高数 01.06极限存在准则

第六节极限存在准则 \color{blue}{第六节极限存在准则}

一、函数极限与数列极限的关系及夹逼准则
二、两个重要极限

一、函数极限与数列极限的关系及夹逼准则 \color{blue}{一、函数极限与数列极限的关系及夹逼准则}

1.函数极限与数列极限的关系
定理1.lim x→x 0 x→∞ f(x)=A⟺∀{x n }:x n ≠x 0 ,f(x n )有定义, 定理1. \lim_{\begin{align}{x \rightarrow x_0}\\ {\color{green}{x \rightarrow \infty}}\end{align}}{f(x)} = A \Longleftrightarrow \forall \{x_n\}:x_n \neq x_0,f(x_n)有定义,
x n →x 0 x n →∞ (n→∞),有lim n→∞ f(x n )=A \qquad {\begin{align}{x_n \rightarrow x_0}\\ {\color{green}{x_n \rightarrow \infty}}\end{align}}{(n \rightarrow \infty)},有\lim_{n \rightarrow \infty}f(x_n) = A
为确定起见，仅讨论x→x 0 的情形. 为确定起见，仅讨论x \rightarrow x_0的情形.
说明：⟹表示必要性⟸表示充分性说明：\\ \Longrightarrow 表示必要性\\ \Longleftarrow 表示充分性\\
证：“⟹”设lim x→x 0 f(x)=A,即∀ε>0,∃δ>0,当0<|x−x 0 |<δ时，有|f(x)−A|<ε. 证：“\Longrightarrow”设\lim_{x \rightarrow x_0}{f(x)} = A,即\forall \varepsilon > 0, \exists \delta > 0, \\ 当0
∀{x n }:x n ≠x 0 ,f(x n )有定义，且x 0 →x 0 (n→∞), \forall \{x_n \}:x_n \neq x_0,f(x_n)有定义，且x_0 \rightarrow x_0(n \rightarrow \infty),
对上述δ,∃N,当n>N时，有0<|x n −x 0 |<δ,于是当n>N时,|f(x n )−A|<ε. 对上述\delta, \exists N,当 n > N时，有0 N时,|f(x_n) - A|
故lim n→∞ f(x n )=A 故 \qquad \lim_{n \rightarrow \infty}f(x_n) = A
“⟸”可用反证法证明(略) “\Longleftarrow”可用反证法证明(略)

说明:此定理常用于判断函数极限不存在.法1找一个数列x n :x n ≠x 0 ，且x n →x 0 (n→∞),使lim n→∞ f(x n )不存在.法2.找两个趋于x 0 的不同数列x n 及x ‘ n ,使lim n→∞ f(x)≠lim n→∞ f(x ‘ n ) {\color {blue} {说明:此定理常用于判断函数极限不存在.\\ 法1 找一个数列{x_n}:x_n \neq x_0，且x_n \rightarrow x_0(n \rightarrow \infty), 使\lim_{n \rightarrow \infty}f(x_n)不存在.\\ 法2.找两个趋于x_0的不同数列 { x_n } 及 { x_n^{‘} },使\lim_{n \rightarrow \infty}f(x) \neq \lim_{n \rightarrow \infty}f(x_n^{‘})}}

例1.证明lim x→0 sin1x 不存在. 例1.证明 \lim_{x \rightarrow 0}{\sin{\dfrac{1}{x}}} 不存在.
证：取两个趋于0的数列
x n =12nπ 及x  ′  n =12nπ+π2  (n=1,2,⋯) x_n = \dfrac{1}{2n \pi }及 x_n^{'} = \dfrac{1}{2n\pi + \frac{\pi}{2}} (n = 1, 2, \cdots)
有lim n→∞ sin1x n  =lim n→∞ sin2nπ=0 有 \lim_{n \rightarrow \infty}{\sin{\dfrac{1}{x_n}}} = \lim_{n \rightarrow \infty}{\sin{2n \pi}} = 0
lim n→∞ sin1x  ′  n  =lim n→∞ sin2nπ+π2 =1 \lim_{n \rightarrow \infty}{\sin{\dfrac{1}{x_n^{'}}}} = \lim_{n \rightarrow \infty}{\sin{2n \pi + \frac{\pi}{2}}} = 1
由定理1知lim x→0 sin1x 不存在. 由定理1知 \lim_{x \rightarrow 0}{\sin{\dfrac{1}{x}}}不存在.

