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 $x=\tan(\theta/2)$.)