Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could you help me evaluating this limit?

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

share|improve this question
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
@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 $$\lim_{x\to 0}\frac{1}{x}\cdot\left[\arccos\left(\frac{1}{x\sqrt{x^{2}-2x\cdot \cos(y)+1}}-\frac{1}{x}\right)-y\right]=-\frac{\frac{3}{2}a^2-\frac{1}{2}}{\sqrt{1-a^2}}$$

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.