<s>[INST] How can I solve this limit without L'Hopital rule? 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?  
 [/INST] 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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