How to solve $\lim_{x \to 0} \dfrac{\arctan(x)}{e^{2x}-1}$? 
$$\lim_{x \to 0} \dfrac{\arctan(x)}{e^{2x}-1}$$

I have no idea how to do this, initially I thought that perhaps $\lim_{x \to 0} \dfrac{e^x-1}{x} = 1$ might be of use but I don't see how I can rewrite this into something easier.

Hint. One may write, as $x \to 0$,
$$
\dfrac{\arctan(x)}{e^{2x}-1}=\frac12\cdot\dfrac{2x}{e^{2x}-1}\cdot \dfrac{\arctan(x)}{x}
$$ observing that
$$
\lim_{x \to 0}\dfrac{\arctan(x)}{x}=\lim_{x \to 0}\dfrac{\arctan(x)-\arctan(0)}{x-0}=\frac1{1+0^2}=1.
$$
