【数学】函数极限（宇哥笔记）

定义及使用

定义

lim⁡x→⋅f(x)=A ⟺ ∀ε>0,x→⋅时,∣f(x)−A∣<ε{x→x0:∃δ>0,0<∣x−x0∣<δx→∞:∃x>0,∣x∣>x\begin{aligned} &\lim_{x\to\cdot}f(x)=A\iff\\ &\forall\varepsilon>0,x\to\cdot时,|f(x)-A|<\varepsilon\\ &\begin{cases}x\to x_0:\exists\delta>0,0<|x-x_0|<\delta\\x\to\infty:\exists x>0,|x|>x\end{cases}\\ \end{aligned} x→⋅limf(x)=A⟺∀ε>0,x→⋅时,∣f(x)−A∣<ε{x→x0:∃δ>0,0<∣x−x0∣<δx→∞:∃x>0,∣x∣>x

性质

唯一性

A唯一：左极限、右极限；左导、右导[例]求lim⁡x→0tan⁡πx∣x∣(x2−1)I+=lim⁡x→0+tan⁡πxx(x2−1)=lim⁡x→0+πxx(−1)=−πI−=lim⁡x→0−πx(−x)(−1)=π ⟹ I不∃[注]如∣x∣,ex,arctan⁡x需要考虑这种情况\begin{aligned} &A唯一：左极限、右极限；左导、右导\\ \color{maroon}[例]&求\lim_{x\to0}\frac{\tan\pi x}{|x|(x^2-1)}\\ &\color{black}I_+=\lim_{x\to0^+}\frac{\tan\pi x}{x(x^2-1)}=\lim_{x\to0^+}\frac{\pi x}{x(-1)}=-\pi\\ &I_-=\lim_{x\to0^-}\frac{\pi x}{(-x)(-1)}=\pi\\ &\implies I不\exists\\ \color{grey}[注]&如|x|,e^x,\arctan x需要考虑这种情况 \end{aligned} [例][注]A唯一：左极限、右极限；左导、右导求x→0lim∣x∣(x2−1)tanπxI+=x→0+limx(x2−1)tanπx=x→0+limx(−1)πx=−πI−=x→0−lim(−x)(−1)πx=π⟹I不∃如∣x∣,ex,arctanx需要考虑这种情况

A是一个数

A是一个数，记lim⁡x→⋅f(x)=A[例]已知lim⁡x→1f(x)存在，f(x)=x−arctan⁡(x−1)−1(x−1)3+2x2ex−1⋅lim⁡x→1f(x)lim⁡x→1f(x)=lim⁡x→1(x−1)−arctan⁡(x−1)(x−1)3+Alim⁡x→12x2ex−1 ⟹ A=lim⁡t→0t−arctan⁡tt3+2A ⟹ A=−13∴f(x)=x−arctan⁡(x−1)−1(x−1)3−23x2ex−1[注]以后会知道，"只要存在"，则有lim⁡x→⋅f(x)=A,lim⁡x→∞xn=Af′′(x0)=A;∫abf(x)dx=A;∬Df(x,y)dσ=A\begin{aligned} &A是一个数，记\lim_{x\to\cdot}f(x)=A\\ \color{maroon}[例]&已知\lim_{x\to1}f(x)存在，f(x)=\frac{x-\arctan(x-1)-1}{(x-1)^3}+2x^2e^{x-1}\cdot\lim_{x\to1}f(x)\\ &\color{black}\lim_{x\to1}f(x)=\lim_{x\to1}\frac{(x-1)-\arctan(x-1)}{(x-1)^3}+A\lim_{x\to1}2x^2e^{x-1}\\ &\implies A=\lim_{t\to0}\frac{t-\arctan t}{t^3}+2A\\ &\implies A=-\frac13\\ &\therefore f(x)=\frac{x-\arctan(x-1)-1}{(x-1)^3}-\frac23x^2e^{x-1}\\ \color{grey}[注]&以后会知道，"只要存在"，则有\lim_{x\to\cdot}f(x)=A,\lim_{x\to\infty}x_n=A\\ &f''(x_0)=A;\int_a^bf(x)dx=A;\iint_Df(x,y)d\sigma=A \end{aligned} [例][注]A是一个数，记x→⋅limf(x)=A已知x→1limf(x)存在，f(x)=(x−1)3x−arctan(x−1)−1+2x2ex−1⋅x→1limf(x)x→1limf(x)=x→1lim(x−1)3(x−1)−arctan(x−1)+Ax→1lim2x2ex−1⟹A=t→0limt3t−arctant+2A⟹A=−31∴f(x)=(x−1)3x−arctan(x−1)−1−32x2ex−1以后会知道，"只要存在"，则有x→⋅limf(x)=A,x→∞limxn=Af′′(x0)=A;∫abf(x)dx=A;∬Df(x,y)dσ=A

有界性

x→⋅,∣f(x)∣≤M[例]若lim⁡x→x0f(x)x−x0=A(存在),求lim⁡x→x0f(x)lim⁡x→x0f(x)=lim⁡x→x0f(x)x−x0⋅(x−x0)=0(前者有界函数，后者无穷小)[注]若增加“f(x)在x0处连续” ⟹ f(x)=lim⁡x→x0f(x)=0\begin{aligned} &x\to\cdot,|f(x)|\leq M\\ \color{maroon}[例]&若\lim_{x\to x_0}\frac{f(x)}{x-x_0}=A(存在),求\lim_{x\to x_0}f(x)\\ &\color{black}\lim_{x\to x_0}f(x)=\lim_{x\to x_0}\frac{f(x)}{x-x_0}\cdot(x-x_0)=0(前者有界函数，后者无穷小)\\ \color{grey}[注]&若增加“f(x)在x_0处连续”\implies f(x)=\lim_{x\to x_0}f(x)=0 \end{aligned} [例][注]x→⋅,∣f(x)∣≤M若x→x0limx−x0f(x)=A(存在),求x→x0limf(x)x→x0limf(x)=x→x0limx−x0f(x)⋅(x−x0)=0(前者有界函数，后者无穷小)若增加“f(x)在x0处连续”⟹f(x)=x→x0limf(x)=0

局部保号性

x→⋅,若A>0, ⟹ f(x)>0(局部保号)(不等式脱帽法)[例]证明：当x→0+时，0<tan⁡2x−x2<x4成立lim⁡x→0+tan⁡2x−x2x4=lim⁡x→0+(tan⁡x+x)(tan⁡x−x)x4=lim⁡x→0+2x⋅13x3x4=23<1故lim⁡x→0+[tan⁡2x−x2x4−1]<0 ⟹ tan⁡2x−x2x4−1<0即tan⁡2x−x2<x4∵x→0+时，tan⁡x>x,故tan⁡2x>x2 ⟹ 0<tan⁡2x−x2<x4\begin{aligned} &x\to\cdot,若A>0,\implies f(x)>0(局部保号)(不等式脱帽法)\\ \color{maroon}[例]&证明：当x\to0^+时，0<\tan^2x-x^2<x^4成立\\ &\color{black}\lim_{x\to0^+}\frac{\tan^2x-x^2}{x^4}=\lim_{x\to0^+}\frac{(\tan x+x)(\tan x-x)}{x^4}=\lim_{x\to0^+}\frac{2x\cdot\frac13x^3}{x^4}=\frac23<1\\ &故\lim_{x\to0^+}[\frac{\tan^2x-x^2}{x^4}-1]<0\implies\frac{\tan^2x-x^2}{x^4}-1<0\\ &即\tan^2x-x^2<x^4\\ &\because x\to0^+时，\tan x>x,故\tan^2x>x^2\implies0<\tan^2x-x^2<x^4 \end{aligned} [例]x→⋅,若A>0,⟹f(x)>0(局部保号)(不等式脱帽法)证明：当x→0+时，0<tan2x−x2<x4成立x→0+limx4tan2x−x2=x→0+limx4(tanx+x)(tanx−x)=x→0+limx42x⋅31x3=32<1故x→0+lim[x4tan2x−x2−1]<0⟹x4tan2x−x2−1<0即tan2x−x2<x4∵x→0+时，tanx>x,故tan2x>x2⟹0<tan2x−x2<x4

