The H is silent again

$\sin x=\dfrac{e^{ix}-e^{-ix}}{2i}$
$\sinh x=\dfrac{e^x-e^{-x}}2$
So $\sinh(ix)=i\sin x$
$\displaystyle\lim_{x\to0}\dfrac{\sinh x}{x}=\lim_{x\to0}\dfrac{\sinh ix}{ix}=\lim_{x\to0}\dfrac{i\sin x}{ix}=1$

I'm going to write·3 proofs, please if someone has some doubt or whishes a longer explanation or see some mistake in any poof, tell me...

1.- $\displaystyle \lim_{x \to 0} \frac{\text{sinh x}}{x} = \lim_{x \to 0} \frac{e^x - e^{-x}}{2x} =$ (Now, use MacLaurin series : $e^x$ ~ $1 + x$ as $x \to 0$ ) $\displaystyle \lim_{x \to 0} \frac{2x}{2x} = 1$ 2.- Use L'Hôpital's Rule : $\displaystyle \lim_{x \to 0} \frac{\text{sinh x}}{x} = \lim_{x \to 0} \frac{\text{cosh x}}{1} = 1$ 3.- Use Taylor polynomial at x = 0 .

Let $x > 0$ , and $f(x) = e^x = \frac{1x^0}{0!} + \frac{1x^1}{1!} + \frac{e^c \cdot x^2}{2!} = 1 + x + \frac{e^c \cdot x^2}{2}$ with $c \in (0,x)$ ( $c$ is a constant depending on $x$ ) . And let $g(x) = e^{-x} = 1 - x + \frac{e^d \cdot x^2}{2}$ with $d \in (-x,0)$ ( $d$ is a constant depending on $x$ ). Then, $\displaystyle \lim_{x \to 0} \frac{\text{sinh x}}{x} = \lim_{x \to 0} \frac{e^x - e^{-x}}{2x} =$ $= \displaystyle \lim_{x \to 0} \frac{1 + x + \frac{e^c \cdot x^2}{2} - 1 + x - \frac{e^d \cdot x^2}{2}}{2x} =$ $= \lim_{x \to 0} \frac{2x + \frac{(e^c - e^d)x^2}{2}}{2x} =$ (Divide now numerator and denomitor by x) $= \lim_{x \to 0} \frac{2 + \frac{(e^c - e^d)x}{2}}{2} = 1$