2. The limit is $- \frac{1}{3}$
$\lim_{x \to 0} \frac{ x - \tan x}{x^3} = \lim_{x \to 0} \frac{ 1 - \sec^{2} x}{3x^{2}} = \lim_{x \to 0} \frac{ -2 \tan x \sec^{2} x}{6x}=$ $\lim_{x \to 0} \frac{-2 \sec^{4} x - 4 \tan^{2} x \sec^{2} x}{6} = -\frac{2}{6} = -\frac {1}{3}$