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

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.