Limit of $\sinh z$ as $z\to i$. I'm trying to understand why 

$$\lim_{z\rightarrow\ i} \sinh(z)= i\sin(1)$$ 

Thanks for any feedback.
I tried using $\sinh(z)=(e^{z}-e^{-z})/2$ but that only seemed to make things more complicated now I'm thinking that L'Hopital may be needed but I don't know how to go about it.
Hint:

\begin{align}
\sin (z)&amp;=\frac{e^{iz}-e^{-iz}}{2i}
\end{align}
Then
\begin{align}
i\sin (1)&amp;=\frac{e^{i}-e^{-i}}{2}
\end{align}
Recall the identity $\sinh(iz)=i\sin z$. Then
$$
\lim_{z \to i} \sinh(z)=\sinh i=\sinh(i\cdot 1)=i \sin(1)
$$