free of charge
definition
If the function f ( x ) f(x)f(x) 满足 lim x → x 0 ( x → ∞ ) f ( x ) = 0 \lim\limits_{\substack{x\rightarrow x_0\\(x\to\infty)}}f(x)=0 x→x0(x→∞)limf(x)=0 , thenf (x) f(x)f(x) 为 x → x 0 x\to x_0 x→x0(orx → ∞ x\to\inftyx→infinitesimalwhen ∞ )
It can be seen from the definition that the essence of infinitesimal is a function. The function value of this function approaches 0 0 infinitely during a certain change of the independent variable.0 . Infinitely is not a very small number
The relationship between infinitesimal and limit
lim x → x 0 ( x → ∞ ) f ( x ) = A \lim\limits_{\substack{x\to x_0\\(x\to\infty)}}f(x)=A x→x0(x→∞)limf(x)=The necessary and sufficient conditions for A are f ( x ) = A + α f(x)=A+\alphaf(x)=A+α , whereα \alphaα 为 x → x 0 x\to x_0 x→x0(orx → ∞ x\to\inftyx→infinitesimal when ∞ )
This theorem is easily proved by the addition rule of the limit
infinitesimal comparison
To x → 0 x\to 0x→0 hours,xxx andx 2 x^2x2 are infinitesimal, butlim x → 0 xx 2 = ∞ \lim\limits_{x\to 0}\frac{x}{x^2}=\inftyx→0limx2x=∞ . Because atx → 0 x\to 0x→0 process,x 2 x^2x2 tends to0 00 is faster thanxxxfast _
According to the different situations of the limits of the ratio of two infinitesimals, it reflects that different infinitesimals tend to 0 00 speed
Assume α \alphaa ,b \betaβ is the infinitesimal value in the same change process of the independent variable. According tolim α β \lim\frac{\alpha}{\beta}limbaThe value of
- 0 0 0 : Higher order infinitesimal
- If there exists k > 0 k>0k>0使lim α β k = c ≠ 0 \lim\frac{\alpha}{\beta^k}=c\neqlimbka=c=0 , thenkkk order infinitesimal
- Not 0 0Constants of 0 : infinitesimals of the same order
- If the constant is 1 11 , then it is equivalent to infinitesimal
- ∞ \infty∞ : low-order infinitesimal
Higher order infinitesimal
若lim α β = 0 \lim\frac{\alpha}{\beta}=0limba=0 is called “α \alphaα is greater thanβ \beta"The infinitesimal of the higher order of β ", denoted asα = o ( β ) \alpha=o(\beta)a=o ( b )
Lower order infinitesimal
若lim α β = ∞ \lim\frac{\alpha}{\beta}=\inftylimba=∞ is called “α \alphaα is greater thanβ \betaThe infinitesimal of the lower order of β ”
If α \alphaα is greater thanβ \betaThe higher order of β is infinitesimal, thenβ \betaβ is greater thanα \alphaThe lower order infinitesimal of α
infinitesimal of the same order
若lim α β = c ≠ 0 \lim\frac{\alpha}{\beta}=c\neqlimba=c=0 is called “α \alphaα与β \betaβ is an infinitesimal of the same order”
k order infinitesimal
若lim α β k = c ≠ 0 , k > 0 \lim\frac{\alpha}{\beta^k}=c\neq 0.~k>0limbka=c=0, k>0 is called “α \alphaα is aboutβ \betaβ的kkInfinitely small of order k ”
Equivalent to infinitesimal
definition
若lim α β = 1 \lim\frac{\alpha}{\beta}=1limba=1 is called “α \alphaα与β \betaβ is equivalent to infinitesimal", denoted asα ∼ β \alpha\sim\betaa∼b
nature
- Reflexivity: α ∼ α \alpha\sim\alphaa∼a
- Symmetry: If α ∼ β \alpha\sim\betaa∼β则β ∼ α \beta\sim\alphab∼a
