Detailed explanation of infinitesimal and infinite

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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 xx0(x)limf(x)=0 , thenf (x) f(x)f(x) x → x 0 x\to x_0 xx0(orx → ∞ x\to\inftyxinfinitesimalwhen ∞ )

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 xx0(x)limf(x)=The necessary and sufficient conditions for A are f ( x ) = A + α f(x)=A+\alphaf(x)=A+α , whereα \alphaα x → x 0 x\to x_0 xx0(orx → ∞ x\to\inftyxinfinitesimal when ∞ )

This theorem is easily proved by the addition rule of the limit

infinitesimal comparison

To x → 0 x\to 0x0 hours,xxx andx 2 x^2x2 are infinitesimal, butlim ⁡ x → 0 xx 2 = ∞ \lim\limits_{x\to 0}\frac{x}{x^2}=\inftyx0limx2x= . Because atx → 0 x\to 0x0 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

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

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

lim ⁡ α β = c ≠ 0 \lim\frac{\alpha}{\beta}=c\neqlimba=c=0 is called “α \alphaαβ \betaβ is an infinitesimal of the same order”

k order infinitesimal

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 \betaβkkInfinitely small of order k ”

Equivalent to infinitesimal
definition

lim ⁡ α β = 1 \lim\frac{\alpha}{\beta}=1limba=1 is called “α \alphaαβ \betaβ is equivalent to infinitesimal", denoted asα ∼ β \alpha\sim\betaab

nature
  • Reflexivity: α ∼ α \alpha\sim\alphaaa
  • Symmetry: If α ∼ β \alpha\sim\betaaββ ∼ α \beta\sim\alphaba
  • Transitivity: If α ∼ β , β ∼ γ \alpha\sim\beta,~\beta\sim\gammaab , bγ∈αγ \alpha\sim\gammaac
Common equivalent infinitesimal

x → 0 x\to 0 x0 o'clock

  • sin ⁡ x ∼ x \sin x\sim x sinxx

    proving process

    As 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)
    Insert image description here

    It can be seen from the figure that when x → 0 x\to 0xWhen 0 , there is always
    S △ AOB < S Sector AOB < S △ AOC S_{\triangle AOB}< S_{Sector AOB}< S_{\triangle AOC}SAOB<SSector A OB<SAOC
    ∴ 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| 21sinx<21x<21tanx
    ∴ ∣ sin ⁡ x ∣ < ∣ x ∣ < ∣ sin ⁡ x ∣ cos ⁡ x ,   cos ⁡ x > 0 \therefore|\sin x|< |x|<\frac{|\sin x|}{\cos x},~\cos x>0 sinx<x<cosxsinx, 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 x0limcosx=1
    Fromthe pinch criterion,lim ⁡ x → 0 sin ⁡ xx = 1 \lim\limits_{x\to 0}\frac{\sin x}{x}=1x0limxsinx=1
    ∴ sin ⁡ x ∼ x \therefore\sin x\sim x sinxx

  • tan ⁡ x ∼ x \tan x\sim x tanxx

    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} x0limxtanx=x0limxcosxsinx=x0limxsinxx0limcosx1=1
    ∴ tan ⁡ x ∼ x \therefore\tan x\sim x tanxx

  • arcsin ⁡ x ∼ x \arcsin x\sim x arcsinxx

    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} x0limxarcsinx=x0lim(x)(arcsinx)=x0lim1x2 1=1
    ∴ arcsin ⁡ x ∼ x \therefore\arcsin x\sim x arcsinxx

  • a x − 1 ∼ x ln ⁡ a a^x-1\sim x\ln a ax1xlna

    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} x0limxlnaax1=x0lim(xlna)(ax1)=x0limlnaaxlna=1
    ∴ a x − 1 ∼ x ln ⁡ a \therefore a^x-1\sim x\ln a ax1xlna

    • ex − 1 ∼ xe^x-1\sim xex1x
    • 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} 1cosx2x2

    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} x0lim21x21cosx=x0lim(21x2)(1cosx)=x0limxsinx=1
    ∴ 1 − cos ⁡ x ∼ 1 2 x 2 \therefore 1-\cos x\sim\frac{1}{2}x^2 1cosx21x2

  • ( 1 + x ) a − 1 ∼ a x (1+x)^a-1\sim ax (1+x)a1ax

    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} x0limax(1+x)a1=x0lim(ax)[(1+x)a1]=x0limaa(1+x)a1=1
    ∴ ( 1 + x ) a − 1 ∼ a x \therefore(1+x)^a-1\sim ax (1+x)a1ax

theorem

Theorem 1: Necessary and sufficient conditions
α ∼ β \alpha\sim\betaaThe 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\betaab , a b 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}x0limx+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^3xx+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 x0limx+x2+x32 x=x0limx2 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 xx0(x)limf(x)= , thenf ( x ) f(x)f(x) x → x 0 x\to x_0 xx0(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 xx0(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 xx0(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 xx0(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

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