AM@由极限的两个存在准则(定理)推导的两个重要极限
文章目录
-
- abstract
- 第一重要极限
-
- sin x < x < tan x \sin{x}
- cos x < sin x x < 1 \cos{x}<\frac{\sin{x}}{x}<1 cosx<xsinx<1
- cos x → 1 ( x → 0 ) \cos{x}\to{1}(x\to{0}) cosx→1(x→0)
- 例
- sin x < x < tan x \sin{x}
- 第二重要极限
-
- 单调性
- 有界性
- 极限存在
- 推广到函数极限
- 形式推广
- 1 ∞ 1^\infin 1∞型幂指函数的极限
-
- 分离常数变形
- 速算结论
-
- 证明
- 例
-
- 传统逐步演算
abstract
- 由极限的两个存在准则(定理)推导的两个重要极限
第一重要极限
-
将任意角 x x x放在单位圆 O O O上讨论 lim x → 0 sin x x \lim\limits_{x\to 0}{\frac{\sin{x}}{x}} x→0limxsinx
- 对于 y = sin x x y=\frac{\sin{x}}{x} y=xsinx对于一切 x ≠ 0 x\neq{0} x=0都有定义
- 设单位圆圆心 O O O和任意角 x x x的始边重合,始边和终边分别交单位圆于 A , B A,B A,B
-
因为任意角总是可以通过诱导公式转换为锐角计算其三角函数值,不妨设角 ∠ A O B = x ( x ∈ ( 0 , π 2 ) ) \angle{AOB}=x(x\in(0,\frac{\pi}{2})) ∠AOB=x(x∈(0,2π))
-
构造正弦线和正切线以及弧度线:
-
正弦线:过点 B B B作 B C ⊥ O A BC\perp{OA} BC⊥OA交于 O A OA OA于 C C C,则 C B CB CB为正弦线: sin x = C B \sin{x}=CB sinx=CB
-
正切线:点 A A A处的切线和 O B OB OB延长线交于 D D D,则AD就是正切线: tan x = A D \tan{x}=AD tanx=AD
-
由角弧度和半径的关系, x = A B ⌢ x=\overset{\huge\frown}{AB} x=AB⌢
-
sin x < x < tan x \sin{x}sinx<x<tanx
- 令 G 1 , G 2 , G 3 G_1,G_2,G_3 G1,G2,G3分别表示:三角形 A O B AOB AOB,扇形 A O B AOB AOB,三角形 A O D AOD AOD,并用 S ( G i ) , i = 1 , 2 , 3 S(G_i),i=1,2,3 S(Gi),i=1,2,3表示图形 G i G_i Gi的面积
- 则 S ( G 1 ) = 1 2 sin x S(G_1)=\frac{1}{2}\sin{x} S(G1)=21sinx; S ( G 2 ) = 1 2 x S(G_2)=\frac{1}{2}x S(G2)=21x; S ( G 3 ) = 1 2 tan x S(G_3)=\frac{1}{2}\tan{x} S(G3)=21tanx
- 显然 S ( G 1 ) < S ( G 2 ) < S ( G 3 ) S(G_1)
- 即 sin x < x < tan x \sin{x}
(0)
cos x < sin x x < 1 \cos{x}<\frac{\sin{x}}{x}<1 cosx<xsinx<1
- 不等号各边同除以 sin x \sin{x} sinx,有 1 < x sin x < 1 cos x 1<\frac{x}{\sin{x}}<\frac{1}{\cos{x}} 1<sinxx<cosx1,从而 cos x < sin x x < 1 \cos{x}<\frac{\sin{x}}{x}<1 cosx<xsinx<1, x ∈ ( 0 , π 2 ) x\in(0,\frac{\pi}{2}) x∈(0,2π)
(1) - cos x , sin x x \cos{x},\frac{\sin{x}}{x} cosx,xsinx都是偶函数,从而 x ∈ ( − π 2 , 0 ) x\in(-\frac{\pi}{2},0) x∈(−2π,0)内, − x ∈ ( 0 , π 2 ) -x\in(0,\frac{\pi}{2}) −x∈(0,2π);不等式