等式脱帽法

f(x)=A+α,lim⁡x→⋅α=0(等式脱帽法)[例]设lim⁡x→0ln⁡[1+f(x)sin⁡x]ax−1=A,a>0,a≠1,求lim⁡x→0f(x)x2ln⁡[1+f(x)sin⁡x]ax−1=A+α ⟹ ln⁡[1+f(x)sin⁡x]=(ax−1)(A+α)1+f(x)sin⁡x=e(ax−1)(A+α)f(x)=[e(ax−1)(A+α)−1]sin⁡x则lim⁡x→0f(x)x2=lim⁡x→0[e(ax−1)(A+α)−1]sin⁡xx2=lim⁡x→0(ax−1)(A+α)x=A⋅ln⁡a\begin{aligned} &f(x)=A+\alpha,\lim_{x\to\cdot}\alpha=0(等式脱帽法)\\ \color{maroon}[例]&设\lim_{x\to0}\frac{\ln[1+\frac{f(x)}{\sin x}]}{a^x-1}=A,a>0,a\neq1,求\lim_{x\to0}\frac{f(x)}{x^2}\\ &\color{black}\frac{\ln[1+\frac{f(x)}{\sin x}]}{a^x-1}=A+\alpha\implies \ln[1+\frac{f(x)}{\sin x}]=(a^x-1)(A+\alpha)\\ &1+\frac{f(x)}{\sin x}=e^{(a^x-1)(A+\alpha)}\\ &f(x)=[e^{(a^x-1)(A+\alpha)}-1]\sin x\\ &则\lim_{x\to0}\frac{f(x)}{x^2}=\lim_{x\to0}\frac{[e^{(a^x-1)(A+\alpha)}-1]\sin x}{x^2}=\lim_{x\to0}\frac{(a^x-1)(A+\alpha)}x=A\cdot\ln a \end{aligned} [例]f(x)=A+α,x→⋅limα=0(等式脱帽法)设x→0limax−1ln[1+sinxf(x)]=A,a>0,a̸=1,求x→0limx2f(x)ax−1ln[1+sinxf(x)]=A+α⟹ln[1+sinxf(x)]=(ax−1)(A+α)1+sinxf(x)=e(ax−1)(A+α)f(x)=[e(ax−1)(A+α)−1]sinx则x→0limx2f(x)=x→0limx2[e(ax−1)(A+α)−1]sinx=x→0limx(ax−1)(A+α)=A⋅lna

计算

七种未定式

分别为：00,∞∞,∞⋅0,∞−∞,∞0,00,1∞技巧：其中前两种用常规方法如泰勒公式、洛必达法则直接求解第三、四种情况遇到分数考虑通分、倒代换等，想办法等价替换、化简后三种情况一律考虑取对数其中遇到根号的情况考虑有理化、通分，遇到无穷大的情况考虑化为无穷小，如除以某个数，倒代换，通分，有理化遇到三角函数考虑无穷小乘以有界量复杂函数可以考虑导数定义法加减中把极限存在（不管是否为0）的部分拆项先算出来，乘除中把极限存在部分先分离出来\begin{aligned} &分别为：\frac00,\frac\infty\infty,\infty\cdot0,\infty-\infty,\infty^0,0^0,1^\infty\\ 技巧：&其中前两种用常规方法如泰勒公式、洛必达法则直接求解\\ &第三、四种情况遇到分数考虑通分、倒代换等，想办法等价替换、化简\\ &后三种情况一律考虑取对数\\ &其中遇到根号的情况考虑有理化、通分，\\ &遇到无穷大的情况考虑化为无穷小，如除以某个数，倒代换，通分，有理化\\ &遇到三角函数考虑无穷小乘以有界量\\ &复杂函数可以考虑导数定义法\\ &加减中把极限存在（不管是否为0）的部分拆项先算出来，乘除中把极限存在部分先分离出来 \end{aligned} 技巧：分别为：00,∞∞,∞⋅0,∞−∞,∞0,00,1∞其中前两种用常规方法如泰勒公式、洛必达法则直接求解第三、四种情况遇到分数考虑通分、倒代换等，想办法等价替换、化简后三种情况一律考虑取对数其中遇到根号的情况考虑有理化、通分，遇到无穷大的情况考虑化为无穷小，如除以某个数，倒代换，通分，有理化遇到三角函数考虑无穷小乘以有界量复杂函数可以考虑导数定义法加减中把极限存在（不管是否为0）的部分拆项先算出来，乘除中把极限存在部分先分离出来

化简先行

等价替换

sin⁡∘∼∘∼tan⁡∘∼arcsin⁡∘∼arctan⁡∘∼ln⁡(1+∘)∼e∘−1a∘−1=e∘ln⁡a−1∼∘ln⁡a(1+∘)k−1=ekln⁡(1+∘)−1∼kln⁡(1+∘)∼k∘1−cos∘=2sin2(∘2)∼2(∘2)2=∘22——————————以下是进阶版——————————x−sin⁡x∼16x3x−arcsin⁡x∼−16x3x−tan⁡x∼−13x3x−arctan⁡x∼13x3x−ln⁡(1+x)∼12x2tan⁡x−sin⁡x∼12x31−cos⁡αx∼α2x2ex−1−x∼12x21+x−1−12x∼−18x2f(x)→1时,ln⁡f(x)∼f(x)−1[补]若α=∘(β)(即lim⁡x→⋅αβ=0),则α+β∼β\begin{aligned} &\sin\circ\sim\circ\sim \tan\circ\sim \arcsin\circ\sim \arctan\circ\sim \ln(1+\circ)\sim e^\circ-1\\ &a^\circ-1=e^{\circ \ln a}-1\sim\circ\ln a\\ &(1+\circ)^k-1=e^{k\ln(1+\circ)}-1\sim k\ln(1+\circ)\sim k\circ\\ &1-cos\circ=2sin^2(\frac\circ2)\sim2(\frac\circ2)^2=\frac{\circ^2}{2}\\ &——————————以下是进阶版——————————\\ &x-\sin x \sim\frac16x^3\qquad x-\arcsin x\sim-\frac16x^3\qquad x-\tan x\sim-\frac13x^3\\ &x-\arctan x\sim\frac13x^3\qquad x-\ln(1+x)\sim\frac12x^2\qquad \tan x-\sin x\sim\frac12x^3\\ &1-\cos^\alpha x\sim\frac\alpha2x^2\qquad e^x-1-x\sim\frac12x^2\qquad\sqrt{1+x}-1-\frac12x\sim-\frac18x^2\\ &f(x)\to1时,\ln f(x)\sim f(x)-1\\ [补]&若\alpha=\circ(\beta)(即\lim_{x\to\cdot}\frac{\alpha}{\beta}=0),则\alpha+\beta\sim\beta\\ \end{aligned} [补]sin∘∼∘∼tan∘∼arcsin∘∼arctan∘∼ln(1+∘)∼e∘−1a∘−1=e∘lna−1∼∘lna(1+∘)k−1=ekln(1+∘)−1∼kln(1+∘)∼k∘1−cos∘=2sin2(2∘)∼2(2∘)2=2∘2——————————以下是进阶版——————————x−sinx∼61x3x−arcsinx∼−61x3x−tanx∼−31x3x−arctanx∼31x3x−ln(1+x)∼21x2tanx−sinx∼21x31−cosαx∼2αx2ex−1−x∼21x21+x−1−21x∼−81x2f(x)→1时,lnf(x)∼f(x)−1若α=∘(β)(即x→⋅limβα=0),则α+β∼β

