Limit with inverse trigonometric functions: $\lim_{x\to 0}\frac{\arctan2x}{\sin3x}$ $$\lim_{x \to 0} \frac{\tan^{-1}2x}{ \sin 3x}$$ 
What is the shortest way to do this please?  
Is there a standard way to solve questions like that with inverse trigonometric functions?
P.S. I apologize to those who answered the earlier,  unedited version of this question which had included an arc tan term.  Since I'm new I made a mistake in the formatting there.  It's fixed now.  This is the one I originally meant.  Any answers please? 
 A: If you know that
$$
\begin{align}
&\lim_{u \to 0} \frac{\sin u}{u}=1,\\
&\lim_{u \to 0} \frac{\arctan u}{u}=1,
\end{align}
$$ then you may write
$$
\lim_{x \to 0} \frac{\arctan2x}{ \sin 3x}=\lim_{x \to 0} \frac{\arctan(2x)}{2x}\times\lim_{x \to 0}\frac{3x}{\sin(3x)}\times \lim_{x \to 0}\frac{2x}{3x}= \color{red}{\frac23}.
$$
A: As the form is $\frac{0}{0}$, we can apply LH rule
$$\implies \lim_{x \to 0} \frac{\arctan2x}{ \sin 3x}$$
$$=\lim_{x \to 0}\frac{2}{(3\cos(3x))(1+4x^2)}$$
$$=\color{blue} {\frac{2}{3}}$$
A: Hint for small $x$ we can assume $$\arctan(x)\approx \sin(x)\approx x$$ which can be proved by Taylor expansion.
A: Using the following equivalent infinitesimals
$$\arctan x\sim x$$
$$\sin x\sim x$$
We have
$$\lim\limits_{x \to 0} \frac{\arctan 2x}{\sin 3x}=\lim\limits_{x \to 0} \frac{2x}{3x}=\frac23$$
A: To show that $\arctan x \sim x$ as $x \to 0$ we can take for granted the limit 
$$\lim_{x \to 0} \frac{\tan x}{x} = 1$$ Now let $x = \arctan t$
Then $$\lim_{t \to 0} \frac{t}{\arctan t} = 1$$
The result $\frac 23$ follows, as shown in other answers. 
