\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\sin 3x}}\) bằng:

* Hướng dẫn giải

Ta có: \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\sin 3x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 2x}}{{2x}}.2x}}{{\dfrac{{\sin 3x}}{{3x}}.3x}} = \dfrac{2}{3}\).

Chọn D.