$\displaystyle\lim_{x \to 0} \dfrac{-2\sin 4x \sin x}{x\sin 2x} \\ =\displaystyle\lim_{x \to 0} \dfrac{\sin 4x}{4x} .\dfrac{\sin x}{x}.\dfrac{2x}{\sin 2x}.(-4)\\ =1.1.1.(-4) =-4$
Chú ý là: $\displaystyle\lim_{x \to 0} \dfrac{\sin a}{a}=\displaystyle\lim_{x \to 0} \dfrac{a}{\sin a}=1$