1.lim⁡x→0(3+2tan⁡x)x−3x3sin⁡2x+x3cos⁡1x由x3cos⁡1x3sin⁡2x=13lim⁡x→0xcos⁡1x=0则I=lim⁡x→03x[(1+23tan⁡x)x−1]3x2=lim⁡x→0exln⁡(1+23tan⁡x)−13x2=lim⁡x→0xln⁡(1+23tan⁡x)3x2=292.lim⁡x→1−ln⁡xln⁡(1−x)=lim⁡x→1−ln⁡(1+x−1)ln⁡(1−x)=lim⁡x→1−(x−1)ln⁡(1−x)t=1−x→−lim⁡t→0+tln⁡t=03.lim⁡x→∞e−x(1+1x)x2=lim⁡x→∞e−xex2ln⁡(1+1x)=lim⁡x→∞ex2ln⁡(1+1x)−x且lim⁡x→∞(x2ln⁡(1+1x)−x)=lim⁡x→∞x2(ln⁡(1+1x)−1x)=lim⁡x→∞x2(−12)(1x)2=−124.lim⁡x→0sin⁡x+x2sin⁡1x(1+cos⁡x)ln⁡(1+x)=lim⁡x→0x2x=12(无穷小乘以有界量)5.lim⁡x→01x3[(2+cos⁡x3)x−1]=lim⁡x→01x3[(1+cos⁡x−13)x−1]=lim⁡x→01x3⋅x⋅cos⁡x−13=−166.lim⁡x→0cos⁡x−cos⁡x3sin⁡2x=lim⁡x→0cos⁡x−1+1−cos⁡x3x2=lim⁡x→0cos⁡x−1x2+lim⁡x→01−cos⁡x3x2=lim⁡x→0−14x2x2+lim⁡x→016x2x2=−112\begin{aligned} 1.&\color{maroon}\lim_{x\to0}\frac{(3+2\tan x)^x-3^x}{3\sin^2x+x^3\cos\frac1x}\\ &由\frac{x^3\cos\frac1x}{3\sin^2x}=\frac13\lim_{x\to0}x\cos\frac1x=0\\ &则I=\lim_{x\to0}\frac{3^x[(1+\frac23\tan x)^x-1]}{3x^2}\\ &=\lim_{x\to0}\frac{e^{x\ln(1+\frac23\tan x)}-1}{3x^2}\\ &=\lim_{x\to0}\frac{x\ln(1+\frac23\tan x)}{3x^2}=\frac29\\ 2.&\color{maroon}\lim_{x\to1^-}\ln x\ln(1-x)\\ =&\lim_{x\to1^-}\ln(1+x-1)\ln(1-x)\\ =&\lim_{x\to1^-}(x-1)\ln(1-x)\underrightarrow{t=1-x}-\lim_{t\to0^+}t\ln t=0\\ 3.&\color{maroon}\lim_{x\to\infty}e^{-x}(1+\frac1x)^{x^2}\\ =&\lim_{x\to\infty}e^{-x}e^{x^2\ln(1+\frac1x)}\\ =&\lim_{x\to\infty}e^{x^2\ln(1+\frac1x)-x}\\ &且\lim_{x\to\infty}(x^2\ln(1+\frac1x)-x)\\ =&\lim_{x\to\infty}x^2(\ln(1+\frac1x)-\frac1x)\\ =&\lim_{x\to\infty}x^2(-\frac12)(\frac1x)^2=-\frac12\\ 4.&\color{maroon}\lim_{x\to0}\frac{\sin x+x^2\sin\frac1x}{(1+\cos x)\ln(1+x)}\\ =&\lim_{x\to0}\frac{x}{2x}=\frac12(无穷小乘以有界量)\\ 5.&\color{maroon}\lim_{x\to0}\frac1{x^3}[(\frac{2+\cos x}{3})^x-1]\\ =&\lim_{x\to0}\frac1{x^3}[(1+\frac{\cos x-1}{3})^x-1]\\ =&\lim_{x\to0}\frac1{x^3}\cdot x\cdot \frac{\cos x-1}3\\ =&-\frac16\\ 6.&\color{maroon}\lim_{x\to0}\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x}\\ =&\lim_{x\to0}\frac{\sqrt{\cos x}-1+1-\sqrt[3]{\cos x}}{x^2}\\ =&\lim_{x\to0}\frac{\sqrt{\cos x}-1}{x^2}+\lim_{x\to0}\frac{1-\sqrt[3]{\cos x}}{x^2}\\ =&\lim_{x\to0}\frac{-\frac14x^2}{x^2}+\lim_{x\to0}\frac{\frac16x^2}{x^2}\\ =&-\frac1{12}\\ \end{aligned} 1.2.==3.====4.=5.===6.====x→0lim3sin2x+x3cosx1(3+2tanx)x−3x由3sin2xx3cosx1=31x→0limxcosx1=0则I=x→0lim3x23x[(1+32tanx)x−1]=x→0lim3x2exln(1+32tanx)−1=x→0lim3x2xln(1+32tanx)=92x→1−limlnxln(1−x)x→1−limln(1+x−1)ln(1−x)x→1−lim(x−1)ln(1−x)t=1−x−t→0+limtlnt=0x→∞lime−x(1+x1)x2x→∞lime−xex2ln(1+x1)x→∞limex2ln(1+x1)−x且x→∞lim(x2ln(1+x1)−x)x→∞limx2(ln(1+x1)−x1)x→∞limx2(−21)(x1)2=−21x→0lim(1+cosx)ln(1+x)sinx+x2sinx1x→0lim2xx=21(无穷小乘以有界量)x→0limx31[(32+cosx)x−1]x→0limx31[(1+3cosx−1)x−1]x→0limx31⋅x⋅3cosx−1−61x→0limsin2xcosx−3cosxx→0limx2cosx−1+1−3cosxx→0limx2cosx−1+x→0limx21−3cosxx→0limx2−41x2+x→0limx261x2−121

恒等变形

提取公因式

换元（倒代换、平移替换）

通分

因式分解公式

取对数

有理化

1.lim⁡x→0(1+x1−e−x−1x)=lim⁡x→0x+x2−1+e−x(1−e−x)⋅x(通分)=lim⁡x→01+2x−e−x2x=1+12=322.lim⁡x→∞e−x(1+1x)x2=lim⁡x→∞e−xex2ln⁡(1+1x)(取对数)=elim⁡x→∞[x2ln⁡(1+1x)−x]令x=1t→elim⁡t→0[ln⁡(1+t)t2−1t](倒代换)=elim⁡t→0ln⁡(1+t)−tt2=e−123.lim⁡x→0+xln⁡(ln⁡x−1ln⁡x+1)=elim⁡x→0+ln⁡(ln⁡x−1ln⁡x+1)⋅ln⁡x=elim⁡x→0+ln⁡(1−2ln⁡x+1)⋅ln⁡x=elim⁡x→0+−2ln⁡x+1⋅ln⁡x=e−24.lim⁡x→0+(sin⁡xx)11−cos⁡x=elim⁡x→0+11−cos⁡x(sin⁡xx−1)=elim⁡x→0+sin⁡x−x12x2⋅x=e−135.lim⁡x→∞(x6+x56−x6−x56)令x=1t→lim⁡t→0+1t6+1t56−1t6−1t56=lim⁡t→0+(1+t6t−1−t6t)=lim⁡t→0+(1+t)16−(1−t)16t=lim⁡t→0+1+16t−(1−16t)+∘(t)t=lim⁡t→0+13t+∘(t)t=13\begin{aligned} 1.&\color{maroon}\lim_{x\to0}(\frac{1+x}{1-e^{-x}}-\frac1x)\\ =&\lim_{x\to0}\frac{x+x^2-1+e^{-x}}{(1-e^{-x})\cdot x}(通分)\\ =&\lim_{x\to0}\frac{1+2x-e^{-x}}{2x}=1+\frac12=\frac32\\ 2.&\color{maroon}\lim_{x\to\infty}e^{-x}(1+\frac1x)^{x^2}\\ =&\lim_{x\to\infty}e^{-x}e^{x^2\ln(1+\frac1x)}(取对数)\\ =&e^{\lim_{x\to\infty}[x^2\ln(1+\frac1x)-x]}\\ \underrightarrow{令x=\frac1t}&e^{\lim_{t\to0}[\frac{\ln(1+t)}{t^2}-\frac1t]}(倒代换)\\ =&e^{\lim_{t\to0}\frac{\ln(1+t)-t}{t^2}}=e^{-\frac12}\\ 3.&\color{maroon}\lim_{x\to0^+}x^{\ln(\frac{\ln x-1}{\ln x+1})}\\ =&e^{\lim_{x\to0^+}\ln(\frac{\ln x-1}{\ln x+1})\cdot\ln x}\\ =&e^{\lim_{x\to0^+}\ln(1-\frac2{\ln x+1})\cdot \ln x}\\ =&e^{\lim_{x\to0^+}\frac{-2}{\ln x+1}\cdot\ln x}=e^{-2}\\ 4.&\color{maroon}\lim_{x\to0^+}(\frac{\sin x}{x})^\frac1{1-\cos x}\\ =&e^{\lim_{x\to0^+}\frac1{1-\cos x}(\frac{\sin x}x-1)}\\ =&e^{\lim_{x\to0^+}\frac{\sin x-x}{\frac12x^2\cdot x}}\\ =&e^{-\frac13}\\ 5.&\color{maroon}\lim_{x\to\infty}(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5})\\ \underrightarrow{令x=\frac1t}&\lim_{t\to0^+}\sqrt[6]{\frac1{t^6}+\frac1{t^5}}-\sqrt[6]{\frac1{t^6}-\frac1{t^5}}\\ =&\lim_{t\to0^+}(\frac{\sqrt[6]{1+t}}{t}-\frac{\sqrt[6]{1-t}}t)\\ =&\lim_{t\to0^+}\frac{(1+t)^\frac16-(1-t)^\frac16}{t}\\ =&\lim_{t\to0^+}\frac{1+\frac16t-(1-\frac16t)+\circ(t)}t\\ =&\lim_{t\to0^+}\frac{\frac13t+\circ(t)}{t}=\frac13\\ \end{aligned} 1.==2.==令x=t1=3.===4.===5.令x=t1====x→0lim(1−e−x1+x−x1)x→0lim(1−e−x)⋅xx+x2−1+e−x(通分)x→0lim2x1+2x−e−x=1+21=23x→∞lime−x(1+x1)x2x→∞lime−xex2ln(1+x1)(取对数)elimx→∞[x2ln(1+x1)−x]elimt→0[t2ln(1+t)−t1](倒代换)elimt→0t2ln(1+t)−t=e−21x→0+limxln(lnx+1lnx−1)elimx→0+ln(lnx+1lnx−1)⋅lnxelimx→0+ln(1−lnx+12)⋅lnxelimx→0+lnx+1−2⋅lnx=e−2x→0+lim(xsinx)1−cosx1elimx→0+1−cosx1(xsinx−1)elimx→0+21x2⋅xsinx−xe−31x→∞lim(6x6+x5−6x6−x5)t→0+lim6t61+t51−6t61−t51t→0+lim(t61+t−t61−t)t→0+limt(1+t)61−(1−t)61t→0+limt1+61t−(1−61t)+∘(t)t→0+limt31t+∘(t)=31

