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I'm trying to calculate the following limit, involving $\arcsin$, where $x$ is getting closer to $0$ from the negative side, so far with no success.

The limit is: $$\lim_{x\rightarrow 0^-}\frac { \arcsin{ \frac {x^2-1}{x^2+1}}-\arcsin{(-1)}}{x}$$ it's part of a bigger question, where I should prove that the same expression, but where $x\rightarrow 0$ has no limit. So, I prove that the limits from both sides are different.

I already know that the solution to the above question is $(-2)$, but have no idea how to get there.

Would appreciate your help!

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Have you tried using L'Hôpital's rule? Also note that, since the numerator does not change sign while the denominator does, if the left sided limit exists and is not $0$, then the right sided limit is the negative, so not the same. –  Gregor Bruns May 27 '12 at 16:19
    
Three answers appeared before anyone up-voted the question, and so far, I'm the only one wh's done so. –  Michael Hardy May 27 '12 at 18:40
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4 Answers

up vote 3 down vote accepted

You can use the mean-value theorem on $f(y) = \arcsin({y^2 - 1 \over y^2 + 1})$ on any interval $[x,0]$ for small $x$, since $f(y)$ is continuous on $[x,0]$ and differentiable on $(x,0)$ (you don't need the derivative to exist at $y = 0$ or $y = x$ to apply the mean value theorem). Thus there is some $y \in (x,0)$ (that depends on $x$) such that $$ f'(y) = {\arcsin({x^2 - 1 \over x^2 + 1}) - \arcsin(-1) \over x} $$ Taking the derivative of $f$ using the chain rule, you get $$f'(y) = {1 \over \sqrt{1 - \big({y^2 - 1 \over y^2 + 1}\big)^2}}\bigg({y^2 - 1 \over y^2 + 1}\bigg)'$$ $$= {y^2 + 1 \over \sqrt{4y^2}}{4y \over (y^2 + 1)^2}$$ $$= -{2 \over y^2 + 1}$$ The minus sign comes from the fact that $y < 0$. Since $y \in (x,0)$, we conclude $f'(y)$ is between $-2$ and $-{2 \over x^2 + 1}$. Thus as $x$ goes to zero, $f'(y)$ must approach $-2$, which therefore is the limit.


Since the asker needs an entirely elementary proof, here's another way. Since $\arcsin(-1) = {-\pi \over 2}$, we are seeking $$\lim_{x \rightarrow 0^-}{\arcsin({x^2 - 1 \over x^2 + 1}) + {\pi \over 2} \over x}$$ Since $\lim_{\theta \rightarrow 0} {\sin(\theta) \over \theta} = 1$, this can be rewritten as $$\lim_{x \rightarrow 0^-} {\sin(\arcsin({x^2 - 1 \over x^2 + 1}) + {\pi \over 2}) \over\arcsin({x^2 - 1 \over x^2 + 1}) + {\pi \over 2}} \times {\arcsin({x^2 - 1 \over x^2 + 1}) + {\pi \over 2} \over x}$$ $$= \lim_{x \rightarrow 0^-}{\sin(\arcsin({x^2 - 1 \over x^2 + 1}) + {\pi \over 2}) \over x}$$ Using that $\sin(\theta + {\pi \over 2}) = \cos(\theta)$, this becomes $$\lim_{x \rightarrow 0^-}{\cos(\arcsin({x^2 - 1 \over x^2 + 1})) \over x}$$ Since $\cos(\arcsin(y)) = \sqrt{1 - y^2}$, the above becomes $$\lim_{x \rightarrow 0^-} {-{2x \over x^2 + 1} \over x}$$ $$= \lim_{x \rightarrow 0^-}-{2 \over x^2 + 1}$$ $$= -2$$ So this is your limit.

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It's a good answer, but as far as I know, I'm not allowed to use derivatives at all here. otherwise there are a lot of ways to solve this question, and they are all pretty simple... but thanks though.. –  RB14 May 27 '12 at 22:17
    
Ok, I added another way of doing it that is elementary and which is also different from the Weierstrass substitution in the other answer. –  Zarrax May 28 '12 at 1:16
    
Strangely I noticed your addition just now. I really love it, that's the kind of solution I expected! thank you. –  RB14 Jun 1 '12 at 16:00
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Let $x=\tan\theta$, where $\theta$ is in the interval $(\pi/2,\pi/2)$. When $x$ is close to $0$ and negative, so is $\theta$.

