求极限lim_(xto0)((1)/(x)-(1)/(sin x))

将原式通分得： \[ \lim_{x \to 0} \frac{\sin x - x}{x \sin x} \] 应用洛必达法则两次： \[ \lim_{x \to 0} \frac{\cos x - 1}{\sin x + x \cos x} \xrightarrow{\text{洛必达}} \lim_{x \to 0} \frac{-\sin x}{2 \cos x - x \sin x} = \frac{0}{2} = 0 \] 或使用泰勒展开： \[ \sin x \approx x - \frac{x^3}{6} \Rightarrow \sin x - x \approx -\frac{x^3}{6}, \quad x \sin x \approx x^2 \] \[ \lim_{x \to 0} \frac{-\frac{x^3}{6}}{x^2} = \lim_{x \to 0} -\frac{x}{6} = 0 \] 答案：$\boxed{0}$