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## 10.5 洛必达法则
- **命题 10.5.1(洛必达法则 1**:设 $X\subseteqq \mathbb R$ 和 $X$ 的聚点 $x_0$,函数 $f:X\to\mathbb R$ 和 $g:X\to\mathbb R$,满足 $f(x_0)=g(x_0)=0$$f$ 和 $g$ 都在 $x_0$ 处可微且 $g'(x_0)\neq 0$。那么:
$$
\lim_{x\to x_0}\frac{f(x)}{g(x)}=\frac{f'(x_0)}{g'(x_0)}
$$
**证明**:正确的顺序是从后往前推:
$$
\begin{aligned}\lim_{x\to x_0}\frac{f(x)}{g(x)}&=\lim_{x\to x_0}\frac{\frac{f(x)-f(x_0)}{x-x_0}}{\frac{g(x)-g(x_0)}{x-x_0}}\\&=\frac{\lim\limits_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}}{\lim\limits_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}}\\&=\frac{f'(x_0)}{g'(x_0)}\end{aligned}
$$