Limit with arctan: $\lim_{x\rightarrow 0} \frac{x\sin 3x}{\arctan x^2}$ $$\lim_{x\rightarrow 0} \frac{x\sin 3x}{\arctan x^2}$$
NB! I haven't learnt about L'Hôpital's rule yet, so I'm still solving limits using common limits.
What I've done so far
$$\lim_{x\rightarrow0}\left[\frac{x}{\arctan x^2}\cdot 3x\cdot \frac{\sin 3x}{3x}\right] = 0\cdot1\cdot\lim_{x\rightarrow0}\left[\frac{x}{\arctan x^2}\right] = 0??$$
Obviously I'm wrong, but I thought I'd show what I tried.

Since $u \sim \tan u$, then $y \sim \arctan y$, so $x^2 \sim \arctan (x^2)$.
This means that 
$$\lim_{x \to 0} \frac{x \sin 3x}{\arctan x^2} = \lim_{x \to 0} \frac{x3x}{x^2} = 3$$
