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