(1)仍然成立 - 所以 cos x < sin x x < 1 \cos{x}<\frac{\sin{x}}{x}<1 cosx<xsinx<1, x ∈ ( − π 2 , 0 ) ∪ ( 0 , π 2 ) x\in(-\frac{\pi}{2},0)\cup(0,\frac{\pi}{2}) x∈(−2π,0)∪(0,2π)或 0 < ∣ x ∣ < π 2 0<|x|<\frac{\pi}{2} 0<∣x∣<2π
cos x → 1 ( x → 0 ) \cos{x}\to{1}(x\to{0}) cosx→1(x→0)
- 0 < ∣ x ∣ < π 2 0<|x|<\frac{\pi}{2} 0<∣x∣<2π时, cos x ∈ ( 0 , 1 ) \cos{x}\in(0,1) cosx∈(0,1),
- 欲证 lim x → 0 cos x = 1 \lim\limits_{x\to0}\cos{x}=1 x→0limcosx=1,可构造 y = cos x − 1 y=\cos{x}-1 y=cosx−1,而证 lim x → 0 ( cos x − 1 ) = 0 \lim\limits_{x\to0}(\cos{x}-1)=0 x→0lim(cosx−1)=0,或 lim x → 0 ( 1 − cos x ) = 0 \lim\limits_{x\to0}(1-\cos{x})=0 x→0lim(1−cosx)=0
- 所以 1 − cos x ∈ ( 0 , 1 ) 1-\cos{x}\in{(0,1)} 1−cosx∈(0,1)
- 0 < ∣ 1 − cos x ∣ = 1 − cos x 0<|1-\cos{x}|=1-\cos{x} 0<∣1−cosx∣=1−cosx= 2 sin 2 x 2 2\sin^{2}{\frac{x}{2}} 2sin22x< 2 ( x 2 ) 2 2(\frac{x}{2})^2 2(2x)2= x 2 2 \frac{x^2}{2} 2x2
- 即 0 < 1 − cos x < x 2 2 0<1-\cos{x}<\frac{x^2}{2} 0<1−cosx<2x2
- 显然 x 2 2 → 0 ( x → 0 ) \frac{x^2}{2}\to{0}(x\to{0}) 2x2→0(x→0),有夹逼准则, lim x → 0 ( 1 − cos x ) = 0 \lim\limits_{x\to0}(1-\cos{x})=0 x→0lim(1−cosx)=0,所以 lim x → 0 cos x = 1 \lim\limits_{x\to0}\cos{x}=1 x→0limcosx=1
- 又因为 cos x < sin x x < 1 \cos{x}<\frac{\sin{x}}{x}<1 cosx<xsinx<1, lim x → 0 cos x = lim x → 0 1 = 1 \lim\limits_{x\to0}\cos{x}=\lim\limits_{x\to0}1=1 x→0limcosx=x→0lim1=1,所以由夹逼准则, lim x → 0 sin x x = 1 \lim\limits_{x\to0}\frac{\sin{x}}{x}=1 x→0limxsinx=1
例
- lim x → 0 arcsin x x = 1 \lim\limits_{x\to0}\frac{\arcsin{x}}{x}=1 x→0limxarcsinx=1
- 令 arcsin x = t \arcsin{x}=t arcsinx=t,从而 x = sin t x=\sin{t} x=sint
- lim x → 0 arcsin x x \lim\limits_{x\to0}\frac{\arcsin{x}}{{x}} x→0limxarcsinx= lim t → 0 t sin t = 1 \lim\limits_{t\to0}\frac{t}{\sin{t}}=1 t→0limsintt=1
第二重要极限
- 设 x n = ( 1 + 1 n ) n x_n=(1+\frac{1}{n})^{n} xn=(1+n1)n,则 { x n } \set{x_n} {xn}单调有界
- ( n i ) \binom{n}{i} (in)= n ! ( n − i ) ! i ! \frac{n!}{(n-i)!i!} (n−i)!i!n!= A n i / i ! A_{n}^{i}/i! Ani/i!