- Transitivity: If α ∼ β , β ∼ γ \alpha\sim\beta,~\beta\sim\gammaa∼b , b∼γ∈α ∼γ \alpha\sim\gammaa∼c
Common equivalent infinitesimal
x → 0 x\to 0 x→0 o'clock
-
sin x ∼ x \sin x\sim x sinx∼x
proving processAs shown in the figure, there is an angle x ∈ ( − π 2 , 0 ) ∪ ( 0 , π 2 ) x\in(-\frac{\pi}{2},0)\cup(0,\frac{\pi }{2})x∈(−2p,0)∪(0,2p)
It can be seen from the figure that when x → 0 x\to 0x→When 0 , there is always
S △ AOB < S Sector AOB < S △ AOC S_{\triangle AOB}< S_{Sector AOB}< S_{\triangle AOC}S△AOB<SSector A OB<S△AOC
∴ 1 2 ∣ sin x ∣ < 1 2 ∣ x ∣ < 1 2 ∣ tan x ∣ \therefore\frac{1}{2}|\sin x|< \frac{1}{2}|x|<\frac{1}{2}|\tan x| ∴21∣sinx∣<21∣x∣<21∣tanx∣
∴ ∣ sin x ∣ < ∣ x ∣ < ∣ sin x ∣ cos x , cos x > 0 \therefore|\sin x|< |x|<\frac{|\sin x|}{\cos x},~\cos x>0 ∴∣sinx∣<∣x∣<cosx∣sinx∣, cosx>0
∴ cos x < ∣ sin x x ∣ < 1 \therefore\cos x<|\frac{\sin x}{x}|< 1 ∴cosx<∣xsinx∣<1
∴ cos x < sin x x < 1 , sin x x > 0 \therefore\cos x<\frac{\sin x}{x}<1,~\frac{\sin x}{x}>0 ∴cosx<xsinx<1, xsinx>0
∵ lim x → 0 cos x = 1 \because\lim\limits_{x\to 0}\cos x=1 ∵x→0limcosx=1
Fromthe pinch criterion,lim x → 0 sin xx = 1 \lim\limits_{x\to 0}\frac{\sin x}{x}=1x→0limxsinx=1
∴ sin x ∼ x \therefore\sin x\sim x ∴sinx∼x -
tan x ∼ x \tan x\sim x tanx∼x
proving process∵ \because ∵
lim x → 0 tan x x = lim x → 0 sin x x cos x = lim x → 0 sin x x ⋅ lim x → 0 1 cos x = 1 \begin{aligned} \lim\limits_{x\to 0}\frac{\tan x}{x}&=\lim\limits_{x\to 0}\frac{\sin x}{x\cos x}\\ &=\lim\limits_{x\to 0}\frac{\sin x}{x}\cdot \lim\limits_{x\to 0}\frac{1}{\cos x}\\ &=1 \end{aligned} x→0limxtanx=x→0limxcosxsinx=x→0limxsinx⋅x→0limcosx1=1
∴ tan x ∼ x \therefore\tan x\sim x ∴tanx∼x -
arcsin x ∼ x \arcsin x\sim x arcsinx∼x
proving process∵ \because∵
lim x → 0 arcsin x x = lim x → 0 ( arcsin x ) ′ ( x ) ′ = lim x → 0 1 1 − x 2 = 1 \begin{aligned} \lim\limits_{x\to 0}\frac{\arcsin x}{x}&=\lim\limits_{x\to 0}\frac{(\arcsin x)'}{(x)'}\\ &=\lim\limits_{x\to 0}\frac{1}{\sqrt{1-x^2}}\\ &=1 \end{aligned} x→0limxarcsinx=x→0lim(x)′(arcsinx)′=x→0lim1−x21=1
∴ arcsin x ∼ x \therefore\arcsin x\sim x ∴arcsinx∼x -
a x − 1 ∼ x ln a a^x-1\sim x\ln a ax−1∼xlna
proving process∵ \because ∵
lim x → 0 a x − 1 x ln a = lim x → 0 ( a x − 1 ) ′ ( x ln a ) ′ = lim x → 0 a x ln a ln a = 1 \begin{aligned} \lim\limits_{x\to 0}\frac{a^x-1}{x\ln a}&=\lim\limits_{x\to 0}\frac{(a^x-1)'}{(x\ln a)'}\\ &=\lim\limits_{x\to 0}\frac{a^x\ln a}{\ln a}\\ &=1 \end{aligned} x→0limxlnaax−1=x→0lim(xlna)′(ax−1)′=x→0limlnaaxlna=1
∴ a x − 1 ∼ x ln a \therefore a^x-1\sim x\ln a ∴ax−1∼xlna- ex − 1 ∼ xe^x-1\sim xex−1∼x
- ln ( 1 + x ) ∼ x \ln(1+x)\sim x ln(1+x)∼x
-
1 − cos x ∼ x 2 2 1-\cos x\sim\frac{x^2}{2} 1−cosx∼2x2
proving process∵ \because ∵
lim x → 0 1 − cos x 1 2 x 2 = lim x → 0 ( 1 − cos x ) ′ ( 1 2 x 2 ) ′ = lim x → 0 sin x x = 1 \begin{aligned} \lim\limits_{x\to 0}\frac{1-\cos x}{\frac{1}{2}x^2}&=\lim\limits_{x\to 0}\frac{(1-\cos x)'}{(\frac{1}{2}x^2)'}\\ &=\lim\limits_{x\to 0}\frac{\sin x}{x}\\ &=1 \end{aligned} x→0lim21x21−cosx=x→0lim(21x2)′(1−cosx)′=x→0limxsinx=1
