# 浅谈等价无穷小量代换原理

## 等价无穷小替换原理

：$$\lim_{x\rightarrow 0}{\frac{\ln{\left(1+3x\right)}}{\sin{2x}}}=\lim_{x\rightarrow 0}{\frac{3x}{2x}}=\frac{3}{2}$$

$1^{\circ}$ 充分性：$$\alpha\sim\alpha_{1},\beta\sim\beta_{1}\Rightarrow \lim{\frac{\alpha}{\alpha_{1}}}=\lim{\frac{\beta}{\beta_{1}}}=1$$ 又 $$\lim{\frac{\alpha}{\beta}}=k\neq 1,\lim{\frac{\alpha_{1}}{\beta_{1}}}=k\neq 1$$ 则 $$\lim{\frac{\alpha-\beta}{\alpha_{1}-\beta_{1}}}=\lim{\frac{\frac{\alpha}{\beta_{1}}-\frac{\beta}{\beta_{1}}}{\frac{\alpha_{1}}{\beta_{1}}-1}}=\frac{k-1}{k-1}=1$$ 即 $$\alpha -\beta\sim \alpha_{1} - \beta_{1}$$

$2^{\circ}$ 必要性：$$\alpha\sim\beta,\alpha_{1}\sim\beta_{1}\Rightarrow \lim{\frac{\alpha-\beta}{\alpha_{1}-\beta_{1}}}=1$$ 即 $$\lim{\left(\frac{\alpha-\beta}{\alpha_{1}-\beta_{1}}-1 \right )}=0$$ 通分得 $$\lim{\frac{\alpha-\alpha_{1}}{\alpha_{1}-\beta_{1}}}-\lim{\frac{\beta-\beta_{1}}{\alpha_{1}-\beta_{1}}}=0$$ 所以 $$\lim{\frac{\frac{\alpha}{\alpha_{1}}-1}{1-\frac{\beta}{\alpha_{1}}}}-\lim{\frac{1-\frac{\beta}{\beta_{1}}}{\frac{\alpha_{1}}{\beta_{1}}-1}}=0$$ 又 $$\lim{\frac{\alpha}{\alpha_{1}}}=1,\lim{\frac{\beta}{\beta_{1}}}=1$$ 所以 $$\lim{\frac{0}{1-\frac{\beta}{\alpha_{1}}}}-\lim{\frac{0}{\frac{\alpha_{1}}{\beta_{1}}-1}}=0$$ 所以 $$\lim{\frac{\beta_{1}}{\alpha_{1}}}=k\neq 1\Rightarrow \lim{\frac{\alpha_{1}}{\beta_{1}}}=k\neq 1$$ 又 $$\lim{\frac{\alpha}{\beta}}=\lim{\frac{\alpha_{1}}{\beta_{1}}}$$ 所以 $$\lim{\frac{\alpha}{\beta}}=k\neq 1,\lim{\frac{\alpha_{1}}{\beta_{1}}}=k\neq 1$$ 由 $1^{\circ},2^{\circ}$ 得，原命题成立。证毕。

：$$1-\cos{x}\sim\frac{x^{2}}{2},-2\sin{x}\sim -2x, 2\arcsin{x}\sim 2x,\sin{x}\sim x\left ( x\rightarrow 0 \right )$$ 所以 $$\lim_{x\rightarrow 0}{\frac{1-\cos{x}}{-2\sin{x}}}=0\neq 1,\lim_{x\rightarrow 0}{\frac{2\arcsin{x}}{\sin{x}}}=2\neq 1$$ 由定理 2 得 $$\lim_{x\rightarrow 0}{\frac{1-\cos{x}+2\sin{x}}{\arcsin{2x}-\sin{x}}}=\lim_{x\rightarrow}{\frac{\frac{x^{2}}{2}+2x}{x}}=2$$

：$$\arctan{2x}\sim 2x,\arcsin{5x}\sim 5x,\sin{3x}\sim 3x\left ( x\rightarrow 0 \right )$$ 又 $$\lim{\frac{\arctan{2x}}{-\arcsin{5x}}}=-\frac{2}{5}\neq 1$$ 由定理 2 得 $$\lim_{x\rightarrow 0}{\frac{\arctan{2x}+\arcsin{5x}}{\sin{3x}}}=\frac{2x+5x}{3x}=\frac{7}{3}$$