单调性
-
由二项式定理:
-
x n = ∑ i = 0 n ( n i ) 1 n i x_n=\sum_{i=0}^{n}\binom{n}{i}\frac{1}{n^{i}} xn=∑i=0n(in)ni1= x n = ∑ i = 0 n A n i i ! 1 n i x_n=\sum_{i=0}^{n}\frac{A_n^{i}}{i!}\frac{1}{n^{i}} xn=∑i=0ni!Anini1
- = 1 + n 1 ⋅ 1 n 1+\frac{n}{1}\cdot{\frac{1}{n}} 1+1n⋅n1+ n ( n − 1 ) 2 ! 1 n 2 \frac{n(n-1)}{2!}\frac{1}{n^2} 2!n(n−1)n21+ n ( n − 1 ) ( n − 2 ) 3 ! 1 n 3 \frac{n(n-1)(n-2)}{3!}\frac{1}{n^3} 3!n(n−1)(n−2)n31+ ⋯ \cdots ⋯+ n ( n − 1 ) ⋯ ( n − ( n − 1 ) ) n ! 1 n n \frac{n(n-1)\cdots{(n-(n-1))}}{n!}\frac{1}{n^n} n!n(n−1)⋯(n−(n−1))nn1
- = 1 + 1 1+1 1+1+ 1 2 ! ( 1 − 1 n ) \frac{1}{2!}(1-\frac{1}{n}) 2!1(1−n1)+ 1 3 ! ( 1 − 1 n ) ( 1 − 2 n ) \frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n}) 3!1(1−n1)(1−n2)+ ⋯ \cdots ⋯+ 1 n ! ( 1 − 1 n ) ( 1 − 2 n ) ⋯ ( 1 − n − 1 n ) \frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{n-1}{n}) n!1(1−n1)(1−n2)⋯(1−nn−1)
-
则 x n + 1 x_{n+1} xn+1= 1 + 1 1+1 1+1+ 1 2 ! ( 1 − 1 n + 1 ) \frac{1}{2!}(1-\frac{1}{n+1}) 2!1(1−n+11)+ 1 3 ! ( 1 − 1 n + 1 ) ( 1 − 2 n + 1 ) \frac{1}{3!}(1-\frac{1}{n+1})(1-\frac{2}{n+1}) 3!1(1−n+11)(1−n+12)+ ⋯ \cdots ⋯+ 1 ( n + 1 ) ! ( 1 − 1 n + 1 ) ( 1 − 2 n + 1 ) ⋯ ( 1 − n − 1 n + 1 ) \frac{1}{(n+1)!}(1-\frac{1}{n+1})(1-\frac{2}{n+1})\cdots(1-\frac{n-1}{n+1}) (n+1)!1(1−n+11)(1−n+12)⋯(1−n+1n−1)
-
x n , x n + 1 x_n,x_{n+1} xn,xn+1的前2项相等,后续的第 i ( i > 2 ) i(i>2) i(i>2)项 x n + 1 x_{n+1} xn+1的总是要大于 x n x_{n} xn的,并且 x n + 1 x_{n+1} xn+1的项数比 x n x_{n} xn的项要多一项大于0的项,所以 x n + 1 > x n x_{n+1}>x_n xn+1>xn;
-
说明 { x n } \set{x_n} {xn}数列是单调递增的
-
有界性
- 通过放缩 x n x_n xn的展开式中的项确定某个上限
- x n x_n xn展开式中 ( 1 − i n ) < 1 (1-\frac{i}{n})<1 (1−ni)<1,所以 x n < T = 1 + 1 + 1 2 ! + ⋯ + 1 n ! x_n
- 又 n ! ⩾ 2 n − 1 ( n ∈ N ) n!\geqslant 2^{n-1}(n\in\mathbb{N}) n!⩾2n−1(n∈N),即 1 n ! < 1 2 n \frac{1}{n!}<\frac{1}{2^{n}} n!1<2n1,( n ! > 2 n − 1 n!>2^{n-1} n!>2n−1, n ∈ N , n ⩾ 3 n\in\mathbb{N},n\geqslant{3} n∈N,n⩾3)
- T < W = 1 + 1 + 1 2 + 1 2 2 + ⋯ + 1 2 n − 1 T
- x n x_n xn展开式中 ( 1 − i n ) < 1 (1-\frac{i}{n})<1 (1−ni)<1,所以 x n < T = 1 + 1 + 1 2 ! + ⋯ + 1 n ! x_n
极限存在
- 根据单调有界极限存在定理(准则), { x n } \set{x_n} {xn}极限存在 ( n → ∞ ) (n\to{\infin}) (n→∞),这个值是个实数,不容易用十进制数表示
- 通常将这个极限记为 e ( e ∈ ( 2 , 3 ) ) e(e\in(2,3)) e(e∈(2,3)),即 lim n → ∞ x n = e \lim\limits_{n\to{\infin}}x_{n}=e n→∞limxn=e,即 lim n → ∞ ( 1 + 1 n ) n = e \lim\limits_{n\to{\infin}}(1+\frac{1}{n})^{n}=e n→∞lim(1+n1)n=e
