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I have found this interesting limit and I'm trying to solve it without use L'Hopital's Rule.

$$\lim\limits_{x\rightarrow 0}\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)}$$

I solved it with L'Hopital's rule and I found that the solution is $1$. But if I try without this rule, I can't solve it. Any ideas?

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  • $\begingroup$ Have you tried Taylor series? $\endgroup$ – Jack Lee Jul 22 '17 at 0:22
  • $\begingroup$ No, I have only tried with algebra and identities $\endgroup$ – Alexis Galois Jul 22 '17 at 0:28
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    $\begingroup$ For your fun (I hope), make the problem as $$\lim\limits_{x\rightarrow 0}\frac{\sin ^{-1}(\sinh (x))-\sinh ^{-1}(\sin (x))}{\sinh (x)-\sin (x)}$$ and using Taylor series, show the the result is $2$. $\endgroup$ – Claude Leibovici Jul 22 '17 at 5:18
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Using the Mean Value Theorem, and the fact that $\frac{\mathrm{d}}{\mathrm{d}x}\sinh^{-1}(x)=\frac1{\sqrt{1+x^2}}$, we get that $$ \frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)}=\frac1{\sqrt{1+\xi^2}}\tag{1} $$ for some $\xi$ between $\sin(x)$ and $\sinh(x)$.

Therefore, since both $\sinh(x)$ and $\sin(x)$ tend to $0$, the $\xi$ in $(1)$ tends to $0$; that is, $$ \begin{align} \lim_{x\to0}\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)} &=\lim_{\xi\to0}\frac1{\sqrt{1+\xi^2}}\\[3pt] &=1\tag{2} \end{align} $$

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  • $\begingroup$ Very clever, but I would use a different notation. For the RHS, I would say $\lim_{x\to0} \frac{\mathrm{d}}{\mathrm{d\xi}}\sinh^{-1}(\xi)\,|_{\xi<|x|}\\[3pt]$ $\endgroup$ – asky Jul 22 '17 at 4:30
  • $\begingroup$ @asky: I worried about the way to denote the derivative. I have changed the presentation a bit to hopefully be better. $\endgroup$ – robjohn Jul 22 '17 at 9:14
  • $\begingroup$ This solution uses the Axiom of Choice, doesn't it? $\endgroup$ – yoann Jul 22 '17 at 15:16
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    $\begingroup$ @PatrickStevens A $\xi$ is chosen for each $x$, and there are (uncountably) infinitely many of such $x$. So it seems to me that a whole function $x\mapsto \xi(x)$ is chosen. Note that it is easy to "choose" a uniquely defined $\xi$ (eg the smallest of them, which exists since the function is continuously differentiable), but I think it's worth explaining. $\endgroup$ – yoann Jul 24 '17 at 6:35
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    $\begingroup$ @yoann: Since $\frac1{\sqrt{1+\xi^2}}$ is a monotonic function (at least for $\xi$ between $\sin(x)$ and $\sinh(x)$), there is only one value of $\xi$ that satisfies $(1)$. No choice needed. $\endgroup$ – robjohn Jul 24 '17 at 16:45
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You can use equivalents and expansion in power series.

Let's begin with the denominator: $$\sinh x-\sin x=x+\frac{x^3}{3!}+o(x^3)-\Bigl(x-\frac{x^3}{3!}+o(x^3)\Bigr)=\frac{x^3}3+o(x^3)\sim_0\frac{x^3}3.$$ Now for the numerator:

First, by definition, $\;\operatorname{arsinh}(\sinh(x))=x$.

Next, $$\operatorname{arsinh} x=x-\frac12\frac{x^3}3+\frac{1\cdot 3}{2\cdot4}\frac{x^5}5-\frac{1\cdot 3\cdot5}{2\cdot4\cdot6}\frac{x^7}7+\dotsm $$

We'll deduce the expansion of $\operatorname{arsinh} (\sin x)$ at order $3$. Remember asymptotic expansions can be composed: \begin{align} \operatorname{arsinh} (\sin x)&=\operatorname{arsinh}\Bigl(x-\frac{x^3}6+o(x^3)\Bigr)=\Bigl(x-\frac{x^3}6\Bigr)-\frac16\Bigl(x-\frac{x^3}6\Bigr)^3+o(x^3)\\ &=x-\frac{x^3}6-\frac16x^3+o(x^3)=x-\frac{x^3}3+o(x^3), \end{align} so that $$\operatorname{arsinh}(\sinh(x))-\operatorname{arsinh}(\sin(x))=x-x+\frac{x^3}3+o(x^3)=\frac{x^3}3+o(x^3)\sim_0\frac{x^3}3.$$

Ultimately, we obtain (if the computation is correct): $$\frac{\operatorname{arsinh}(\sinh(x))-\operatorname{arsinh}(\sin(x))}{\sinh x-\sin x}\sim_0\frac{\dfrac{x^3}3}{\dfrac{x^3}3}=1. $$

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  • $\begingroup$ Thank you for you help. Your solution is interesting. $\endgroup$ – Alexis Galois Jul 23 '17 at 23:56
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Let $\sin x = y $ , $ \sinh x = y + z$

Then $L = \lim_{x\rightarrow 0}\frac{\sinh^{-1}(y+z) - \sinh^{-1}(y)}{z} = {\sinh^{-1}}' (0) = 1 $

($z \rightarrow 0 $ as $ x \rightarrow 0$)

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