2.函数极限存在的夹逼准则
定理2.当x∈U ° (x 0 ,δ)(|x|>X>0) 时，g(x)≤f(x)≤h(x),且定理2.当{\begin{align}{x \in \mathring{U}(x_0, \delta)}\\ {\color{green}{(|x| > X > 0)}}\end{align}}时，g(x) \leq f(x) \leq h(x),且 \\\\
lim x→x 0 x→∞ g(x)= {\lim_{\begin{align} {x \rightarrow x_0} \\ {\color{green}{x \rightarrow \infty}}\end{align} } {g(x)}} = lim x→x 0 x→∞ h(x)=A, {\lim_{ \begin{align} {x \rightarrow x_0} \\ {\color{green} {x \rightarrow \infty} } \end{align} } {h(x)}= A}, \\\\
⟹lim x→x 0 x→∞ f(x)=A {\Longrightarrow \lim_{\begin{align}{x \rightarrow x_0}\\{\color{green}{x \rightarrow \infty}}\end{align}}{f(x)} = A}
(利用定理1及数列的加倍准则可证)

二、两个重要的极限 \color{blue}{二、两个重要的极限}

1.lim x→0 sinxx =1 1.\lim_{x \rightarrow 0}{\dfrac{\sin{x}}{x}} = 1
证：当x∈(0,π2 )时，ΔAOB的面积<圆扇形AOB的面积<ΔAOD的面积证：当x \in (0, \dfrac{\pi}{2})时， \Delta AOB的面积
即12 sinx<12 x<12 tanx 即 \dfrac{1}{2}\sin{x}
故有1<xsinx <1cosx (0<x<π2 ) 故有 1
显然有cosx<sinxx <1(0<|x|<π2 ) 显然有 \cos{x}
∵lim x→0 cosx=1, 注 ∴lim x→0 sinxx =1 \because \lim_{x \rightarrow 0}{\cos{x}} = 1, ^{\color{green} 注} \quad \therefore \lim_{x \rightarrow 0}{\dfrac{\sin{x}}{x}} = 1

例2.求lim x→0 tanxx . 例2.求\lim_{x \rightarrow 0}{\dfrac{\tan{x}}{x}}.
解：lim x→0 tanxx =lim x→0 (sinxx 1cosx )=lim x→0 sinxx ⋅lim x→0 1cosx =1 解：\lim_{x \rightarrow 0}{\dfrac { \tan{x}}{x}} = \lim_{x \rightarrow 0}{\big( \dfrac{\sin{x}}{x} \dfrac{1}{\cos{x}} \big)} = \lim_{x \rightarrow 0}{\dfrac{\sin{x}}{x}} \cdot \lim_{x \rightarrow 0}{\dfrac {1}{\cos{x}}} = 1

例3.求lim x→0 arcsinxx . 例3.求\lim_{x \rightarrow 0}{\dfrac{\arcsin{x}}{x}}.
解：令t=arcsinx,x=sint原式=lim t→0 tsint =lim t→0 1sintt =1 解：令 t = arcsin{x}, x = \sin{t} \\ 原式 = \lim_{t \rightarrow 0}{\dfrac {t}{\sin{t}}} \\ \qquad = lim_{t \rightarrow 0}\dfrac {1}{\dfrac{\sin{t}}{t}} = 1