6.lim⁡x→01+tan⁡x−1+sin⁡xxln⁡(1+x)−x2=lim⁡x→01−12x3⋅tan⁡x−sin⁡x1+tan⁡x+1+sin⁡x=−127.lim⁡x→0cos⁡x−cos⁡x3sin⁡2x[分析]令cos⁡x6=t,原式=lim⁡t→1−t3−t21−t12=−lim⁡t→1−t−1t12−1=−lim⁡t→1−t−1(t−1)(t11+t10+…+1)=−112[注]an−bn=(a−b)(an−1+an−2b+…+abn−2+bn−1)8.lim⁡x→+∞(x+1+x2)1x=elim⁡x→+∞1xln⁡(x+1+x2)=elim⁡x→+∞1x+1+x2(1+12(1+x2)−12⋅2x)=elim⁡x→+∞11+x2=e0=19.lim⁡x→0(1+x2)(1−cos⁡2x)−2x2x4(拆)=lim⁡x→01−cos⁡2x−2x2x4+lim⁡x→0x2(1−cos⁡2x)x4=lim⁡x→01−(1−12(2x)2+124(2x)4+∘(x4))−2x2x4+2=lim⁡x→0−1624x4x4+2=4310.lim⁡x→01−x2sin⁡2x−tan⁡2xx2[ln⁡(1+x)]2=lim⁡x→01−x2sin⁡2x−sin⁡2x+sin⁡2x−tan⁡2xx4=lim⁡x→0sin⁡x(1−x2−1)x4+lim⁡x→0sin⁡2x−tan⁡2xx4=−12+lim⁡x→0(sin⁡x+tan⁡x)(sin⁡x+tan⁡)x4=−12+lim⁡x→0(sin⁡x+tan⁡x)(−12x3)x4=−12+(−12)(1+1)=−32\begin{aligned} 6.&\color{maroon}\lim_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x\ln(1+x)-x^2}\\ =&\lim_{x\to0}\frac{1}{-\frac12x^3}\cdot\frac{\tan x-\sin x}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}=-\frac12\\ 7.&\color{maroon}\lim_{x\to0}\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x}\\ [分析]&令\sqrt[6]{\cos x}=t,原式=\lim_{t\to1^-}\frac{t^3-t^2}{1-t^{12}}=-\lim_{t\to1^-}\frac{t-1}{t^{12}-1}\\ =&-\lim_{t\to1^-}\frac{t-1}{(t-1)(t^{11}+t^{10}+\ldots+1)}=-\frac1{12}\\ [注]&\color{red}a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})\\ 8.&\color{maroon}\lim_{x\to+\infty}(x+\sqrt{1+x^2})^\frac1x\\ =&e^{\lim_{x\to+\infty}\frac1x\ln(x+\sqrt{1+x^2})}=e^{\lim_{x\to+\infty}\frac{1}{x+\sqrt{1+x^2}}(1+\frac12(1+x^2)^{-\frac12}\cdot2x)}\\ =&e^{\lim_{x\to+\infty}\frac{1}{\sqrt{1+x^2}}}=e^0=1\\ 9.&\color{maroon}\lim_{x\to0}\frac{(1+x^2)(1-\cos2x)-2x^2}{x^4}(拆)\\ =&\lim_{x\to0}\frac{1-\cos2x-2x^2}{x^4}+\lim_{x\to0}\frac{x^2(1-\cos2x)}{x^4}\\ =&\lim_{x\to0}\frac{1-(1-\frac12(2x)^2+\frac1{24}(2x)^4+\circ(x^4))-2x^2}{x^4}+2\\ =&\lim_{x\to0}\frac{-\frac{16}{24}x^4}{x^4}+2=\frac43\\ 10.&\color{maroon}\lim_{x\to0}\frac{\sqrt{1-x^2}\sin^2x-\tan^2x}{x^2[\ln(1+x)]^2}\\ =&\lim_{x\to0}\frac{\sqrt{1-x^2}\sin^2x-\sin^2x+\sin^2x-\tan^2x}{x^4}\\ =&\lim_{x\to0}\frac{\sin^x(\sqrt{1-x^2}-1)}{x^4}+\lim_{x\to0}\frac{\sin^2x-\tan^2x}{x^4}\\ =&-\frac12+\lim_{x\to0}\frac{(\sin x+\tan x)(\sin x+\tan )}{x^4}\\ =&-\frac12+\lim_{x\to0}\frac{(\sin x+\tan x)(-\frac12x^3)}{x^4}\\ =&-\frac12+(-\frac12)(1+1)=-\frac32\\ \end{aligned} 6.=7.[分析]=[注]8.==9.===10.=====x→0limxln(1+x)−x21+tanx−1+sinxx→0lim−21x31⋅1+tanx+1+sinxtanx−sinx=−21x→0limsin2xcosx−3cosx令6cosx=t,原式=t→1−lim1−t12t3−t2=−t→1−limt12−1t−1−t→1−lim(t−1)(t11+t10+…+1)t−1=−121an−bn=(a−b)(an−1+an−2b+…+abn−2+bn−1)x→+∞lim(x+1+x2)x1elimx→+∞x1ln(x+1+x2)=elimx→+∞x+1+x21(1+21(1+x2)−21⋅2x)elimx→+∞1+x21=e0=1x→0limx4(1+x2)(1−cos2x)−2x2(拆)x→0limx41−cos2x−2x2+x→0limx4x2(1−cos2x)x→0limx41−(1−21(2x)2+241(2x)4+∘(x4))−2x2+2x→0limx4−2416x4+2=34x→0limx2[ln(1+x)]21−x2sin2x−tan2xx→0limx41−x2sin2x−sin2x+sin2x−tan2xx→0limx4sinx(1−x2−1)+x→0limx4sin2x−tan2x−21+x→0limx4(sinx+tanx)(sinx+tan)−21+x→0limx4(sinx+tanx)(−21x3)−21+(−21)(1+1)=−23

11.lim⁡x→∞x2(a1x+a−1x−2),其中常数a>0令1x=tI=lim⁡t→01t2(at+a−t−2)=lim⁡t→0atln⁡a+a−t⋅ln⁡a−12t=ln⁡2a12.lim⁡x→0(cos⁡xcos⁡2x)1x2I=eA,其中A=lim⁡x→01x2(cos⁡xcos⁡2x−1)=lim⁡x→0cos⁡x−cos⁡2xx2cos⁡2x=lim⁡x→0cos⁡x−cos⁡2xx2=lim⁡x→0cos⁡x−1+1−cos⁡2xx2=lim⁡x→0cos⁡x−1x2+lim⁡x→01−cos⁡2xx2=−12+lim⁡x→012(2x)2x2=32故I=e3213.lim⁡x→+∞[(x3+x2−tan⁡1x)e1x−1+x6]=lim⁡x→+∞[(x3+x2)e1x−1+x6]−lim⁡x→+∞tan⁡1x⋅e1xt=1x→lim⁡t→0+[(1t3+12t)et−1+1t6]−0=lim⁡t→0+(2+t2)et−21+t6+2−22t3=lim⁡t→0+(2+t2)et−22t3−2lim⁡t→0+1+t6−12t3=lim⁡t→0+(2+t2)(1+t+12t2+16t3+∘(t3))−22t3−0=lim⁡t→0+2t+2t2+43t32t3=∞14.lim⁡x→0[ax−(1x2−a2)ln⁡(1+ax)].其中a≠0=lim⁡x→0[ax−1x2ln⁡(1+ax)]+a2lim⁡x→0ln⁡(1+ax)=lim⁡x→0ax−ln⁡(1+ax)x2+0=lim⁡x→012(ax)2x2=12a215.lim⁡x→0(1+x)1x−(1+2x)12xsin⁡x=lim⁡x→0eln⁡(1+x)/x−eln⁡(1+2x)/2xx=lim⁡x→0eln⁡(1+2x)/2x(eln⁡(1+x)/x−ln⁡(1+2x)/2x−1)x=elim⁡x→0ln⁡(1+x)/x−ln⁡(1+2x)/2xx=elim⁡x→02ln⁡(1+x)−ln⁡(1+2x)2x2=e⋅2(x−12x2)+∘(x2)−(2x−12(2x)2)−∘(x2)2x2=e⋅12=e2\begin{aligned} 11.&\color{maroon}\lim_{x\to\infty}x^2(a^{\frac1x}+a^{-\frac1x}-2),其中常数a>0\\ &令\frac1x=t\\ I&=\lim_{t\to0}\frac1{t^2}(a^t+a^{-t}-2)\\ &=\lim_{t\to0}\frac{a^t\ln a+a^{-t}\cdot\ln a^{-1}}{2t}\\ &=\ln^2a\\ 12.&\color{maroon}\lim_{x\to0}(\frac{\cos x}{\cos2x})^{\frac1{x^2}}\\ I&=e^A,其中A=\lim_{x\to0}\frac1{x^2}(\frac{\cos x}{\cos2x}-1)\\ &=\lim_{x\to0}\frac{\cos x-\cos2x}{x^2\cos2x}\\ &=\lim_{x\to0}\frac{\cos x-\cos2x}{x^2}\\ &=\lim_{x\to0}\frac{\cos x-1+1-\cos2x}{x^2}\\ &=\lim_{x\to0}\frac{\cos x-1}{x^2}+\lim_{x\to0}\frac{1-\cos2x}{x^2}\\ &=-\frac12+\lim_{x\to0}\frac{\frac12(2x)^2}{x^2}\\ &=\frac32\quad 故I=e^{\frac32}\\ 13.&\color{maroon}\lim_{x\to+\infty}[(x^3+\frac x2-\tan\frac1x)e^{\frac1x}-\sqrt{1+x^6}]\\ =&\lim_{x\to+\infty}[(x^3+\frac x2)e^{\frac1x}-\sqrt{1+x^6}]-\lim_{x\to+\infty}\tan\frac1x\cdot e^{\frac1x}\\ \underrightarrow{t=\frac1x}&\lim_{t\to0^+}[(\frac1{t^3}+\frac1{2t})e^t-\sqrt{1+\frac1{t^6}}]-0\\ =&\lim_{t\to0^+}\frac{(2+t^2)e^t-2\sqrt{1+t^6}+2-2}{2t^3}\\ =&\lim_{t\to0^+}\frac{(2+t^2)e^t-2}{2t^3}-2\lim_{t\to0^+}\frac{\sqrt{1+t^6}-1}{2t^3}\\ =&\lim_{t\to0^+}\frac{(2+t^2)(1+t+\frac12t^2+\frac16t^3+\circ(t^3))-2}{2t^3}-0\\ =&\lim_{t\to0^+}\frac{2t+2t^2+\frac43t^3}{2t^3}=\infty\\ 14.&\color{maroon}\lim_{x\to0}[\frac ax-(\frac1{x^2}-a^2)\ln(1+ax)].其中a\neq0\\ =&\lim_{x\to0}[\frac ax-\frac1{x^2}\ln(1+ax)]+a^2\lim_{x\to0}\ln(1+ax)\\ =&\lim_{x\to0}\frac{ax-\ln(1+ax)}{x^2}+0\\ =&\lim_{x\to0}\frac{\frac12(ax)^2}{x^2}=\frac12a^2\\ 15.&\color{maroon}\lim_{x\to0}\frac{(1+x)^{\frac1x}-(1+2x)^{\frac1{2x}}}{\sin x}\\ =&\lim_{x\to0}\frac{e^{\ln(1+x)/x}-e^{\ln(1+2x)/2x}}{x}\\ =&\lim_{x\to0}\frac{e^{\ln(1+2x)/2x}(e^{\ln(1+x)/x-\ln(1+2x)/2x}-1)}x\\ =&e\lim_{x\to0}\frac{\ln(1+x)/x-\ln(1+2x)/2x}{x}\\ =&e\lim_{x\to0}\frac{2\ln(1+x)-\ln(1+2x)}{2x^2}\\ =&e\cdot\frac{2(x-\frac12x^2)+\circ(x^2)-(2x-\frac12(2x)^2)-\circ(x^2)}{2x^2}\\ =&e\cdot\frac12=\frac e2\\ \end{aligned} 11.I12.I13.=t=x1====14.===15.======x→∞limx2(ax1+a−x1−2),其中常数a>0令x1=t=t→0limt21(at+a−t−2)=t→0lim2tatlna+a−t⋅lna−1=ln2ax→0lim(cos2xcosx)x21=eA,其中A=x→0limx21(cos2xcosx−1)=x→0limx2cos2xcosx−cos2x=x→0limx2cosx−cos2x=x→0limx2cosx−1+1−cos2x=x→0limx2cosx−1+x→0limx21−cos2x=−21+x→0limx221(2x)2=23故I=e23x→+∞lim[(x3+2x−tanx1)ex1−1+x6]x→+∞lim[(x3+2x)ex1−1+x6]−x→+∞limtanx1⋅ex1t→0+lim[(t31+2t1)et−1+t61]−0t→0+lim2t3(2+t2)et−21+t6+2−2t→0+lim2t3(2+t2)et−2−2t→0+lim2t31+t6−1t→0+lim2t3(2+t2)(1+t+21t2+61t3+∘(t3))−2−0t→0+lim2t32t+2t2+34t3=∞x→0lim[xa−(x21−a2)ln(1+ax)].其中a̸=0x→0lim[xa−x21ln(1+ax)]+a2x→0limln(1+ax)x→0limx2ax−ln(1+ax)+0x→0limx221(ax)2=21a2x→0limsinx(1+x)x1−(1+2x)2x1x→0limxeln(1+x)/x−eln(1+2x)/2xx→0limxeln(1+2x)/2x(eln(1+x)/x−ln(1+2x)/2x−1)ex→0limxln(1+x)/x−ln(1+2x)/2xex→0lim2x22ln(1+x)−ln(1+2x)e⋅2x22(x−21x2)+∘(x2)−(2x−21(2x)2)−∘(x2)e⋅21=2e

