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.

You should put in
$$
\frac{x^2}{\arctan x^2}
$$
that has limit $1$:
$$
\lim_{x\to0}\frac{x\sin3x}{\arctan x^2}=
\lim_{x\to0}3\frac{\sin3x}{3x}\frac{x^2}{\arctan x^2}=\dots
$$
If you don't know the limit above, just substitute $t=\arctan x^2$, so $x^2=\tan t$ and the limit is
$$
\lim_{x\to0}\frac{x^2}{\arctan x^2}=\lim_{t\to0}\frac{\tan t}{t}
$$
that you should be able to manage.
