(3)lim_(xto0)(tan x-x)/(x-sin x);

使用泰勒展开或洛必达法则求解： **泰勒展开：** 近似 $\tan x \approx x + \frac{x^3}{3}$，$\sin x \approx x - \frac{x^3}{6}$，则 \[ \lim_{x \to 0} \frac{\tan x - x}{x - \sin x} = \lim_{x \to 0} \frac{\frac{x^3}{3}}{\frac{x^3}{6}} = 2. \] **洛必达法则：** 连续应用法则，最终化简为 \[ \lim_{x \to 0} \frac{2 \sec^2 x \tan x}{\sin x} = \lim_{x \to 0} \frac{2}{\cos^3 x} = 2. \] **答案：** \[ \boxed{2} \]