及时提出极限=c≠0的因式

1.lim⁡x→3+cos⁡xln⁡(x−3)ln⁡(ex−e3)=cos⁡3lim⁡x→3+ln⁡(x−3)ln⁡(ex−e3)=cos⁡3lim⁡x→3+1x−3⋅ex−e3ex=cos⁡3e3lim⁡x→3+ex−e3x−3=cos⁡3e3lim⁡x→3+ex1=cos⁡3\begin{aligned} 1.&\color{maroon}\lim_{x\to3^+}\frac{\cos x\ln(x-3)}{\ln(e^x-e^3)}\\ =&\cos3\lim_{x\to3^+}\frac{\ln(x-3)}{\ln(e^x-e^3)}\\ =&\cos3\lim_{x\to3^+}\frac1{x-3}\cdot\frac{e^x-e^3}{e^x}\\ =&\frac{\cos3}{e^3}\lim_{x\to3^+}\frac{e^x-e^3}{x-3}\\ =&\frac{\cos3}{e^3}\lim_{x\to3^+}\frac{e^x}1=\cos3\\ \end{aligned} 1.====x→3+limln(ex−e3)cosxln(x−3)cos3x→3+limln(ex−e3)ln(x−3)cos3x→3+limx−31⋅exex−e3e3cos3x→3+limx−3ex−e3e3cos3x→3+lim1ex=cos3

洛必达法则

1.lim⁡x→⋅f(x)g(x)=lim⁡x→⋅f′(x)g′(x)2.lim⁡x→a∫axf(t)dt∫axg(t)dt=lim⁡x→af(x)g(x)3.lim⁡x→a∫aφ(x)f(t)dt∫aϕ(x)g(t)dt=lim⁡x→af[φ(x)]⋅φ′(x)g[ϕ(x)]⋅ϕ′(x)[注]1.00型2.可导3.结果为0,c≠0,∞\begin{aligned} 1.&\lim_{x\to\cdot}\frac{f(x)}{g(x)}=\lim_{x\to\cdot}\frac{f'(x)}{g'(x)}\\ 2.&\lim_{x\to a}\frac{\int_a^xf(t)dt}{\int_a^xg(t)dt}=\lim_{x\to a}\frac{f(x)}{g(x)}\\ 3.&\lim_{x\to a}\frac{\int_a^\varphi(x)f(t)dt}{\int_a^\phi(x)g(t)dt}=\lim_{x\to a}\frac{f[\varphi(x)]\cdot\varphi'(x)}{g[\phi(x)]\cdot\phi'(x)}\\ [注]&1.\frac00型\quad2.可导\quad3.结果为0,c\neq0,\infty\\ \end{aligned} 1.2.3.[注]x→⋅limg(x)f(x)=x→⋅limg′(x)f′(x)x→alim∫axg(t)dt∫axf(t)dt=x→alimg(x)f(x)x→alim∫aϕ(x)g(t)dt∫aφ(x)f(t)dt=x→alimg[ϕ(x)]⋅ϕ′(x)f[φ(x)]⋅φ′(x)1.00型2.可导3.结果为0,c̸=0,∞

1.lim⁡x→0∫0xsin⁡2t4+t2∫0x(t+1−1)dtdt=lim⁡x→0∫0xsin⁡2t4+t2dt∫0x(t+1−1)dt=lim⁡x→0sin⁡2x4+x2x+1−1=lim⁡x→02x2(12x)=22.lim⁡x→∞∫1x[t2(e1t−1)−t]dtx2ln⁡(1+1x)=lim⁡x→∞=x2(e1x−1)−x1x=1t→lim⁡t→0+(et−1t2−1t)=lim⁡t→0+et−1−tt2=12\begin{aligned} 1.&\color{maroon}\lim_{x\to0}\int_0^x\frac{\sin2t}{\sqrt{4+t^2}\int_0^x(\sqrt{t+1}-1)dt}dt\\ =&\lim_{x\to0}\frac{\int_0^x\frac{\sin2t}{\sqrt{4+t^2}}dt}{\int_0^x(\sqrt{t+1}-1)dt}\\ =&\lim_{x\to0}\frac{\frac{\sin2x}{\sqrt{4+x^2}}}{\sqrt{x+1}-1}\\ =&\lim_{x\to0}\frac{2x}{2(\frac12x)}=2\\ 2.&\color{maroon}\lim_{x\to\infty}\frac{\int_1^x[t^2(e^\frac1t-1)-t]dt}{x^2\ln(1+\frac1x)}\\ =&\lim_{x\to\infty}=\frac{x^2(e^\frac1x-1)-x}{1}\underrightarrow{x=\frac1t}\lim_{t\to0^+}(\frac{e^t-1}{t^2}-\frac1t)\\ =&\lim_{t\to0^+}\frac{e^t-1-t}{t^2}=\frac12\\ \end{aligned} 1.===2.==x→0lim∫0x4+t2∫0x(t+1−1)dtsin2tdtx→0lim∫0x(t+1−1)dt∫0x4+t2sin2tdtx→0limx+1−14+x2sin2xx→0lim2(21x)2x=2x→∞limx2ln(1+x1)∫1x[t2(et1−1)−t]dtx→∞lim=1x2(ex1−1)−xx=t1t→0+lim(t2et−1−t1)t→0+limt2et−1−t=21

