$\lim\limits_{x\to 0}\dfrac{\tan2x}{3x}$
$=\lim\limits_{x\to 0}\dfrac{\sin2x}{3x.\cos3x}$
$=\lim\limits_{x\to 0}\Big[ \dfrac{\sin2x}{2x}.\dfrac{2x}{3x\cos2x}\Big]$
$=\lim\limits_{x\to 0}\Big[ \dfrac{\sin2x}{2x}.\dfrac{2}{3cos2x}\Big]$
$=1.\dfrac{2}{3.\cos0}$
$=\dfrac{2}{3}$