Advanced Mathematics (Seventh Edition) Tongji University exercises 1-7 personal solutions

Advanced mathematics （ The seventh edition ） Tongji University exercises 1-7

 

$\begin{array}{rlr}& 1.Whenx\to 0when,2x-x^{2}And{x}^2-x^{3}phaseThan,whichOneindividualyeshighranknothingpoorSmall？& \end{array}$

Explain ：

$\begin{array}{rlr}& becauseby\underset{x\to 0}{\lim }(2x-x^2)=0,\underset{x\to 0}{\lim }(x^2-x^3)=0,\underset{x\to 0}{\lim }\frac{x^2-x^3}{2x-x^2}=\underset{x\to 0}{\lim }\frac{x-x^2}{2-x}=0,& \\ & & \\ & theWithWhenx\to 0when,x^2-x^{3}yesThan2x-x^{2}highrankOfnothingpoorSmall.& \end{array}$

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$\begin{array}{rlr}& 2.Whenx\to 0when,(1-cosx{)}^{2}Andsi{n}^{2}xphaseThan,whichOneindividualyeshighranknothingpoorSmall？& \end{array}$

Explain ：

$\begin{array}{rlr}& becauseby\underset{x\to 0}{\lim }(1-cosx{)}^2=0,\underset{x\to 0}{\lim }si{n}^{2}x=0,\underset{x\to 0}{\lim }\frac{(1-cosx{)}^2}{si{n}^{2}x}=\underset{x\to 0}{\lim }\frac{{\left(\frac{1}{2}{x}^2\right)}^2}{x^2}=\underset{x\to 0}{\lim }\frac{1}{4}{x}^2=0,& \\ & & \\ & theWithWhenx\to 0when,(1-cosx{)}^{2}yesThansi{n}^{2}xhighrankOfnothingpoorSmall.& \end{array}$

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$\begin{array}{rlr}& 3.Whenx\to 1when,nothingpoorSmall1-xand(1)1-x^3,(2)\frac{1}{2}(1-x^2)yesnoSame\; asrank,yesnoetc.price？& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)becauseby\underset{x\to 1}{\lim }\frac{1-x}{1-x^3}=\underset{x\to 1}{\lim }\frac{1}{1+x+x^2}=\frac{1}{3},theWithWhenx\to 1when,nothingpoorSmall1-xAnd1-x^{3}Same\; asrank.& \\ & & \\ & (2)becauseby\underset{x\to 1}{\lim }\frac{1-x}{\frac{1}{2}(1-x^2)}=\underset{x\to 1}{\lim }\frac{2}{1+x}=1,theWithWhenx\to 1when,nothingpoorSmall1-xAnd\frac{1}{2}(1-x^2)etc.price.& \end{array}$

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$\begin{array}{rlr}& 4.Provebright：Whenx\to 0when,Yes& \end{array}$

$\begin{array}{rlr}& (1)arctanx\sim x;(2)secx-1\sim \frac{x^2}{2}& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)Makex=tant,t=arctanx,Whenx\to 0when,t\to 0.becauseby\underset{x\to 0}{\lim }\frac{arctanx}{x}=\underset{t\to 0}{\lim }\frac{t}{tant}=1,& \\ & & \\ & theWithWhenx\to 0when,arctanx\sim x.& \\ & & \\ & (2)becauseby\underset{x\to 0}{\lim }\frac{secx-1}{\frac{x^2}{2}}=\underset{x\to 0}{\lim }\frac{\frac{1-cosx}{cosx}}{\frac{1}{2}{x}^2}=\underset{x\to 0}{\lim }\frac{1}{cosx}=1,& \\ & & \\ & theWithWhenx\to 0when,secx-1\sim \frac{x^2}{2}.& \end{array}$

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$\begin{array}{rlr}& 5.benefituseetc.pricenothingpoorSmallOfsexquality,seekNextColumnextremelylimit：& \end{array}$

$\begin{array}{rlr}& (1)\underset{x\to 0}{\lim }\frac{tan3x}{2x};(2)\underset{x\to 0}{\lim }\frac{sin(x^n)}{(sinx{)}^m}(n,mbyjustwholeCount);& \\ & & \\ & (3)\underset{x\to 0}{\lim }\frac{tanx-sinx}{si{n}^{3}x};(4)\underset{x\to 0}{\lim }\frac{sinx-tanx}{(\sqrt[3]{1+x^2}-1)(\sqrt{1+sinx}-1)}& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)\underset{x\to 0}{\lim }\frac{tan3x}{2x}=\underset{x\to 0}{\lim }\frac{sin3x}{2cos3x\cdot x}=\underset{x\to 0}{\lim }\left(\frac{3}{2}\cdot \frac{1}{cos3x}\cdot \frac{sin3x}{3x}\right)=\frac{3}{2}& \\ & & \\ & (2)\underset{x\to 0}{\lim }\frac{sin(x^n)}{(sinx{)}^m}=\underset{x\to 0}{\lim }\frac{x^n}{x^m}=\{\begin{array}{l}0,n>m,\\ \\ 1,n=m,\\ \\ \infty ,n<m.\end{array}& \\ & & \\ & (3)\underset{x\to 0}{\lim }\frac{tanx-sinx}{si{n}^{3}x}=\underset{x\to 0}{\lim }\frac{sinx-sinx\cdot cosx}{cosx\cdot si{n}^{3}x}=\underset{x\to 0}{\lim }\frac{1-cosx}{cosx\cdot si{n}^{2}x}=\underset{x\to 0}{\lim }\frac{1}{2cosx\cdot \frac{si{n}^{2}x}{x^2}}=\frac{1}{2}& \\ & & \\ & (4)\underset{x\to 0}{\lim }\frac{sinx-tanx}{(\sqrt[3]{1+x^2}-1)(\sqrt{1+sinx}-1)}=\underset{x\to 0}{\lim }\frac{sinx(1-secx)}{\frac{1}{3}{x}^2\cdot \frac{1}{2}sinx}=\underset{x\to 0}{\lim }\frac{-\frac{1}{2}{x}^2}{\frac{1}{6}{x}^2}=-3& \end{array}$

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$\begin{array}{rlr}& 6.ProvebrightnothingpoorSmallOfetc.priceTurn\; offsystemhaveYesNextColumnsexquality：& \end{array}$

$\begin{array}{rlr}& (1)\alpha \sim \alpha （sincebacksex）;(2)if\alpha \sim \beta ,be\beta \sim \alpha （Yescallsex）;& \\ & & \\ & (3)if\alpha \sim \beta ,\beta \sim \gamma ,be\alpha \sim \gamma （Pass\; onDeliversex）& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)becauseby\lim \frac{\alpha }{\alpha }=1,theWith\alpha \sim \alpha ;& \\ & & \\ & (2)becauseby\alpha \sim \beta ,namely\lim \frac{\alpha }{\beta }=1,theWith\lim \frac{\beta }{\alpha }=1,namely\beta \sim \alpha ;& \\ & & \\ & (3)becauseby\alpha \sim \beta ,\beta \sim \gamma ,namely\lim \frac{\alpha }{\beta }=1,\lim \frac{\beta }{\gamma }=1,theWith\lim \frac{\alpha }{\gamma }=\lim \left(\frac{\alpha }{\beta }\cdot \frac{\beta }{\gamma }\right)=\lim \frac{\alpha }{\beta }\cdot \lim \frac{\beta }{\gamma }=1,namely\alpha \sim \gamma & \end{array}$