泰勒公式

公式

泰勒发现，任何可导函数都可以写成∑anxn,这样，任何可导函数都具有了统一美，也可以统一计算当x→0时，就如以下常见的函数泰勒展开式，最后的∘(x3)被称作佩亚诺余项sin⁡x=x−16x3+15!x5−x77!+⋯+(−1)nx2n+1(2n+1)!+⋯=∑n=0∞(−1)n⋅x2n+1(2n+1)!ex=1+x+x22+x36+⋯+xnn!+⋯=∑n=0∞xnn!cos⁡x=1−12x2+124x4−x66!+⋯+(−1)nx2n(2n)!+⋯=∑n=0∞(−1)n⋅x2n(2n)!ln⁡(1+x)=x−12x2+13x3−⋯+(−1)n−1xnn+⋯=∑n=1∞(−1)n−1⋅xnn(1+x)α=1+αx+α(α−1)2x2+∘(x2)tan⁡x=x+13x3+∘(x3)arcsin⁡x=x+16x3+1120x5+∘(x5)arctan⁡x=x−13x3+∘(x3)\begin{aligned} &泰勒发现，任何可导函数都可以写成\sum a_nx^n,这样，任何可导函数都具有了统一美，也可以统一计算\\ &当x\to0时，就如以下常见的函数泰勒展开式，最后的\circ(x^3)被称作佩亚诺余项\\ &\sin x=x-\frac16x^3+\frac{1}{5!}x^5-\frac{x^7}{7!}+\cdots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+\cdots=\sum_{n=0}^\infty(-1)^n\cdot\frac{x^{2n+1}}{(2n+1)!}\\ &e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots+\frac{x^n}{n!}+\cdots=\sum_{n=0}^{\infty}\frac{x^n}{n!}\\ &\cos x=1-\frac12x^2+\frac1{24}x^4-\frac{x^6}{6!}+\cdots+(-1)^n\frac{x^{2n}}{(2n)!}+\cdots=\sum_{n=0}^\infty(-1)^n\cdot\frac{x^{2n}}{(2n)!}\\ &\ln(1+x)=x-\frac12x^2+\frac13x^3-\cdots+(-1)^{n-1}\frac{x^n}n+\cdots=\sum_{n=1}^\infty(-1)^{n-1}\cdot\frac{x^n}n\\ &(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2}x^2+\circ(x^2)\quad\tan x=x+\frac13x^3+\circ(x^3)\\ &\arcsin x=x+\frac16x^3+\frac1{120}x^5+\circ(x^5)\quad\arctan x=x-\frac13x^3+\circ(x^3)\\ \end{aligned} 泰勒发现，任何可导函数都可以写成∑anxn,这样，任何可导函数都具有了统一美，也可以统一计算当x→0时，就如以下常见的函数泰勒展开式，最后的∘(x3)被称作佩亚诺余项sinx=x−61x3+5!1x5−7!x7+⋯+(−1)n(2n+1)!x2n+1+⋯=n=0∑∞(−1)n⋅(2n+1)!x2n+1ex=1+x+2x2+6x3+⋯+n!xn+⋯=n=0∑∞n!xncosx=1−21x2+241x4−6!x6+⋯+(−1)n(2n)!x2n+⋯=n=0∑∞(−1)n⋅(2n)!x2nln(1+x)=x−21x2+31x3−⋯+(−1)n−1nxn+⋯=n=1∑∞(−1)n−1⋅nxn(1+x)α=1+αx+2α(α−1)x2+∘(x2)tanx=x+31x3+∘(x3)arcsinx=x+61x3+1201x5+∘(x5)arctanx=x−31x3+∘(x3)

展开原则

1.AB型，上下同阶2.A−B型，幂次最低，即将A,B分别展开至系数不相等的x的最低次幂为止\begin{aligned} &1.\frac AB型，上下同阶\\ &2.A-B型，幂次最低，即将A,B分别展开至系数不相等的x的最低次幂为止\\ \end{aligned} 1.BA型，上下同阶2.A−B型，幂次最低，即将A,B分别展开至系数不相等的x的最低次幂为止

1.lim⁡x→0arctan⁡x−xx3=lim⁡x→0x−13x3−xx3=−132.x→0时，cos⁡x−e−x22与axb为等价无穷小，求a,bcos⁡x=1−12x2+124x4+∘(x4)e−x22=1−x22+x48+∘(x4)I=−112x4+∘(x4)∴a=−112,b=43.lim⁡x→01+12x2−1+x2(cos⁡x−ex22)sin⁡x22=lim⁡x→018x4−x42=−144.lim⁡x→0∑n=1∞2nn!x2narctan⁡x2=lim⁡x→02x2+222!x4+…x2=25.lim⁡x→0ex+ln⁡(1−x)−1x−arctan⁡x=lim⁡x→0ex+ln⁡(1−x)−113x3=lim⁡x→01+12x2+x+16x3+∘(x3)−x−12x2−13x3+∘(x3)−113x3=lim⁡x→0−16x313x3=−12\begin{aligned} 1.&\color{maroon}\lim_{x\to0}\frac{\arctan x-x}{x^3}\\ =&\lim_{x\to0}\frac{x-\frac13x^3-x}{x^3}=-\frac13\\ 2.&\color{maroon}x\to0时，\cos x-e^{-\frac{x^2}{2}}与ax^b为等价无穷小，求a,b\\ &\cos x=1-\frac12x^2+\frac1{24}x^4+\circ(x^4)\ \ \ \ e^{-\frac{x^2}{2}}=1-\frac{x^2}{2}+\frac{x^4}{8}+\circ(x^4)\\ &I=-\frac1{12}x^4+\circ(x^4)\ \ \ \therefore a=-\frac1{12},b=4\\ 3.&\color{maroon}\lim_{x\to0}\frac{1+\frac12x^2-\sqrt{1+x^2}}{(\cos x-e^{\frac{x^2}2})\sin\frac{x^2}2}\\ =&\lim_{x\to0}\frac{\frac18x^4}{-\frac{x^4}2}=-\frac14\\ 4.&\color{maroon}\lim_{x\to0}\frac{\sum_{n=1}^\infty\frac{2^n}{n!}x^{2n}}{\arctan x^2}\\ =&\lim_{x\to0}\frac{2x^2+\frac{2^2}{2!}x^4+\ldots}{x^2}=2\\ 5.&\color{maroon}\lim_{x\to0}\frac{e^x+\ln(1-x)-1}{x-\arctan x}\\ =&\lim_{x\to0}\frac{e^x+\ln(1-x)-1}{\frac13x^3}\\ =&\lim_{x\to0}\frac{1+\frac12x^2+x+\frac16x^3+\circ(x^3)-x-\frac12x^2-\frac13x^3+\circ(x^3)-1}{\frac13x^3}\\ =&\lim_{x\to0}\frac{-\frac16x^3}{\frac13x^3}=-\frac12\\ \end{aligned} 1.=2.3.=4.=5.===x→0limx3arctanx−xx→0limx3x−31x3−x=−31x→0时，cosx−e−2x2与axb为等价无穷小，求a,bcosx=1−21x2+241x4+∘(x4) e−2x2=1−2x2+8x4+∘(x4)I=−121x4+∘(x4) ∴a=−121,b=4x→0lim(cosx−e2x2)sin2x21+21x2−1+x2x→0lim−2x481x4=−41x→0limarctanx2∑n=1∞n!2nx2nx→0limx22x2+2!22x4+…=2x→0limx−arctanxex+ln(1−x)−1x→0lim31x3ex+ln(1−x)−1x→0lim31x31+21x2+x+61x3+∘(x3)−x−21x2−31x3+∘(x3)−1x→0lim31x3−61x3=−21

中值定理

拉格朗日中值定理：设f(x){[a,b]连续(a,b)可导 ⟹ f(b)−f(a)=f′(ξ)(b−a),∃ξ∈(a,b)\begin{aligned} &拉格朗日中值定理：设f(x)\begin{cases}[a,b]连续\\(a,b)可导\end{cases}\implies f(b)-f(a)=f'(\xi)(b-a),\exists\xi\in(a,b)\\ \end{aligned} 拉格朗日中值定理：设f(x){[a,b]连续(a,b)可导⟹f(b)−f(a)=f′(ξ)(b−a),∃ξ∈(a,b)

