# 微积分学/极限/一些极限性质的证明

 ← 极限/极限与连续 微积分学 极限/习题 → 极限/一些极限性质的证明
${\displaystyle a}$${\displaystyle b}$均为常数，则${\displaystyle \lim _{x\to a}b=b}$

${\displaystyle a}$为常数，则${\displaystyle \lim _{x\to a}x=a}$

${\displaystyle \lim _{x\to c}f(x)=L}$${\displaystyle \lim _{x\to c}g(x)=M}$，则${\displaystyle \lim _{x\to c}{\Big [}f(x)+g(x){\Big ]}=\lim _{x\to c}f(x)+\lim _{x\to c}g(x)=L+M}$

${\displaystyle \delta _{fg}(\varepsilon )}$${\displaystyle \delta _{f}({\tfrac {\varepsilon }{2}})}$${\displaystyle \delta _{g}({\tfrac {\varepsilon }{2}})}$二者中较小者，则${\displaystyle \lim _{x\to c}{\Big [}f(x)+g(x){\Big ]}}$的定义中的${\displaystyle \delta }$即为${\displaystyle \delta _{fg}(\varepsilon )}$，求出值为${\displaystyle L+M}$，证毕。

${\displaystyle \lim _{x\to c}f(x)=L}$${\displaystyle \lim _{x\to c}g(x)=M}$，则${\displaystyle \lim _{x\to c}{\Big [}f(x)-g(x){\Big ]}=\lim _{x\to c}f(x)-\lim _{x\to c}g(x)=L-M}$

${\displaystyle h(x)=-g(x)}$，则${\displaystyle \lim _{x\to c}h(x)=-M}$，故${\displaystyle \lim _{x\to c}{\Big [}f(x)-g(x){\Big ]}=\lim _{x\to c}{\Big [}f(x)+h(x){\Big ]}=L-M}$，证毕。

${\displaystyle \lim _{x\to c}f(x)=L}$${\displaystyle \lim _{x\to c}g(x)=M}$，则${\displaystyle \lim _{x\to c}{\Big [}f(x)g(x){\Big ]}=\lim _{x\to c}f(x)\lim _{x\to c}g(x)=LM}$

${\displaystyle \varepsilon }$为任意正数，则必有${\displaystyle \delta _{1},\delta _{2},\delta _{3}}$，使得

1. ${\displaystyle 0<|x-c|<\delta _{1}}$时，${\displaystyle {\Big |}f(x)-L{\Big |}<{\frac {\varepsilon }{2(1+|M|)}}}$
2. ${\displaystyle 0<|x-c|<\delta _{2}}$时，${\displaystyle {\Big |}g(x)-M{\Big |}<{\frac {\varepsilon }{2(1+|L|)}}}$
3. ${\displaystyle 0<|x-c|<\delta _{3}}$时，${\displaystyle {\Big |}g(x)-M{\Big |}<1}$

${\displaystyle \lim _{x\to c}f(x)=L}$${\displaystyle \lim _{x\to c}g(x)=M\neq 0}$，则${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}={\frac {\lim \limits _{x\to c}f(x)}{\lim \limits _{x\to c}g(x)}}={\frac {L}{M}}}$

${\displaystyle \lim _{x\to c}{\frac {1}{g(x)}}={\frac {1}{M}}}$，则可令${\displaystyle h(x)={\frac {1}{g(x)}}}$，运用积规则可证商规则。下证${\displaystyle \lim _{x\to c}{\frac {1}{g(x)}}={\frac {1}{M}}}$

${\displaystyle \varepsilon }$为任意正数，则必有${\displaystyle \delta _{1},\delta _{2}}$，使得

1. ${\displaystyle 0<|x-c|<\delta _{1}}$时，${\displaystyle {\Big |}g(x)-M{\Big |}<\varepsilon |M|{\Big |}|M|-1{\Big |}}$
2. ${\displaystyle 0<|x-c|<\delta _{2}}$时，${\displaystyle {\Big |}g(x)-M{\Big |}<1}$

${\displaystyle 0<|x-c|<\min\{\delta _{1},\delta _{2}\}}$时，有

{\displaystyle {\begin{aligned}\left|{\frac {1}{g(x)}}-{\frac {1}{M}}\right|&=\left|{\frac {M-g(x)}{Mg(x)}}\right|\\&=\left|{\frac {g(x)-M}{Mg(x)}}\right|\\&=\left|{\frac {1}{g(x)}}\right|\left|{\frac {g(x)-M}{M}}\right|\\&<\left|{\frac {1}{|M|-1}}\right|\left|{\frac {g(x)-M}{M}}\right|\\&<\left|{\frac {1}{|M|-1}}\right|\left|{\frac {\varepsilon |M|{\Big |}|M|-1{\Big |}}{M}}\right|\\&=\varepsilon \end{aligned}}}，证毕。

${\displaystyle \lim _{x\to c}g(x)=\lim _{x\to c}h(x)=L}$，且在${\displaystyle c}$的某个去心邻域内有${\displaystyle g(x)\leq f(x)\leq h(x)}$，则${\displaystyle \lim _{x\to c}f(x)=L}$

 ← 极限/极限与连续 微积分学 极限/习题 → 极限/一些极限性质的证明