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.
 A: 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
$$
\lim_{x \to 0}\dfrac{\arctan(x)}{x}=\lim_{x \to 0}\dfrac{\arctan(x)-\arctan(0)}{x-0}=\frac1{1+0^2}=1.
$$
A: $1)$ 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
$$\lim _{x\to \:0}\left(\frac{\arctan \left(x\right)}{e^{2x}-1}\right)= \lim _{x\to \:0}\left(\frac{1}{e^{2x}2\left(x^2+1\right)}\right)= \frac{1}{2}$$
A: Note $\arctan x \sim x$ for small $x$. Hence we can evaluate 
$$\lim_{x \to 0} \frac{1}{\frac{e^{2x} - 1}{x}} =  \frac{1}{\lim_{x \to 0}\frac{e^{2x} - 1}{x}} = \frac 12$$
The denominator can be identified as the derivative of $x \mapsto \exp 2x$ at $x=0$.
A: Use equivalents:
$\arctan x\sim_0 x$, $\;\mathrm e^u-1\sim_0u$, hence
$$\frac{\arctan x}{\mathrm e^{2x}-1}\sim_0\frac x{2x}=\frac12.$$