1.lim⁡x→0(1+x)2x−e2[1−ln⁡(1+x)]x=lim⁡x→0(1+x)2x−e2x+e2lim⁡x→0ln⁡(1+x)x=lim⁡x→0e2ln⁡(1+x)/x−e2x+e2=lim⁡x→0eξ(2ln⁡(1+x)x−2)x+e2=2e2lim⁡x→0ln⁡(1+x)x−1x+e2=2e2lim⁡x→0ln⁡(1+x)−xx2+e2=2e2lim⁡x→0−12x2x2+e2=02.lim⁡x→∞(x6+x56−x6−x56)=lim⁡x→∞f′(ξ)(2x5)=lim⁡x→∞16ξ−56(2x5)=lim⁡x→∞16⋅x6(−56)(2x5)=lim⁡x→∞16x−5(2x5)=13\begin{aligned} 1.&\color{maroon}\lim_{x\to0}\frac{(1+x)^{\frac2x}-e^2[1-\ln(1+x)]}{x}\\ =&\lim_{x\to0}\frac{(1+x)^{\frac2x}-e^2}{x}+e^2\lim_{x\to0}\frac{\ln(1+x)}{x}\\ =&\lim_{x\to0}\frac{e^{2\ln(1+x)/x}-e^2}{x}+e^2\\ =&\lim_{x\to0}\frac{e^\xi(\frac{2\ln(1+x)}x-2)}{x}+e^2\\ =&2e^2\lim_{x\to0}\frac{\frac{\ln(1+x)}{x}-1}{x}+e^2\\ =&2e^2\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}+e^2\\ =&2e^2\lim_{x\to0}\frac{-\frac12x^2}{x^2}+e^2=0\\ 2.&\color{maroon}\lim_{x\to\infty}(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5})\\ =&\lim_{x\to\infty}f'(\xi)(2x^5)\\ =&\lim_{x\to\infty}\frac16\xi^{-\frac56}(2x^5)\\ =&\lim_{x\to\infty}\frac16\cdot x^{6(-\frac56)}(2x^5)\\ =&\lim_{x\to\infty}\frac16x^{-5}(2x^5)=\frac13\\ \end{aligned} 1.======2.====x→0limx(1+x)x2−e2[1−ln(1+x)]x→0limx(1+x)x2−e2+e2x→0limxln(1+x)x→0limxe2ln(1+x)/x−e2+e2x→0limxeξ(x2ln(1+x)−2)+e22e2x→0limxxln(1+x)−1+e22e2x→0limx2ln(1+x)−x+e22e2x→0limx2−21x2+e2=0x→∞lim(6x6+x5−6x6−x5)x→∞limf′(ξ)(2x5)x→∞lim61ξ−65(2x5)x→∞lim61⋅x6(−65)(2x5)x→∞lim61x−5(2x5)=31

无穷小比阶

定义

lim⁡x→⋅f(x)g(x)00→{c≠0,同阶无穷小(c=1,等价无穷小)0,高阶无穷小∞,低阶无穷小如{lim⁡x→0sin⁡xx=1lim⁡x→0x2x=0lim⁡x→0xx2=∞\begin{aligned} &\lim_{x\to\cdot}\frac{f(x)}{g(x)}\underrightarrow{\quad\frac00\quad}\begin{cases}c\neq0,同阶无穷小(c=1,等价无穷小)\\0,高阶无穷小\\\infty,低阶无穷小\end{cases}\\ &\qquad\qquad\quad如\begin{cases}\lim_{x\to0}\frac{\sin x}x=1\\\lim_{x\to0}\frac{x^2}x=0\\\lim_{x\to0}\frac x{x^2}=\infty\end{cases} \end{aligned} x→⋅limg(x)f(x)00⎩⎪⎨⎪⎧c̸=0,同阶无穷小(c=1,等价无穷小)0,高阶无穷小∞,低阶无穷小如⎩⎪⎨⎪⎧limx→0xsinx=1limx→0xx2=0limx→0x2x=∞

反问题、求未知参数

1.当x→0+时，下列（）与x1同阶A.1+x−1B.ln⁡(1+x)−xC.cos⁡(sin⁡x)−1D.xx−1[分析]A.(1+x)12−1∼12x1B.ln⁡(1+x)−x∼−12x2C.cos⁡(sin⁡x)−1∼−12(sin⁡x)2D.xx−1=exln⁡x−1∼xln⁡x是比x低阶的无穷小2.当x→0+时，比较α,β,γ的阶α=∫0xcos⁡t2dtβ=∫0x2tan⁡tdtγ=∫0xsin⁡t3dtlim⁡x→0+γα=lim⁡x→0+sin⁡x32⋅12xcos⁡x2=12lim⁡x→0+x32x12=0 ⟹ γ=∘(α)lim⁡x→0+βγ=lim⁡x→0+tan⁡x−2xsin⁡x32⋅12x=0 ⟹ β=∘(γ) ⟹ 由高阶至低阶：β→γ→α\begin{aligned} 1.&\color{maroon}当x\to0^+时，下列（\quad）与x^1同阶\\ &A.\sqrt{1+x}-1\quad B.\ln(1+x)-x\\ &C.\cos(\sin x)-1\quad D.x^x-1\\ [分析]&A.(1+x)^\frac12-1\sim\frac12x^1\\ &B.\ln(1+x)-x\sim-\frac12x^2\\ &C.\cos(\sin x)-1\sim-\frac12(\sin x)^2\\ &D.x^x-1=e^{x\ln x}-1\sim x\ln x是比x低阶的无穷小\\ 2.&\color{maroon}当x\to0^+时，比较\alpha,\beta,\gamma的阶\\ &\alpha=\int_0^x\cos t^2dt\quad\beta=\int_0^{x^2}\tan\sqrt{t}dt\quad\gamma=\int_0^{\sqrt x}\sin t^3dt\\ &\lim_{x\to0^+}\frac{\gamma}{\alpha}=\lim_{x\to0^+}\frac{\sin x^{\frac32}\cdot\frac1{2\sqrt x}}{\cos x^2}\\ &=\frac12\lim_{x\to0^+}\frac{x^{\frac32}}{x^{\frac12}}=0\\ &\implies\gamma=\circ(\alpha)\\ &\lim_{x\to0^+}\frac{\beta}{\gamma}=\lim_{x\to0^+}\frac{\tan x-2x}{\sin x^{\frac32}\cdot\frac1{2\sqrt x}}=0\\ &\implies \beta=\circ(\gamma)\\ &\implies 由高阶至低阶：\beta\to\gamma\to\alpha\\ \end{aligned} 1.[分析]2.当x→0+时，下列（）与x1同阶A.1+x−1B.ln(1+x)−xC.cos(sinx)−1D.xx−1A.(1+x)21−1∼21x1B.ln(1+x)−x∼−21x2C.cos(sinx)−1∼−21(sinx)2D.xx−1=exlnx−1∼xlnx是比x低阶的无穷小当x→0+时，比较α,β,γ的阶α=∫0xcost2dtβ=∫0x2tantdtγ=∫0xsint3dtx→0+limαγ=x→0+limcosx2sinx23⋅2x1=21x→0+limx21x23=0⟹γ=∘(α)x→0+limγβ=x→0+limsinx23⋅2x1tanx−2x=0⟹β=∘(γ)⟹由高阶至低阶：β→γ→α

存在性

具体型，但洛必达失效

用夹逼准则[例]记S(x)=∫0x∣sin⁡t∣dt,(1)证明当nπ≤x<(n+1)π时，2n≤S(x)<2(n+1)(2)求lim⁡x→+∞∫0x∣sin⁡t∣dtx(1)S(nπ)=∫0nπ∣sin⁡t∣dt=2nS((n+1)π)=∫0(n+1)π∣sin⁡t∣dt=2(n+1)又S′(x)=∣sin⁡x∣≥0 ⟹ S(x)单调递增故当nπ≤x<(n+1)π时，2n=S(nπ)≤S(x)<S((n+1)π)=2(n+1)(2)2n(n+1)π≤∫0x∣sin⁡t∣dtx<2(n+1)nπx→+∞ ⟹ n→∞ ⟹ 由夹逼准则得：2π[注]1.lim⁡x→+∞∫0x∣sin⁡t∣dtx2.lim⁡x→+∞∫0x∣cos⁡t∣dtx3.lim⁡x→+∞∫0x(t−[t])dtx\begin{aligned} &用夹逼准则\\ [例]&\color{maroon}记S(x)=\int_0^x\mid \sin t\mid dt,\\ &\color{maroon}(1)证明当n\pi\leq x<(n+1)\pi时，2n\leq S(x)<2(n+1)\\ &\color{maroon}(2)求\lim_{x\to+\infty}\frac{\int_0^x\mid\sin t\mid dt}{x}\\ &(1)S(n\pi)=\int_0^{n\pi}\mid \sin t\mid dt=2n\\ &S((n+1)\pi)=\int_0^{(n+1)\pi}\mid\sin t\mid dt=2(n+1)\\ &又S'(x)=\mid\sin x\mid\geq0\implies S(x)单调递增\\ &故当n\pi\leq x<(n+1)\pi时，2n=S(n\pi)\leq S(x)<S((n+1)\pi)=2(n+1)\\ &(2)\frac{2n}{(n+1)\pi}\leq\frac{\int_0^x\mid\sin t\mid dt}{x}<\frac{2(n+1)}{n\pi}\\ &x\to+\infty\implies n\to\infty\implies 由夹逼准则得：\frac2{\pi}\\ [注]&1.\lim_{x\to+\infty}\frac{\int_0^x\mid\sin t\mid dt}x\\ &2.\lim_{x\to+\infty}\frac{\int_0^x\mid\cos t\mid dt}x\\ &3.\lim_{x\to+\infty}\frac{\int_0^x(t-[t])dt}x \end{aligned} [例][注]用夹逼准则记S(x)=∫0x∣sint∣dt,(1)证明当nπ≤x<(n+1)π时，2n≤S(x)<2(n+1)(2)求x→+∞limx∫0x∣sint∣dt(1)S(nπ)=∫0nπ∣sint∣dt=2nS((n+1)π)=∫0(n+1)π∣sint∣dt=2(n+1)又S′(x)=∣sinx∣≥0⟹S(x)单调递增故当nπ≤x<(n+1)π时，2n=S(nπ)≤S(x)<S((n+1)π)=2(n+1)(2)(n+1)π2n≤x∫0x∣sint∣dt<nπ2(n+1)x→+∞⟹n→∞⟹由夹逼准则得：π21.x→+∞limx∫0x∣sint∣dt2.x→+∞limx∫0x∣cost∣dt3.x→+∞limx∫0x(t−[t])dt

