L'Hopital's Rule!

Because this is an indeterminant form zero over zero, we can use L'Hopital's rule. We evaluate this limit by taking the derivatives of the top and the bottom.

$\lim _{ x\rightarrow 0 }{ \frac { \frac { d }{ dx } { (e }^{ x }-{ e }^{ -x }) }{ \frac { d }{ dx } (\sin { x } ) } } =\quad \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-(-{ e }^{ -x }) }{ \cos { x } } } =\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }+{ e }^{ -x } }{ \cos { x } } }$

This simplifies to $2$

You can also see the solution by visualizing that $\sinh (x) = \frac{e^x - e^{-x}}{2}$ so $\lim_{x\to0}\frac{e^x - e^{-x}}{\sin(x)} = 2\lim_{x\to0}\frac{\sinh(x)}{\sin(x)} = 2\lim_{x\to0} \frac{\sinh(x)}{x} \frac{x}{\sin(x)} = 2$

by the fundamental trigonometric limit $\frac{\sin(x)}{x}$ and $\frac{\sinh(x)}{x}$ when $x \to 0$ iquals 1.