Limit $lim_{uto0} frac{3u}{tan 2u}$

I’m currently stuck trying to evaluate this limit,
$$
\lim_{u\to0} \frac{3u}{\tan(2u)},
$$

without using L’Hôpital’s rule. I’ve tried both substituting for $\tan(2u)=\dfrac{2\tan u}{1-(\tan u)^2}$, and $\tan 2u=\dfrac{\sin 2u}{\cos 2u}$ without success. Am I on the right path to think trig sub?