TITLE: How do you find the limit of $\lim\limits_{x\rightarrow 0^-}\frac { \arcsin{ \frac {x^2-1}{x^2+1}}-\arcsin{(-1)}}{x}$?
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: 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.