第14章§6 - 12

$$\begin{array}{rl}\underset{x\to 0}{lim}\frac{\sin x-x}{x^2}& =\underset{x\to 0}{lim}\frac{1}{x^2}\{(x-\frac{x^3}{3!}+\cdots )-x\}\\ & =\underset{x\to 0}{lim}(-\frac{x}{3!}+\cdots )\\ & =0\end{array}$$