We express $\frac{x^2-1}{x^2+1}$ in terms of $\theta$. Start maybe with $x^2+1=\tan^2\theta+1=\sec^2\theta$. After a short while we find that $$\frac{x^2-1}{x^2+1}=\sin^2\theta-\cos^2\theta=-\cos 2\theta.$$

Take the $\arcsin$ of this. We get $-\arcsin(\cos 2\theta)$. Now we have to be careful. When $\theta$ is small and positive, $\arcsin(\cos 2\theta)$ is simply $\pi/2-2\theta$. But when $\theta$ is small negative, then $\arcsin(\cos 2\theta)$ is not equal to $\pi/2-2\theta$, it is $\pi/2+ 2\theta$.

In all cases when $\theta$ is not far from $0$, we have $$\arcsin(\cos 2\theta)=\pi/2-2|\theta|,$$ and therefore $$\arcsin\left(\frac{x^2-1}{x^2+1}\right)-\arcsin(-1)=-(\pi/2-2|\theta|)+ \pi/2=2|\theta|.$$

So it all comes down to finding $$\lim_{\theta\to 0^-} \frac{2|\theta|}{\tan\theta}.$$ This is easy. if we know about the behaviour of $\frac{\sin x}{x}$ near $0$.

Remark: The $2$ in the limits comes from a double-angle identity. The discontinuous behaviour comes from the kinky behaviour of $\arcsin(\cos\phi)$ when $\phi$ is close to $0$.

There is less to this solution than meets the eye. The expression $\frac{1-x^2}{1+x^2}$ is easy to recognize from the standard rational parametrization of the circle. It also comes up in calculus when we do the so-called Weierstrass substitution. (If I were sentimental I would have said let $x=\tan(\theta/2)$.)

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That's a lovely answer! it's a bit long, but it does the job. thanks for you efforts! –  RB14 May 27 '12 at 22:13
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Since the sine function is odd and $\cos(\pi/2-u)=\sin(u)$, the numerator is also $$ n(x)=\arccos\left(\frac{1-x^2}{1+x^2}\right). $$ When $x\to0$, $\dfrac{1-x^2}{1+x^2}=1-2x^2+o(x^2)$. Since $\cos(u)=1-\frac12u^2+o(u^2)$ when $u\to0$, $\arccos(1-v)\sim\sqrt{2v}$ when $v\to0^+$. Applying this to $v=2x^2$ yields $n(x)\sim\sqrt{4x^2}=2|x|$. Finally, $\dfrac{n(x)}x\to2$ when $x\to0^+$ and $\dfrac{n(x)}x\to-2$ when $x\to0^-$.

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What you can do is to make use of L'hopital's rule (since upper limit gives zero and lower limit also gives zero)

Differentiating the denominator gives 1. Now differentiate the numerator and obtain; $$\frac{(x^2+1)2x-(x^2-1)2x}{(x^2+1)^2}\frac{1}{\sqrt {1-\frac{(x^2-1)^2}{(x^2+1)^2}}}$$ and simplify to obtain $$\frac{2\sqrt {x^2}}{x(x^2+1)}=\frac{2|x|}{x(x^2+1)}=\frac{2\operatorname{sgn}(x)}{x^2+1}$$ Now taking the limit gives the desired result.

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Well, actually, I didn't mention it but i'm not allowed to use L'hopital's rule, since I'm trying to prove the function is not differentiable, and I do this by proving that the limit from both sides does not exists. so it's illegal for me to differentiate the function. I'm trying to prove that using the most basic concepts of differentiation of a function. but maybe I'm going to wrong way here. anyway the function is: $arcsin{\frac{x^2-1}{x^2+1}}$ and I should prove that it's not differentiable at point $x=0$ –  RB14 May 27 '12 at 16:26
    
The function is differentiable on all $\mathbb{R}\setminus\{0\}$, so you can differentiate it there in order to use L'hopital for the one-sided limits. –  Gregor Bruns May 27 '12 at 16:29
    
well, I can't since we did not learn in my class yet. (but I know all about it from another class I'm taking). so it's not allowed in this exercise –  RB14 May 27 '12 at 16:33
    
smanoos, the correct derivative is $$\frac{2\sqrt{x^2}}{x (x^2+1)} = \frac{2 \operatorname{sign}(x)}{x^2+1}.$$ –  Antonio Vargas May 27 '12 at 16:52
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@RB14, please clarify, what can you actually use? Are you allowed to use differentiation rules, but not L'hopital, or none of both? Without power series and/or differentiation rules this seems pretty hard. –  Gregor Bruns May 27 '12 at 17:40
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