一、四则运算#

定理：若 $\displaystyle \lim_{x\to x_0}f(x)=A$，$\displaystyle \lim_{x\to x_0}g(x)=B$，则有

• $\displaystyle \lim_{x\to x_0}[f(x)\pm g(x)]=\lim_{x\to x_0}f(x)\pm \lim_{x\to x_0}g(x)=A\pm B$
• $\displaystyle \lim_{x\to x_0}[f(x)g(x)]=\lim_{x\to x_0}f(x)\cdot \lim_{x\to x_0}g(x) = A\cdot B$
• $\displaystyle \lim_{x\to x_0}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{\displaystyle \lim_{x\to x_0}f(x)}{\displaystyle \lim_{x\to x_0}g(x)}=\dfrac{A}{B}(B\ne 0)$

四则运算定理证明

1. 加法定则 $\displaystyle \lim_{x\to x_0}[f(x)+ g(x)]=\lim_{x\to x_0}f(x)+ \lim_{x\to x_0}g(x)$

   证：原命题可化为证明 $f(x)+g(x)-(A+B)$ 无穷小 $f(x)=A+\alpha(x) \quad g(x)=B+\beta(x)$ 原式 $=\alpha(x)+\beta(x)$，是无穷小量，故 $\displaystyle \lim_{x\to x_0}[f(x)+g(x)-A-B]=0$ 即 $\displaystyle \lim_{x\to x_0}[f(x)+ g(x)]=A+B$，减法定则同理

2. 乘法定则 $\displaystyle \lim_{x\to x_0}[f(x)g(x)]=\displaystyle \lim_{x\to x_0}f(x)\cdot \displaystyle \lim_{x\to x_0}g(x)$

   证：$f(x)\cdot g(x)=AB+\underbrace{B\alpha(x)+A\beta(x)+\alpha(x)\beta(x)}_{\text{三个无穷小量}}$ $\because \displaystyle \lim_{x\to x_0}[B\alpha(x)+A\beta(x)+\alpha(x)\beta(x)]=0$ $\therefore \displaystyle \lim_{x\to x_0}[f(x)g(x)]=AB=\lim_{x\to x_0}f(x)\cdot \lim_{x\to x_0}g(x)$

3. 除法定则 $\displaystyle \lim_{x\to x_0}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{\displaystyle \lim_{x\to x_0}f(x)}{\displaystyle \lim_{x\to x_0}g(x)}(\lim_{x\to x_0}g(x)\ne 0)$

   证：$\dfrac{f(x)}{g(x)}-\dfrac{A}{B}=\dfrac{A+\alpha}{B+\beta}-\dfrac{A}{B}=\dfrac{1}{B(B+\beta)}(B\alpha - A\beta)$ $\forall \epsilon \gt 0 \quad \exists \delta \quad |\beta-0|\lt \epsilon$ 取 $\epsilon=\dfrac{|B|}{2}$，则 $|\beta| \lt \dfrac{|B|}{2}$ $|B+\beta |\ge |B| -|\beta |\gt |B|-\dfrac{|B|}{2}=\dfrac{|B|}{2}=\epsilon$ $\left|\dfrac{1}{B(B+\beta)}\right|\le \dfrac{2}{|B|^2}$，即 $\dfrac{1}{B(B+\beta)}$ 为有界量 $\therefore \displaystyle \lim_{x\to x_0}[\frac{f(x)}{g(x)}-\frac{A}{B}]=0 \Rightarrow \lim_{x\to x_0}\left[\frac{f(x)}{g(x)}\right]=\frac{A}{B}$

• 有用的推论

  1. $\displaystyle \lim_{x\to x_0}f(x)=A \Rightarrow \lim_{x\to x_0}c\cdot f(x)=c\cdot \lim_{x\to x_0}f(x)=c\cdot A$ ($c$ 为常数)
  2. $\displaystyle \lim_{x\to x_0}f(x)=A \Rightarrow \lim_{x\to x_0}f(x)^n=[\lim_{x\to x_0}f(x)]^n=A^n$ ($n$ 为正整数)
  3. $\displaystyle \lim_{x\to x_0}f(x)=A \Rightarrow \lim_{x\to x_0}\root n \of {f(x)}=\root n \of {\lim_{x\to x_0}f(x)}=\root n \of A$ ($A\gt 0$，$f(x)\gt 0$，$n$ 为正整数)

