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.

$1)$ Asymptotic expansion
$arctan(x)\sim_0 x$
$e^{2x}-1 \sim_0 2x$
$$\lim _{x\to 0}\left(\frac{arctan(x)}{e^{2x}-1}\right)\approx \lim _{x\to 0}\left(\frac{x}{2x}\right) = \frac{1}{2}$$
$2)$ 
Apply L'Hopital's Rule
$$\lim _{x\to \:0}\left(\frac{\arctan \left(x\right)}{e^{2x}-1}\right)= \lim _{x\to \:0}\left(\frac{1}{e^{2x}2\left(x^2+1\right)}\right)= \frac{1}{2}$$
