How to find the following limit? $ \lim_{x \to 0} \frac{x}{\sin(2x)\cos(3x)}$ I am stuck with this limit problem

$$\lim_{x \to 0} \frac{x}{\sin(2x)\cos(3x)} $$

Any hints are appreciated. Also, I can't use L'Hopital's.
We know that: $\cos(3x)\to1$ as $x\to0$, so the only difficulty you're left with is to prove that: $$\lim_{x\to0}\dfrac{x}{\sin(2x)}=\dfrac12$$ and as a hint you can use: $$\lim_{x\to0}\dfrac{{x}}{\sin(2x)}=\lim_{x\to0}\dfrac12\dfrac{2{x}}{\sin(2x)}\quad\color{grey}{\sf and}\quad\lim_{x\to0}\dfrac{x}{\sin(x)}=1.$$
After l'Hôpital you get $$\lim_{x\to0}\frac{1}{2\cos(2x)\cos(3x)-3\sin(2x)\sin(3x)}=\frac{1}{2-0}=\frac{1}{2}$$.