Excuse me if I'm being dense, but how do you derive $$\lim_{x \to -\infty} \sinh^{-1} (x) = -\infty$$
I have $$ \sinh^{-1}(x) = \log \left(x + \sqrt{x ^ 2 + 1}\right) = \log \left(x \left(1 + \sqrt{1 + x ^ {-2}}\right)\right) = \log \left( x \left(1 + 1 + \mathcal O (x ^ {-2})\right)\right) \sim \log (2x) $$ which isn't even real. It's trivially true the other way $$\lim_{x \to -\infty} \sinh(x) = \frac{e^{-\infty} - e^{\infty}}{2} = -\infty$$ so I don't get what I'm missing.
Being more careful with the signs, we have for $x<0$ that $x=-|x|$, so \begin{align} x+\sqrt{x^2+1}=-|x|+|x|\sqrt{1+\frac{1}{x^2}}=|x|\left(\sqrt{1+\frac{1}{x^2}}-1\right) \end{align} For large $|x|$, this is roughly $|x|\left(\left(1+\frac{1}{2}\cdot\frac{1}{x^2}\right)-1\right)=\frac{1}{2|x|}$, and the logarithm of this is $-\ln(2|x|)\to -\infty$ as $x\to -\infty$.