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?
On
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): $$\frac{\operatorname{arsinh}(\sinh(x))-\operatorname{arsinh}(\sin(x))}{\sinh x-\sin x}\sim_0\frac{\dfrac{x^3}3}{\dfrac{x^3}3}=1. $$
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, $$ \begin{align} \lim_{x\to0}\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)} &=\lim_{\xi\to0}\frac1{\sqrt{1+\xi^2}}\\[3pt] &=1\tag{2} \end{align} $$