第14章§6 - 11

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