa+x^2(\frac{3}{2}a^2-\frac{1}{2})+O(x^3)$$ so $$\frac{1}{x\sqrt{x^{2}-2xa+1}}-\frac{1}{x}=a+x(\frac{3}{2}a^2-\frac{1}{2})+O(x^2)$$ Now using $$\arccos(a+\alpha x)=\arccos(a)-\frac{\alpha}{\sqrt{1-a^2}}x+O(x^2)$$ we have $$\arccos(\frac{1}{x\sqrt{x^{2}-2xa+1}}-\frac{1}{x})=\arccos(a)-\frac{\frac{3}{2}a^2-\frac{1}{2}}{\sqrt{1-a^2}}x+O(x^2)$$ so if we suppose that $y\in[-\frac{\pi}{2},\frac{\pi}{2}]$ then $$\lim_{x\to 0}\frac{1}{x}\cdot\left[\arccos\left(\frac{1}{x\sqrt{x^{2}-2x\cdot \cos(y)+1}}-\frac{1}{x}\right)-y\right]=-\frac{\frac{3}{2}a^2-\frac{1}{2}}{\sqrt{1-a^2}}$$

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I am officially impressed :) –  BinaryBurst Aug 12 '13 at 18:59
    
Actually, this is fairly standard once you have mastered Tayoler series and big-O notation. But +1 anyhow, for actually doing it. It is fairly laborious, if standard. One nitpick, though: The result is false for $y<0$, because for such $y$, $\arccos\cos y\ne y$. –  Harald Hanche-Olsen Aug 12 '13 at 19:25
    
There is trouble for $y=0$ as well, for then you're taking arccos of a number greater than 1 for $x$ small and positive. The limit from the left might be okay, though – I haven't checked, and have no time right now. But you have to be more careful because of the singularity of arccos at 1. –  Harald Hanche-Olsen Aug 12 '13 at 19:32
    
I will remember that :D –  BinaryBurst Aug 12 '13 at 19:34
    
...and limits! ${}{}{}$ ;-) –  amWhy Apr 14 at 14:44

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