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?  

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,
$$
\begin{align}
\lim_{x\to0}\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)}
&=\lim_{\xi\to0}\frac1{\sqrt{1+\xi^2}}\\[3pt]
&=1\tag{2}
\end{align}
$$