推广到函数极限
- lim x → ∞ ( 1 + 1 x ) x = e \lim\limits_{x\to{\infin}}(1+\frac{1}{x})^{x}=e x→∞lim(1+x1)x=e
形式推广
-
上述是 x → ∞ x\to{\infin} x→∞的过程的极限,利用变量代换, t = 1 x t=\frac{1}{x} t=x1,即 x = 1 t x=\frac{1}{t} x=t1, x = 1 t → ∞ x=\frac{1}{t}\to{\infin} x=t1→∞过程对应 t → 0 t\to{0} t→0的过程
-
则 lim x → ∞ ( 1 + 1 x ) x \lim\limits_{x\to{\infin}}(1+\frac{1}{x})^{x} x→∞lim(1+x1)x= lim t → 0 ( 1 + t ) 1 t = e \lim\limits_{t\to{0}}(1+t)^{\frac{1}{t}}=e t→0lim(1+t)t1=e
-
更一般的,在自变量的某个变化过程中 ( x → ∗ ) (x\to{*}) (x→∗),若 α ( x ) \alpha(x) α(x)是无穷小量,即 α ( x ) → 0 ( x → ∗ ) \alpha(x)\to{0}(x\to{*}) α(x)→0(x→∗),则 lim x → ∗ ( 1 + α ( x ) ) 1 α ( x ) = e \lim\limits_{x\to{*}}(1+\alpha(x))^{\frac{1}{\alpha(x)}}=e x→∗lim(1+α(x))α(x)1=e,简记为 lim ( 1 + α ( x ) ) 1 α ( x ) = e \lim(1+\alpha(x))^{\frac{1}{\alpha(x)}}=e lim(1+α(x))α(x)1=e
1 ∞ 1^\infin 1∞型幂指函数的极限
分离常数变形
- 有时,需要使用分离常数的技巧将函数的形式转换为 f ( x ) = ( 1 + α ( x ) ) β ( x ) {f(x)=(1+\alpha (x))^{\beta(x)}} f(x)=(1+α(x))β(x)的形式,
- 例如: ( x + 1 x − 3 ) x = ( x − 3 + 3 + 1 x − 3 ) x = ( 1 + 4 x − 3 ) x (\frac{x+1}{x-3})^x=(\frac{x-3+3+1}{x-3})^x=(1+\frac{4}{x-3})^{x} (x−3x+1)x=(x−3x−3+3+1)x=(1+x−34)x
速算结论
-
如果判断出 f ( x ) = ( 1 + α ( x ) ) β ( x ) {f(x)=(1+\alpha (x))^{\beta(x)}} f(x)=(1+α(x))β(x)的某个过程的极限属于 1 ∞ 1^\infin 1∞型的情况下求极限 S = lim f ( x ) S=\lim f(x) S=limf(x),则可以按如下步骤求解
-
先计算出 A = lim ( α ( x ) β ( x ) ) A=\lim(\alpha(x)\beta(x)) A=lim(α(x)β(x))
-
那么: S = e A S=e^A S=eA,也即是说,结果是 e e e的幂的形式
-
证明
- 设 α ( x ) , β ( x ) \alpha(x),\beta(x) α(x),β(x)分别极限过程 x → ∗ x\to{*} x→∗的无穷小量和无穷大量,即 lim α ( x ) = 0 \lim{\alpha(x)}=0 limα(x)=0, lim β ( x ) = ∞ \lim\beta(x)=\infin limβ(x)=∞
- γ ( x ) = α ( x ) β ( x ) \gamma(x)=\alpha(x)\beta(x) γ(x)=α(x)β(x); β ( x ) = 1 α ( x ) α ( x ) β ( x ) \beta(x)=\frac{1}{\alpha(x)}\alpha(x)\beta(x) β(x)=α(x)1α(x)β(x)= 1 α ( x ) γ ( x ) \frac{1}{\alpha(x)}\gamma(x) α(x)1γ(x)
- S = lim ( 1 + α ( x ) ) β ( x ) S=\lim(1+\alpha(x))^{\beta(x)} S=lim(1+α(x))β(x)= lim ( 1 + α ( x ) ) 1 α ( x ) α ( x ) β ( x ) \lim(1+\alpha(x))^{\frac{1}{\alpha(x)}\alpha(x)\beta(x)} lim(1+α(x))α(x)1α(x)β(x)= lim ( ( ( 1 + α ( x ) ) 1 α ( x ) ) γ ( x ) \lim{(((1+\alpha(x))^\frac{1}{\alpha(x)}})^{\gamma(x)} lim(((1+α(x))α(x)1)γ(x)= [ lim ( ( ( 1 + α ( x ) ) 1 α ( x ) ) ] γ ( x ) [\lim{(((1+\alpha(x))^\frac{1}{\alpha(x)}})]^{\gamma(x)} [lim(((1+α(x))α(x)1)]γ(x)= e γ ( x ) e^{\gamma(x)} eγ(x)
