第14章§6 - 5

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