$$\lim_{x\to \infty} {\cosh^{-1}(x^{3}) + \coth^{-1}(\sqrt{x^{2}+1}) - 3\sinh^{-1}(x)}$$
Honestly, I don't really know how to approach this. I know the logarithmic formulae for the inverse hyperbolic functions, but that gives me a very complicated expression. Is there an identity I can use or a rule? Any tips are welcome.
Thank you.
We can directly use the expression of inverse hyperbolic functions namely \begin{align} \sinh^{-1}x &= \log(x + \sqrt{x^{2} + 1})\notag\\ \cosh^{-1}x &= \log(x + \sqrt{x^{2} - 1})\notag\\ \coth^{-1}x &= \frac{1}{2}\log\left(\frac{x + 1}{x - 1}\right)\notag \end{align} From the above we see that $\coth^{-1}x \to 0$ as $x \to \infty$. It follows by the same logic that $\coth^{-1}\sqrt{x^{2} + 1} \to 0$ so we need to consider the only first and last terms of the expression given in question. Clearly we have \begin{align} \cosh^{-1}x^{3} - 3\sinh^{-1}x &= \log(x^{3} + \sqrt{x^{6} - 1}) - 3\log(x + \sqrt{x^{2} + 1})\notag\\ &= \log\left(\frac{x^{3} + \sqrt{x^{6} - 1}}{(x + \sqrt{x^{2} + 1})^{3}}\right)\notag\\ &= \log\left(\frac{1 + \sqrt{1 - 1/x^{6}}}{(1 + \sqrt{1 + 1/x^{2}})^{3}}\right)\notag\\ &\to \log 1 = 0\notag \end{align} The desired limit is $0$.