Find $\lim_{x\to0}{\frac{\arctan (2x)}{3x}}$ without using '0/0=1' Find limit as 
$$
\lim_{x \to 0} \frac{\arctan (2x)}{3x}
$$
without using $\frac{0}{0} = 1$.
I wanted to use 
$$
\frac{2}{3} \cdot \frac{0}{0} = \frac{2}{3} \cdot 1,
$$ 
but our teacher considers $\frac{0}{0}$ a "dangerous case" and we are not allowed to use this method.
We have not studied integrals yet, so I cannot use integral formulas. We have only studied a bit of derivative taking...
I tried many substitutions, but failed :( Any suggestion?

Let $u=\arctan(2x)$ Then as $ x \rightarrow 0$ we have $u \rightarrow 0$ So $\lim_{x \rightarrow 0} \frac{\arctan(2x)}{3x}=\lim_{u \rightarrow 0} \frac{u}{3 \frac{tan(u)}{2}}=\frac{2}{3}\lim_{u \rightarrow 0 } \cos(x) \cdot \frac{u}{\sin(u)}=\frac{2}{3} \lim_{u \rightarrow 0} \frac{u}{\sin(u)} \cdot \lim_{u \rightarrow 0} \cos(u) $
