Ta có $A=\dfrac{(1-\cos 3x)\cos 2x}{2\sin^2 x \cos x}=\dfrac{2\sin^2 \dfrac{3x}{2} \cos 2x}{2\sin^2 x \cos x}$
$=\dfrac{\dfrac{9}{4}.\dfrac{\sin^2 \dfrac{3x}{2}}{\dfrac{9x^2}{4}}. \cos 2x}{\dfrac{\sin^2 x}{x^2}. \cos x}$
Vậy $\lim_{x\to 0}A= \lim_{x\to 0} \dfrac{9 \cos 2x}{4\cos x}=\dfrac{9}{4}$