第14章§6 - 9

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