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.
 A: Starting as you did, we want to find the limit of 
$$\frac{3x^2}{\arctan(x^2)}\cdot\frac{\sin(3x)}{3x}.$$
Only the first term gives any trouble. Let $x$ be not too large, and let $x^2=\tan w$, You want to find 
$$\lim_{w\to 0} \frac{3\tan w}{w},$$
which is not difficult. Replace $\tan w$ by $\frac{\sin w}{cos w}$.
A: 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.
A: Since $u \sim \tan u$, then $y \sim \arctan y$, so $x^2 \sim \arctan (x^2)$.
This means that 
$$\lim_{x \to 0} \frac{x \sin 3x}{\arctan x^2} = \lim_{x \to 0} \frac{x3x}{x^2} = 3$$
A: Using the Asymptotic expansion:
$\sin(x)\approx_0 x$
$\arctan(x^2) \approx_0 x^2$
So
$$\lim _{x\rightarrow \:0}\:\frac{x\sin \:3x}{\arctan \:x^2}=\lim \:_{x\rightarrow 0}\:\frac{3x^2}{x^2}=\color{red}{3}$$
