1
$\begingroup$

$$\lim_{x\to \infty} {\cosh^{-1}(x^{3}) + \coth^{-1}(\sqrt{x^{2}+1}) - 3\sinh^{-1}(x)}$$

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.

$\endgroup$
  • $\begingroup$ $\coth\big(0^+\big)=\infty,$ so $\text{arccoth }(\infty)=0.$ $\endgroup$ – Lucian Nov 4 '15 at 23:44
1
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ Interesting. Thank you. $\endgroup$ – Daniel Waleniak Nov 5 '15 at 3:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.