math_ Proving common equivalent infinitesimal & Case & substitution

Proving the common equivalent infinitesimal

The latter half is less commonly used than the first part

Proving the common equivalent infinitesimal

$\sin (x)\sim x$

$$\underset{x\to 0}{\lim }\frac{sin(x)}{x}=1;The\; firstOneheavywantextremelylimit$$

$tan(x)\sim x$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{tan(x)}{x}=1 \\ \underset{x\to 0}{\lim }\frac{sinx}{xcos(x)}=\underset{x\to 0}{\lim }\frac{sinx}{x}\frac{1}{cosx}=\underset{x\to 0}{\lim }(1\cdot \frac{1}{cosx})=\frac{\underset{x\to 0}{\lim }1}{\underset{x\to 0}{\lim }cos(x)}=1 \end{gathered}$$

$arcsin(x)\sim x$

$$\begin{gathered} Maket=arcsin(x),x=sin(t),t\to 0(x\to 0) \\ \underset{x\to 0}{\lim }\frac{arcsin(x)}{x}=\underset{t\to 0}{\lim }\frac{t}{sin(t)}=1 \end{gathered}$$

$arctan(x)\sim x$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{arctan(x)}{x}=1 \\ Maket=arctan(x), \\ x=tan(t); \\ \Rightarrow \underset{x\to 0}{\lim }\frac{arctan(x)}{x}=\underset{t\to 0}{\lim }\frac{t}{tan(t)}=1 \end{gathered}$$

$ln(1+x)\sim x$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{ln(1+x)}{x}=1 \\ benefituseYesCountsexqualityandThe\; firstTwoheavywantextremelylimitProvebright \\ \underset{x\to 0}{\lim }\frac{ln(1+x)}{x}=\underset{x\to 0}{\lim }\frac{1}{x}ln(1+x)=\underset{x\to 0}{\lim }ln(1+x{)}^{\frac{1}{x}}=\underset{x\to 0}{\lim }ln(e)=1 \end{gathered}$$

$lo{g}_a(1+x)\sim \frac{1}{ln(a)}x$

$$\begin{gathered} moreOnelikeOf,canYes\underset{x\to 0}{\lim }\frac{lo{g}_a(1+x)}{\frac{1}{ln(a)}x}=1 \\ rootAccording\; to\; theinAt\; the\; end\; ofMaletype(changebase) \\ {\log }_{a}e=\frac{ln(e)}{ln(a)}=\frac{1}{ln(a)} \\ \underset{x\to 0}{\lim }\frac{lo{g}_a(1+x)}{x}=\underset{x\to 0}{\lim }\frac{1}{x}lo{g}_a(1+x)=\underset{x\to 0}{\lim }lo{g}_a((1+x{)}^{\frac{1}{x}})=lo{g}_a(e)=\frac{1}{ln(a)} \\ \therefore lo{g}_a(1+x)\sim \frac{1}{ln(a)}x \end{gathered}$$

$e^x-1\sim x$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{e^x-1}{x}=1 \\ inelementLaw:Maket=e^x-1; \\ t=(e^x-1)\to 0(x\to 0)namely,Yes\underset{x\to 0}{\lim }x=\underset{t\to 0}{\lim }t=0 \\ bex=ln(t+1) \\ \underset{x\to 0}{\lim }\frac{e^x-1}{x}=\underset{t\to 0}{\lim }\frac{t}{ln(t+1)}=1 \\ {e}^x-1\sim x \end{gathered}$$

$(a^x-1)\sim x\ln a$

A more general , Can have
$$\begin{gathered} (a^x-1)\sim x\ln a \\ original=\underset{x\to 0}{\lim }\frac{a^x-1}{x\ln a} \\ Maket=a^x-1; \\ t\to 0(x\to 0); \\ x=lo{g}_a(t+1) \\ lo{g}_a(1+t)\sim \frac{1}{ln(a)}t \\ original=\underset{t\to 0}{\lim }\frac{t}{{\log }_a(t+1)}=\underset{t\to 0}{\lim }\frac{t}{\frac{1}{\ln a}t}=\ln a \\ \therefore (a^x-1)\sim x\ln a \end{gathered}$$

$1-cos(x)\sim \frac{1}{2}{x}^2$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{1-cos(x)}{\frac{1}{2}{x}^2}=1 \\ 3、\; ...\; andhornLetterCounttimeshornMaletypecosx=co{s}^2(\frac{x}{2})-si{n}^2(\frac{x}{2}) \\ =1-2{\sin }^2(\frac{x}{2}),(cosx=2co{s}^2(\frac{x}{2})-1;sinshapetypemoreheavywant,ThanaPick\; upnear(Rongeasysenduse)The\; firstOneheavywantextremelylimit) \\ \underset{x\to 0}{\lim }\frac{1-cos(x)}{x^2}=\underset{x\to 0}{\lim }\frac{2si{n}^2(\frac{x}{2})}{x^2}=\underset{x\to 0}{\lim }\frac{2si{n}^2(\frac{x}{2})}{4(\frac{x}{2}{)}^2} \\ =\underset{x\to 0}{\lim }\frac{1}{2}(\frac{sin(\frac{x}{2})}{\frac{x}{2}}{)}^2=\frac{1}{2} \\ \therefore \underset{x\to 0}{\lim }\frac{1-cos(x)}{\frac{1}{2}{x}^2}=1 \end{gathered}$$