∴ 1 − cos x ∼ 1 2 x 2 \therefore 1-\cos x\sim\frac{1}{2}x^2 ∴1−cosx∼21x2 -
( 1 + x ) a − 1 ∼ a x (1+x)^a-1\sim ax (1+x)a−1∼ax
proving process∵ \because ∵
lim x → 0 ( 1 + x ) a − 1 a x = lim x → 0 [ ( 1 + x ) a − 1 ] ′ ( a x ) ′ = lim x → 0 a ( 1 + x ) a − 1 a = 1 \begin{aligned} \lim\limits_{x\to 0}\frac{(1+x)^a-1}{ax}&=\lim\limits_{x\to 0}\frac{[(1+x)^a-1]'}{(ax)'}\\ &=\lim\limits_{x\to 0}\frac{a(1+x)^{a-1}}{a}\\ &=1 \end{aligned} x→0limax(1+x)a−1=x→0lim(ax)′[(1+x)a−1]′=x→0limaa(1+x)a−1=1
∴ ( 1 + x ) a − 1 ∼ a x \therefore(1+x)^a-1\sim ax ∴(1+x)a−1∼ax
theorem
Theorem 1: Necessary and sufficient conditions
α ∼ β \alpha\sim\betaa∼The necessary and sufficient conditions for β are β = α + o ( α ) \beta=\alpha+o(\alpha)b=a+o ( a )
Theorem 2: Equivalent substitution
If α ∼ β , α ~ ∼ β ~ \alpha\sim\beta,~\widetilde\alpha\sim\widetilde\betaa∼b , a
∼b
则lim α β = lim α ~ β = lim α β ~ = lim α ~ β ~ \lim\frac{\alpha}{\beta}=\lim\frac{\widetilde\alpha}{\beta }=\lim\frac{\alpha}{\widevalue\beta}=\lim\frac{\widevalue\alpha}{\widevalue\beta}limba=limba
=limb
a=limb
a
Using the above two theorems is very helpful for finding the limit, such as finding the limit lim x → 0 2 xx + x 2 + x 3 \lim\limits_{x\to 0}\frac{2x}{x+x^ 2+x^3}x→0limx+x2+x32 x
Because x = x + o ( x ) x=x+o(x)x=x+o ( x ) , whereo ( x ) = x 2 + x 3 o(x)=x^2+x^3o(x)=x2+x3So
x ∼ x + x 2 + x3 x\sim x+x^2+x^3x∼x+x2+x3
所以 lim x → 0 2 x x + x 2 + x 3 = lim x → 0 2 x x = 2 \lim\limits_{x\to 0}\frac{2x}{x+x^2+x^3}=\lim\limits_{x\to 0}\frac{2x}{x}=2 x→0limx+x2+x32 x=x→0limx2 x=2
Great size
If the function f ( x ) f(x)f(x) 满足 lim x → x 0 ( x → ∞ ) f ( x ) = ∞ \lim\limits_{\substack{x\rightarrow x_0\\(x\to\infty)}}f(x)=\infty x→x0(x→∞)limf(x)=∞ , thenf ( x ) f(x)f(x) 为 x → x 0 x\to x_0 x→x0(orx → ∞ x\to\inftyx→∞ ) infinitywhen
注意: lim x → x 0 ( x → ∞ ) f ( x ) = ∞ \lim\limits_{\substack{x\rightarrow x_0\\(x\to\infty)}}f(x)=\infty x→x0(x→∞)limf(x)=∞(①式)是 lim x → x 0 ( x → ∞ ) f ( x ) = + ∞ \lim\limits_{\substack{x\rightarrow x_0\\(x\to\infty)}}f(x)=+\infty x→x0(x→∞)limf(x)=+∞(②式)和 lim x → x 0 ( x → ∞ ) f ( x ) = − ∞ \lim\limits_{\substack{x\rightarrow x_0\\(x\to\infty)}}f(x)=-\infty x→x0(x→∞)limf(x)=− ∞ (formula ③) is the general name. If one of the formulas ②③ is satisfied, it must satisfy the formula ①
It can be seen from the definition that the essence of infinity is a function. The function value of this function approaches infinitely to ∞ during a certain change of the independent variable \infty∞ . Infinity is not a very big number
The relationship between infinitesimal and infinite
In the same change process of the independent variable,
- If f ( x ) f(x)f ( x ) is infinitesimal, then1 f ( x ) \frac{1}{f(x)}f(x)1is infinity
- If f ( x ) f(x)f ( x ) is infinite, then1 f ( x ) \frac{1}{f(x)}f(x)1is infinitesimal
reference
[1] Higher Mathematics, Department of Mathematics, Tongji University, Higher Education Press, Volume 1