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.

Using the Asymptotic expansion:
$\sin(x)\approx_0 x$
$\arctan(x^2) \approx_0 x^2$
So
$$\lim _{x\rightarrow \:0}\:\frac{x\sin \:3x}{\arctan \:x^2}=\lim \:_{x\rightarrow 0}\:\frac{3x^2}{x^2}=\color{red}{3}$$
