# 夾擠定理

## 定义

${\displaystyle I}$為包含某點${\displaystyle a}$區間${\displaystyle f,g,h}$為定義在${\displaystyle I}$上的函數。若對於所有屬於${\displaystyle I}$而不等於${\displaystyle a}$${\displaystyle x}$，有：

• ${\displaystyle g(x)\leq f(x)\leq h(x)}$
• ${\displaystyle \lim _{x\to a}g(x)=\lim _{x\to a}h(x)=L}$

${\displaystyle \lim _{x\to a}f(x)=L}$

${\displaystyle g(x)}$${\displaystyle h(x)}$分別稱為${\displaystyle f(x)}$下界上界

${\displaystyle a}$若在${\displaystyle I}$的端點，上面的極限是左極限或右極限。 對於${\displaystyle x\to \infty }$，這個定理還是可用的。

## 例子

### 有关正弦函数的极限

${\displaystyle -1\leq \sin {\frac {1}{x}}\leq 1}$
${\displaystyle -x^{2}\leq x^{2}\sin {\frac {1}{x}}\leq x^{2}}$
${\displaystyle g(x)\leq f(x)\leq h(x)}$

${\displaystyle \lim _{x\to 0}\ g(x)=\lim _{x\to 0}\ h(x)=0}$，根據夾擠定理

${\displaystyle \lim _{x\to 0}f(x)=0}$

${\displaystyle AC
${\displaystyle \sin x
${\displaystyle {\frac {\sin x}{x}}<1}$
${\displaystyle arcAD
${\displaystyle x<\tan x}$
${\displaystyle \cos x<{\frac {\sin x}{x}}}$

${\displaystyle \lim _{x\to 0^{+}}{\frac {\sin x}{x}}=1}$

### 高斯函數

${\displaystyle (2I)^{2}=\left[2\int _{0}^{a}e^{-x^{2}}dx\right]^{2}=\left[\int _{-a}^{a}e^{-x^{2}}dx\right]^{2}=\int _{-a}^{a}\int _{-a}^{a}e^{-(x^{2}+y^{2})}dxdy}$

${\displaystyle \int _{0}^{2\pi }\int _{0}^{a}re^{-r^{2}}\,dr\,d\theta \leq (2I)^{2}\leq \int _{0}^{2\pi }\int _{0}^{a{\sqrt {2}}}re^{-r^{2}}\,dr\,d\theta }$
${\displaystyle \pi (1-e^{-a^{2}})\leq (2I)^{2}\leq \pi (1-e^{-2a^{2}})}$
${\displaystyle \lim _{a\to \infty }\pi \left(1-e^{-a^{2}}\right)=\lim _{a\to \infty }\pi \left(1-e^{-2a^{2}}\right)=\pi \vdash [2I(\infty )]^{2}=\pi }$
${\displaystyle \lim _{a\to \infty }(2I)^{2}=\pi }$
${\displaystyle I(\infty )={\frac {\sqrt {\pi }}{2}}}$

## 證明

### 極限為0的情況

${\displaystyle \forall x\in \mathbb {R} }$${\displaystyle g(x)=0}$，而且${\displaystyle \lim _{x\to a}h(x)=0}$

${\displaystyle \varepsilon >0}$，根據函數的極限的定義，存在${\displaystyle \delta >0}$使得：若${\displaystyle 0<|x-a|<\delta }$，則${\displaystyle |h(x)|<\varepsilon }$

${\displaystyle 0<|x-a|<\delta }$，則${\displaystyle |f(x)|\leq |h(x)|<\varepsilon }$。於是，${\displaystyle \lim _{x\to a}f(x)=0}$

### 一般情況

${\displaystyle g(x)\leq f(x)\leq h(x)}$

${\displaystyle 0\leq f(x)-g(x)\leq h(x)-g(x)}$

${\displaystyle x\to a}$

${\displaystyle h(x)-g(x)\to L-L=0}$

${\displaystyle f(x)=[f(x)-g(x)]+g(x)\to 0+L=L}$

## 参考

1. ^ Stewart, James. Chapter 15.2 Limits and Continuity. Multivariable Calculus (6th ed.). 2008: 909–910. ISBN 978-0495011637.