例4.求lim x→0 1−cosxx 2 . 例4.求\lim_{x \rightarrow 0}{\dfrac{1 - \cos{x}}{x^2}}.
原式=lim x→0 2sin 2 x2 x 2 =12 lim x→0 [sinx2 x2 ] 2 =12 ⋅1 2 =12 原式 = \lim_{x \rightarrow 0}{\dfrac{2 \sin^2{\dfrac{x}{2}}}{x^2}} = \dfrac{1}{2} \lim_{x \rightarrow 0}{\Bigg[ \dfrac{\sin{\dfrac{x}{2}}}{\dfrac{x}{2}} \Bigg]^2} = \dfrac{1}{2} \cdot 1^2 = \dfrac{1}{2}

例5.已知圆内接正n边形面积为A n =nR 2 sinπn cosπn 例5.已知圆内接正n边形面积为 A_n = nR^2\sin{\dfrac{\pi}{n}}\cos{\dfrac{\pi}{n}}
证明：lim n→∞ A n =πR 2 . 证明：\lim_{n \rightarrow \infty}{A_n} = \pi R^2.
证：
lim n→∞ A n =lim n→∞ πR 2 sinπn πn cosπn =πR 2 \lim_{n \rightarrow \infty}{A_n} = \lim_{n \rightarrow \infty}{\pi R^2}\dfrac{\sin{\dfrac{\pi}{n}}}{\dfrac{\pi}{n}} \cos{\dfrac{\pi}{n}} = \pi R^2
说明：计算中注意利用lim ϕ(x)→0 sinϕ(x)ϕ(x) =1 说明：计算中注意利用\lim_{\phi(x) \rightarrow 0}{\dfrac{\sin{\phi(x)}}{\phi(x)}} = 1

2.lim x→∞ (1+1x ) x =e 2.\lim_{x \rightarrow \infty}(1 + \dfrac {1}{x})^x = e
证：当x>0时，设n≤x<n+1,则证：当x > 0时，设n \leq x
(1+1n+1 ) n <(1+1x ) x <(1+1n ) n+1 (1 + \dfrac{1}{n + 1})^n
lim n→∞ (1+1n+1 ) n =lim n→∞ (1+1n+1 ) n+1 1+1n+1 =e \lim_{n \rightarrow \infty}(1 + \dfrac {1}{n + 1})^n = \lim_{n \rightarrow \infty}{\dfrac{(1 + \dfrac{1}{n + 1})^{n + 1}}{1 + \dfrac{1}{n + 1}}} = e
lim n→∞ (1+1n ) n+1 =lim n→∞ [(1+1n ) n (1+1n )]=e \lim_{n \rightarrow \infty}{(1 + \dfrac{1}{n})^{n + 1}} = \lim_{n \rightarrow \infty}[(1 + \dfrac{1}{n})^n(1 + \dfrac{1}{n})] = e
∴lim x→+∞ (1+1x ) x =e \therefore \lim_{x \rightarrow +\infty}{(1 + \dfrac{1}{x})^x} = e

当x→−∞时，令x=−(t+1),则t→+∞,从而有当x \rightarrow -\infty时，令 x = -(t + 1), 则t \rightarrow +\infty,从而有
lim x→−∞ (1+1x ) x =lim t→+∞ (1−1t+1 ) −(t+1) =lim t→+∞ (tt+1 ) −(t+1) =lim t→+∞ (1+1t ) t+1 =lim t→+∞ [(1+1t ) t ⋅(1+1t )]=e⋅1=e \lim_{x \rightarrow -\infty}{(1 + \dfrac{1}{x})^x} \\ = \lim_{t \rightarrow +\infty}{(1 - \dfrac{1}{t + 1})^{-(t + 1)}} \\ = \lim_{t \rightarrow +\infty}{(\dfrac{t}{t + 1})^{-(t + 1)}} \\ =\lim_{t \rightarrow +\infty}{(1 + \dfrac{1}{t})^{t + 1}} \\ =\lim_{t \rightarrow +\infty}{[(1 + \dfrac{1}{t})^t \cdot (1 + \dfrac{1}{t})}] \\ = e \cdot 1 \\ = e
故lim x→∞ (1+1x ) x =e 故 \qquad \lim_{x \rightarrow \infty}{(1 + \dfrac{1}{x})^x} = e
说明：此极限也可写为lim z→0 (1+z) 1z =e 说明：此极限也可写为 \lim_{z \rightarrow 0}{(1 + z)^{\frac{1}{z}}} = e

