For small \(x\): \(\tan x = x + \frac{x^3}{3} + \dots\), so \(\tan x - x \approx \frac{x^3}{3}\), and \(x^2\tan x \approx x^3\).

\(\lim_{x \to 0} \frac{\tan x - x}{x^2\tan x} = \lim_{x \to 0} \frac{x^3/3}{x^3} = \frac{1}{3}\)

Answer: (D) \(\frac{1}{3}\)