Slightly more complex equivalent infinitesimal

$(1+x{)}^a-1\sim ax$

• Replace the two equivalent infinitesimals proved above , To prove a slightly more complex equivalent infinitesimal

According to the meaning of logarithm & nature :

$$\begin{gathered} {a}^{lo{g}_{a}b}=b \\ ln{x}^n=n\cdot ln(x) \\ (1+x{)}^a=e^{lo{g}_e(1+x{)}^a}=e^{ln(1+x{)}^a}=e^{a\cdot ln(1+x)} \end{gathered}$$

• Thus the proposition that needs to be proved becomes :

$$\begin{gathered} e^{a\cdot ln(1+x)}-1\sim x \\ orpersonsay:(1+x{)}^a-1=e^{a\cdot \ln (x+1)}-1\sim a\cdot \ln (x+1) \end{gathered}$$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{e^{a\cdot ln(1+x)}-1}{x} \\ *benefitusefrontNoodlesProvebrightOfln(x+1)\sim x,takebranchmotherInto\; theThat\text{'}s\; okOn\; behalf\; ofin(etc.pricenothingpoorSmallOn\; behalf\; ofinsetThe\; reason\; is) \\ fromandhave\; toToshapeSuch\; asanotherOneindividualetc.pricenothingpoorSmallOfshapetype: \\ \underset{x\to 0}{\lim }\frac{e^x-1}{x}=1 \\ *orperson,On\; behalf\; ofinbranchSon(branchSonwholebodyyesoperatorclose{e}^x-1(e^x-1\sim xOfshapetype)), \\ thisinxtakevaluebysurfacereachtypex=a\cdot \ln (x+1),fromand: \\ (1+x{)}^a-1=e^{a\cdot \ln (x+1)}-1\sim a\cdot \ln (x+1) \\ presentstay,\underset{x\to 0}{\lim }\frac{(1+x{)}^a-1}{x}=\underset{x\to 0}{\lim }\frac{a\cdot \ln (x+1)}{x}=\underset{x\to 0}{\lim }a\frac{ln(x+1)}{x}=a \\ fromand: \\ \underset{x\to 0}{\lim }\frac{(x+1{)}^a-1}{ax}=1 \end{gathered}$$

Or use yuan exchange + The method of matching

$$\begin{gathered} Maket=(1+x{)}^a-1;namely,(1+x{)}^a=t+1 \\ be\ln (1+x{)}^a=\ln (t+1) \\ =\underset{x\to 0}{\lim }\frac{(1+x{)}^a-1}{x}\frac{ln(1+x{)}^a}{ln(1+x{)}^a} \\ =\underset{x\to 0}{\lim }\frac{(1+x{)}^a-1}{x}\frac{a\cdot ln(1+x)}{ln(1+x{)}^a} \\ =\underset{x\to 0}{\lim }\frac{(1+x{)}^a-1}{ln(1+x{)}^a}\frac{a\cdot ln(1+x)}{x} \\ =\underset{t\to 0}{\lim }\frac{t}{ln(t+1)}\cdot \underset{x\to 0}{\lim }\frac{a\cdot ln(1+x)}{x} \\ =1\times a=a \end{gathered}$$

Comprehensive example

• Comparison between compound functions and infinitesimal quantities

  • The equivalent infinitesimal used includes :

    $$\begin{gathered} cosx-1\sim \frac{-1}{2}{x}^2;(1-cosx\sim \frac{1}{2}{x}^2) \\ sin(x)\sim x;(classlikeOfsin(\alpha (x))\sim \alpha (x)) \end{gathered}$$

  • Compound functions need to consider the constraints between the definition domain of the outer function and the value domain of the inner function

  • In this question

$$\begin{gathered} rootAccording\; to\; theetc.pricenothingpoorSmall,\underset{x\to 0}{\lim }h(x)=\underset{x\to 0}{\lim }sin(\alpha (x))=-\frac{1}{2}, \\ h(x)=sin(\alpha (x))andy=xyesx\to 0OfSame\; asranknothingpoorSmallfromand,\underset{x\to 0}{\lim }h(x)=0 \\ |\alpha (x)|<\frac{\pi }{2}by了moreAddthroughcustomOfThe\; reason\; isExplainThestripPieces\; of,Go\; tofallmostYesvaluehave\; toTo: \\ -\frac{\pi }{2}<\alpha (x)<\frac{\pi }{2} \\ fingerOut了LetterCount\alpha (x)OftakevalueFanaround \\ remember:u=\alpha (x) \\ \underset{x\to 0}{\lim }u=u_0 \\ rootAccording\; to\; the3、\; ...\; andhornsitmarksinglepositionroundcanknow\underset{u\to k\pi }{\lim }sin(u)= \\ \underset{x\to 0}{\lim }sin(\alpha (x))=\underset{u\to u_0}{\lim }sin(u)=0 \\ \because \underset{u\to k\pi }{\lim }sin(u)=0 \\ \therefore u_0=k\pi \\ rootAccording\; to\; the\alpha (x)OfvalueDomain,canknow,u_0⩽\frac{\pi }{2} \\ \therefore k=0 \\ (u_0=0,namely\underset{x\to 0}{\lim }\alpha (x)=u_0=0), \end{gathered}$$