例6.求lim x→∞ (1−1x ) x . 例6.求\lim_{x \rightarrow \infty}{(1 - \dfrac{1}{x})^x}.
解：令t=−x，则lim x→∞ (1−1x ) x =lim t→−∞ (1+1t ) −t =lim t→−∞ 1(1+1t ) t =1e 解：令 t = -x，则 \lim_{x \rightarrow \infty}{(1 - \dfrac{1}{x})^x} = \lim_{t \rightarrow -\infty}{(1 + \dfrac{1}{t})^{-t}} = \lim_{t \rightarrow -\infty}{\dfrac{1}{(1 + \dfrac{1}{t})^t}} = \dfrac{1}{e}

说明：若利用lim ϕ(x)→∞ (1+1ϕ(x) ) ϕ(x) =e,则原式=lim −x→∞ [(1+1−x ) −x ] − 1=e −1 说明：若利用\lim_{\phi(x) \rightarrow \infty}{(1 + \dfrac{1}{\phi(x)})^{\phi(x)}} = e,则原式 = \lim_{-x \rightarrow \infty}{[(1 + \dfrac{1}{-x})^{-x}]^-1} = e^{-1}

内容小结
1.函数极限与数列极限关系的应用
(1)利用数列极限判断函数极限不存在
法1找一个数列{x n }:x n ≠x 0 ,且x n →x 0 (n→∞)使lim n→∞ f(x n )不存在. 法1 找一个数列\{x_n\}:x_n \neq x_0,且x_n \rightarrow x_0(n \rightarrow \infty)使\lim_{n \rightarrow \infty}f(x_n)不存在.
法2.找两个趋于x 0 的不同数列{x n }及{x ′ n },使lim n→∞ f(x)≠lim n→∞ f(x ′ n ) 法2.找两个趋于x_0的不同数列\{x_n\}及\{x_n^{'}\},使\lim_{n \rightarrow \infty}f(x) \neq \lim_{n \rightarrow \infty}{f(x_n^{'})}

(2)数列极限存在的夹逼准则⟹函数的极限存在夹逼准则 (2)数列极限存在的夹逼准则 \Longrightarrow 函数的极限存在夹逼准则

2.两个重要的极限
(1)lim x→0 sinxx =1 (1)\lim_{x \rightarrow 0}{\dfrac{sin{x}}{x}} = 1
(2)lim x→∞ (1+1x ) x =e (2)\lim_{x \rightarrow \infty}{(1 + \dfrac{1}{x})^x} = e
或lim x→0 (1+x) 1x =3 或\lim_{x \rightarrow 0}{(1 + x)^{\frac{1}{x}}} = 3

思考与练习
填空题(1~4)
1.lim x→∞ sinxx = 0  − −  ; 1.\lim_{x \rightarrow \infty} {\dfrac{\sin{x}}{x}} = \underline{\ 0 \ };
2.lim x→∞ xsin1x = 1  − −   2.\lim_{x \rightarrow \infty}{x \sin{\dfrac{1}{x}} } = \underline{ \ 1 \ } ;
3.lim x→0 xsin1x = 0  − −   3.\lim_{x \rightarrow 0}{x \sin{\dfrac{1}{x}}} = \underline{\ 0 \ };
4.lim n→∞ (1−1n ) n = e −1   − − − −   4.\lim_{n \rightarrow \infty}{(1 - \dfrac{1}{n})^n} = \underline{\ e^{-1} \ };