How do you find the limit of $\lim\limits_{x\rightarrow 0^-}\frac { \arcsin{ \frac {x^2-1}{x^2+1}}-\arcsin{(-1)}}{x}$? 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!
 A: 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 $x=\tan(\theta/2)$.)
A: 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 $x\to0^-$.
A: 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.
A: 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. 