  类似性质在数列中也同样适用($\{x_n\},\{y_n\},\displaystyle \lim_{n\to \infty}x_n=A,\lim_{n\to \infty}y_n=B$)

  1. $\displaystyle \lim_{n\to \infty}(x_n\pm y_n)=\lim_{n\to \infty}x_n\pm \lim_{n\to \infty}y_n=A\pm B$
  2. $\displaystyle \lim_{n\to \infty}(x_n\cdot y_n)=\lim_{n\to \infty}x_n\cdot \lim_{n\to \infty}y_n=A B$
  3. $\displaystyle \lim_{n\to \infty}\dfrac{x_n}{y_n}=\dfrac{\displaystyle \lim_{n\to \infty}x_n}{\displaystyle \lim_{n\to \infty}y_n} = \dfrac{A}{B}$，其中 $B\ne 0$，$y_n\ne 0(n=1,2,\cdots)$

例 1. $P_n(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0 \quad(a_n\ne 0, n\in N^*)$，求 $\displaystyle \lim_{x\to x_0}P_n(x)$ 解：

$$\begin{align} 原式&=\lim_{x\to x_0}a_nx^n+\lim_{x\to x_0}a_{n-1}x^{n-1}+\cdots +\lim_{x\to x_0}a_1x+\lim_{x\to x_0}a_0\nonumber \\&=a_n[\lim_{x\to x_0}x]^n+a_{n-1}[\lim_{x\to x_0}x]^{n-1}+\cdots +a_1[\lim_{x\to x_0}x]+a_0\nonumber\\ &=a_nx_0^n+a_{n-1}x_0^{n-1}+\cdots +a_1x_0+a_0\nonumber\\ &=P_n(x_0)\nonumber \end{align}$$

证明 $\displaystyle \lim_{x\to \infty}P_n(x)=\infty$

$$\begin{align} 考虑\lim_{x\to \infty}\frac{1}{P_n(x)}&=\lim_{x\to x_0}\frac{1}{a_nx^n+\lim_{x\to x_0}a_{n-1}x^{n-1}+\cdots +\lim_{x\to x_0}a_1x+\lim_{x\to x_0}a_0}\nonumber \\ &=\lim_{x\to \infty}\frac{\frac{1}{x^n}}{a_n+a_{n-1}\frac{1}{x}+\cdots +a_1\frac{1}{x^{n-1}}+a_0\frac{1}{x^n}}\nonumber\\ &=\frac{0}{a_n}=0\nonumber \end{align}$$

$\therefore \displaystyle \lim_{x\to \infty}P_n(x)=\infty$

例2. (1) $\displaystyle \lim_{x\to 0}\dfrac{x-3}{x^2-5x+6}=\dfrac{\displaystyle \lim_{x\to 0}(x-3)}{\displaystyle \lim_{x\to 0}(x^2-5x+6)}=\dfrac{-3}{6}=-\dfrac{1}{2}$

(2) $\displaystyle \lim_{x\to 3}\dfrac{x-3}{x^2-5x+6}=\lim_{x\to 3}\dfrac{1}{x-2}=1$

(3) $\displaystyle \lim_{x\to \infty}\dfrac{x-3}{x^2-5x+6}=\lim_{x\to \infty}\dfrac{1}{x-2}=0$

(4) $\displaystyle \lim_{x\to \infty}\dfrac{x+3}{x^2-5x+6}=\lim_{x\to \infty}\dfrac{\frac{1}{x}+\frac{3}{x^2}}{1-\frac{5}{x}+\frac{6}{x^2}}=\dfrac{0}{1}=0$

(5)

$$\lim_{x\to \infty}\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots +a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots +b_1x+b_0}=\begin{cases}0 \quad m\lt n\\ \frac{a_m}{b_n} \quad m=n\\ \infty \quad m\gt n\end{cases}$$

(6) [分子有理化]$\displaystyle \lim_{x\to +\infty}x(\sqrt{x^2+1}-\sqrt{x^2-1})=\lim_{x\to +\infty}\frac{2x}{\sqrt{x^2+1}+\sqrt{x^2-1}}=\lim_{x\to +\infty}\frac{2}{\sqrt{\frac{x^2+1}{x^2}}+\sqrt{\frac{x^2-1}{x^2}}}=\frac{2}{1+1}=1$

二、复合运算#

定理：