- 记 A = lim γ ( x ) A=\lim{\gamma(x)} A=limγ(x),则 S = e A S=e^{A} S=eA
例
- 以下3个的 1 ∞ 1^\infin 1∞型极限都可以用 e A e^A eA模型法来计算,先确定 α ( x ) 和 β ( x ) \alpha{(x)}和\beta{(x)} α(x)和β(x)
-
S 1 = lim x → ∞ ( 1 − 1 x ) x S_1=\lim\limits_{x\to \infin}(1-\frac{1}{x})^x S1=x→∞lim(1−x1)x
-
S 2 = lim x → ∞ ( 1 + a x ) b x S_2=\lim\limits_{x\to \infin}{(1+\frac{a}{x})^{bx}} S2=x→∞lim(1+xa)bx
-
S 3 = lim x → ∞ ( 1 + a x ) b x + c S_3=\lim\limits_{x\to \infin}(1+\frac{a}{x})^{bx+c} S3=x→∞lim(1+xa)bx+c
-
- 分别计算 A 1 , A 2 , A 3 A_1,A_2,A_3 A1,A2,A3
-
A 1 = lim x → ∞ − 1 x x = − 1 A_1=\lim\limits_{x\to \infin} \frac{-1}{x}x=-1 A1=x→∞limx−1x=−1
-
A 2 = lim x → ∞ a x b x = a b A_2=\lim\limits_{x\to \infin} \frac{a}{x}bx=ab A2=x→∞limxabx=ab
-
A 3 = lim x → ∞ a x ( b x + c ) = a b A_3=\lim\limits_{x\to \infin} \frac{a}{x}(bx+c)=ab A3=x→∞limxa(bx+c)=ab
-
- S 1 = e − 1 S_1=e^{-1} S1=e−1, S 2 = e a b S_2=e^{ab} S2=eab, S 3 = e a b S_3=e^{ab} S3=eab
传统逐步演算
-
lim x → ∞ ( 1 − 1 x ) x \lim\limits_{x\to \infin}{(1-\frac{1}{x})}^x x→∞lim(1−x1)x= lim x → ∞ ( 1 − 1 x ) − ( − x ) \lim\limits_{x\to \infin}{(1-\frac{1}{x})}^{-(-x)} x→∞lim(1−x1)−(−x)= lim x → ∞ 1 ( 1 − 1 x ) − x \lim\limits_{x\to \infin}\frac{1}{{{(1-\frac{1}{x})}^{-x}}} x→∞lim(1−x1)−x1= lim x → ∞ 1 lim x → ∞ ( 1 − 1 x ) − x \frac{\lim\limits_{x\to \infin}1}{\lim\limits_{x\to \infin}(1-\frac{1}{x})^{-x}} x→∞lim(1−x1)−xx→∞lim1= 1 e \frac{1}{e} e1
-
lim x → ∞ ( 1 + a x ) b x \lim\limits_{x\to \infin}{(1+\frac{a}{x})^{bx}} x→∞lim(1+xa)bx= lim x → ∞ ( 1 + a x ) x a a b \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{\frac{x}{a}ab} x→∞lim(1+xa)axab= lim x → ∞ \lim\limits_{x\to \infin} x→∞lim ( ( 1 + a x ) x a ) a b \left ({(1+\frac{a}{x})}^{\frac{x}{a}}\right)^{ab} ((1+xa)ax)ab= e a b e^{ab} eab
-
lim x → ∞ ( 1 + a x ) b x + c \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{bx+c} x→∞lim(1+xa)bx+c= lim x → ∞ ( 1 + a x ) b x \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{bx} x→∞lim(1+xa)bx ⋅ \cdot ⋅ lim x → ∞ ( 1 + a x ) c \lim\limits_{x\to \infin}{(1+\frac{a}{x})}^{c} x→∞lim(1+xa)c= e a b ⋅ 1 c e^{ab}\cdot 1^c eab⋅1c= e a b ⋅ 1 e^{ab}\cdot 1 eab⋅1= e a b e^{ab} eab