How to find the limit of $\frac{\tan(2x)}{\sin(3x)}$ as $x$ approaches $0$ analytically? I'm taking a calculus 1 course, and I'm running into issues when trig gets added into the mix. I keep trying to simplify $\frac{\tan(2x)}{\sin(3x)}$ so that I can either plug in $0$ or find one of the special limits, but I always end up at a dead end.
I've looked for existing help, but everything seems to use L'Hopital's rule. We haven't gotten to derivatives yet, so that wouldn't be an acceptable answer to turn in.
 A: The computation can be done simply as follows
\begin{eqnarray*}
\frac{\tan 2x}{\sin 3x} &=&\frac{\sin 2x}{\sin 3x\cos 2x} \\
&=&\frac{2\sin x\cos x}{\left[ \sin x\color{red}{(2\cos 2x+1)}\right] \cos 2x} \\
&=&\frac{2\cos x}{\color{red}{(2\cos 2x+1)}\cos 2x}\rightarrow \frac{2\cdot 1}{(2\cdot
1+1)\cdot 1}=\dfrac{2}{3}.
\end{eqnarray*}
No L'Hospitale's rule, no Taylor series, no limit of  $\dfrac{\sin x}{x}$  too!
(Thanks to P.S. :) )
$\bf EDIT$ The  misprint is now corrected. Thanks to Blue who pointed out to $\color{red}{it}$.
A: We just need to recall that
$$ \lim_{x\to 0}\frac{\sin(ax)}{x} = a, $$
hence:
$$ \lim_{x\to 0}\frac{\tan(2x)}{\sin(3x)} = \lim_{x\to 0}\frac{1}{\cos(2x)}\cdot\frac{x}{\sin(3x)}\cdot\frac{\sin(2x)}{x} = \frac{2}{3}.$$
A: Hint:
$$\frac{\tan2x}{\sin3x}=\frac23\frac{\sin2x}{2x}\frac{1}{\cos2x}\Big/
  \frac{\sin3x}{3x}\ .$$
You should see some known "special limits" here.
A: when trying to find the limit of $\dfrac{\tan 2x}{\sin 3x} = \dfrac{\sin 2x}{\cos 2x\sin 3x}$  one can see that the $\cos 3x $ in the numerator is harmless can be evaluated to $1.$ finding the original limit is reduced to finding the limit of $\dfrac{\sin(2x)}{\sin(3x)}$ here we make further simplification that $\sin(small) = small + \cdots$ so that when the argument to $\sin$ is small, you can erase the word $\sin$ leaving the small argument. in this case finding the limit is reduced to finding $\lim_{x \to 0}\dfrac{2x}{3x}.$ that is clearly $\dfrac{2}{3}$
A: With equivalents: $\, \sin ax\sim_0 ax,\enspace \cos ax\sim_01$, hence $\tan ax \sim_0 \dfrac{ax}1=ax$. Hence:
$$\frac{\sin 2x}{\tan 3x}\sim_0\frac{2x}{3x}=\frac23,$$ 
which proves the limit is $\dfrac23$.
