Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$. 
Prove that $ \lim\limits_{x \to 0} \frac{\sinh x}{x} =1.$

I am having some trouble proving this without derivative. Some help would be much appreciate!
 A: $$\frac{\sinh x}{x}=\frac{e^x-e^{-x}}{2x}=\frac{1+x+\frac{x^2}{2!}+...-\left(1-x+\frac{x^2}{2!}_...\right)}{2x}=\frac{2x+\frac{2x^3}{3!}+..}{2x}=1+\frac{x^2}
{3!}+...$$
Takong $\lim$ on both sides we get $1$
${{{{}}}}$
A: Rewrite using known limit for $e$:
$$
\lim_{x \rightarrow 0} \frac{\sinh(x)}{x} = \lim_{x \rightarrow 0} \frac{e^x - e^{-x}}{2x} = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{-x}}{2x} = \frac{1}{2} \lim \frac{e^x - 1}{x} - \frac{1}{2} \lim_{x \rightarrow 0} \frac{e^{-x}-1}{x}
$$
$$
= \frac{1}{2} \lim \frac{e^x - 1}{x} + \frac{1}{2} \lim_{x \rightarrow 0} \frac{e^x-1}{x} = \lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1
$$
A: Hint: $\displaystyle\sin ⁡x=\frac{e^{ix}-e^{-ix}}{2i}$ and $\displaystyle\sinh ⁡x=\frac{\sin⁡(ix)}{i}$
A: You could use the squeeze theorem with bounding functions 1 and $e^{|x|}$ 
A: No reason to shy away from complex numbers. Let $x = iu$.
$$\begin{align}\lim_{x\to 0} \frac{\sinh x}{x} &= \lim_{u\to 0} \frac{\sinh (iu)}{iu} \\ &= \lim_{u\to 0} \frac{i\sin (u)}{iu} \\ &= \lim_{u\to 0} \frac{\sin (u)}{u} \\ &=1\end{align}$$