抽象型——单调有界准则

若f(x)单调递增（递减），且有上（下）界 ⟹ lim⁡x→+∞f(x)=∃[例]设x≥0,f(x)满足f′(x)=1x2+f2(x),f(0)=1,证明(1)f′(x)≤11+x2,x≥0(2)lim⁡x→+∞f(x)存在且其值小与1+π2(1)f′(x)>0 ⟹ f(x)单调递增 ⟹ f(x)≥f(0)=1故f′(x)=1x2+f2(x)≤1x2+1(2)f(x)=f(a)+∫axf′(x)dt,故f(x)=f(0)+∫1x1t2+f2(t)dt≤1+∫0x1t2+1dt=1+arctan⁡x<1+π2,故f(x)有上界由单调有界准则,得lim⁡x→+∞f(x)存在且lim⁡x→+∞f(x)=1+∫0+∞1t2+f2(t)dt<1+∫0+∞1t2+1dt=1+π2\begin{aligned} &若f(x)单调递增（递减），且有上（下）界\implies \lim_{x\to+\infty}f(x)=\exists\\ [例]&\color{maroon}设x\geq0,f(x)满足f'(x)=\frac1{x^2+f^2(x)},f(0)=1,证明\\ &\color{maroon}(1)f'(x)\leq\frac1{1+x^2},x\geq0\\ &\color{maroon}(2)\lim_{x\to+\infty}f(x)存在且其值小与1+\frac{\pi}2\\ &(1)f'(x)>0\implies f(x)单调递增\implies f(x)\geq f(0)=1\\ &故f'(x)=\frac1{x^2+f^2(x)}\leq\frac1{x^2+1}\\ &(2)f(x)=f(a)+\int_a^xf'(x)dt,故f(x)=f(0)+\int_1^x\frac1{t^2+f^2(t)}dt\leq1+\int_0^x\frac1{t^2+1}dt\\ &=1+\arctan x<1+\frac{\pi}2,故f(x)有上界\\ &由单调有界准则,得\lim_{x\to+\infty}f(x)存在且\lim_{x\to+\infty}f(x)=1+\int_0^{+\infty}\frac1{t^2+f^2(t)}dt<1+\int_0^{+\infty}\frac1{t^2+1}dt=1+\frac{\pi}2\\ \end{aligned} [例]若f(x)单调递增（递减），且有上（下）界⟹x→+∞limf(x)=∃设x≥0,f(x)满足f′(x)=x2+f2(x)1,f(0)=1,证明(1)f′(x)≤1+x21,x≥0(2)x→+∞limf(x)存在且其值小与1+2π(1)f′(x)>0⟹f(x)单调递增⟹f(x)≥f(0)=1故f′(x)=x2+f2(x)1≤x2+11(2)f(x)=f(a)+∫axf′(x)dt,故f(x)=f(0)+∫1xt2+f2(t)1dt≤1+∫0xt2+11dt=1+arctanx<1+2π,故f(x)有上界由单调有界准则,得x→+∞limf(x)存在且x→+∞limf(x)=1+∫0+∞t2+f2(t)1dt<1+∫0+∞t2+11dt=1+2π

应用–连续与间断

1.由于“一切初等函数在其定义区域内必连续”，故只研究两类特殊的点{无定义点(间断)分段函数的分段点2.连续(1)内点处：lim⁡x→x0+f(x)=lim⁡x→x0−f(x)=f(x) ⟹ f(x)在x0处连续(2)端点处：{lim⁡x→a+f(x)=f(a)(左端点右连续)lim⁡x→b−f(x)=f(b)(右端点左连续)f(x)在(a,b)内连续称f(x)在[a,b]上连续3.间断:(前提:f(x)在x=x0左右两侧均有定义)(1)若lim⁡x→x0+f(x),lim⁡x→x0−f(x)均存在但lim⁡x→x0+f(x)̸=lim⁡x→x0−f(x) ⟹ x0为跳跃间断点(2)lim⁡x→x0+f(x),lim⁡x→x0−f(x)均存在且lim⁡x→x0+f(x)=lim⁡x→x0−f(x)，但̸=f(x) ⟹ x0为可去间断点以上统称第一类间断点(3)lim⁡x→x0+f(x),lim⁡x→x0−f(x)至少一个不存在，且不存在=∞ ⟹ x0为无穷间断点(4)lim⁡x→x0+f(x),lim⁡x→x0−f(x)至少一个不存在且不存在为振荡不存在 ⟹ x0为振荡间断点(3)(4)属于第二类间断点\begin{aligned} 1.&由于“一切初等函数在其定义区域内必连续”，故只研究两类特殊的点\begin{cases}无定义点(间断)\\分段函数的分段点\end{cases}\\ 2.&连续\\ (1)&内点处：\lim_{x\to x_0^+}f(x)=\lim_{x\to x_0^-}f(x)=f(x)\implies f(x)在x_0处连续\\ (2)&端点处：\begin{cases}\lim_{x\to a^+}f(x)=f(a)(左端点右连续)\\\lim_{x\to b^-}f(x)=f(b)(右端点左连续)\\f(x)在(a,b)内连续\end{cases}称f(x)在[a,b]上连续\\ 3.&间断:(前提:f(x)在x=x_0左右两侧均有定义)\\ (1)&若\lim_{x\to x_0^+}f(x),\lim_{x\to x_0^-}f(x)均存在但\lim_{x\to x_0^+}f(x)\not=\lim_{x\to x_0^-}f(x)\implies x_0为跳跃间断点\\ (2)&\lim_{x\to x_0^+}f(x),\lim_{x\to x_0^-}f(x)均存在且\lim_{x\to x_0^+}f(x)=\lim_{x\to x_0^-}f(x)，但\not=f(x)\implies x_0为可去间断点\\ &以上统称第一类间断点\\ (3)&\lim_{x\to x_0^+}f(x),\lim_{x\to x_0^-}f(x)至少一个不存在，且不存在=\infty\implies x_0为无穷间断点\\ (4)&\lim_{x\to x_0^+}f(x),\lim_{x\to x_0^-}f(x)至少一个不存在且不存在为振荡不存在\implies x_0为振荡间断点\\ &(3)(4)属于第二类间断点\\ \end{aligned} 1.2.(1)(2)3.(1)(2)(3)(4)由于“一切初等函数在其定义区域内必连续”，故只研究两类特殊的点{无定义点(间断)分段函数的分段点连续内点处：x→x0+limf(x)=x→x0−limf(x)=f(x)⟹f(x)在x0处连续端点处：⎩⎪⎨⎪⎧limx→a+f(x)=f(a)(左端点右连续)limx→b−f(x)=f(b)(右端点左连续)f(x)在(a,b)内连续称f(x)在[a,b]上连续间断:(前提:f(x)在x=x0左右两侧均有定义)若x→x0+limf(x),x→x0−limf(x)均存在但x→x0+limf(x)̸=x→x0−limf(x)⟹x0为跳跃间断点x→x0+limf(x),x→x0−limf(x)均存在且x→x0+limf(x)=x→x0−limf(x)，但̸=f(x)⟹x0为可去间断点以上统称第一类间断点x→x0+limf(x),x→x0−limf(x)至少一个不存在，且不存在=∞⟹x0为无穷间断点x→x0+limf(x),x→x0−limf(x)至少一个不存在且不存在为振荡不存在⟹x0为振荡间断点(3)(4)属于第二类间断点

[例]当x∈(−12,1]时，确定f(x)=tan⁡πx∣x∣(x2−1)的间断点并判断其类型1.{lim⁡x→0+tan⁡πxx(x−1)=−πlim⁡x→0−tan⁡πx−x(x−1)=π ⟹ x=0为跳跃间断点2.lim⁡x→1tan⁡πxx(x+1)(x−1)=12lim⁡x→1tan⁡πxx−1=π2 ⟹ x=1为可去间断点3.lim⁡x→12tan⁡πxx(x2−1)=∞ ⟹ x=12为无穷间断点\begin{aligned} \ [例]&\color{maroon}当x\in(-\frac12,1]时，确定f(x)=\frac{\tan \pi x}{\mid x\mid(x^2-1)}的间断点并判断其类型\\ &1.\begin{cases}\lim_{x\to0^+}\frac{\tan\pi x}{x(x^-1)}=-\pi\\\lim_{x\to0^-}\frac{\tan\pi x}{-x(x^-1)}=\pi\end{cases}\implies x=0为跳跃间断点\\ &2.\lim_{x\to1}\frac{\tan\pi x}{x(x+1)(x-1)}=\frac12\lim_{x\to1}\frac{\tan\pi x}{x-1}=\frac{\pi}2\implies x=1为可去间断点\\ &3.\lim_{x\to\frac12}\frac{\tan\pi x}{x(x^2-1)}=\infty\implies x=\frac12为无穷间断点\\ \end{aligned} [例]当x∈(−21,1]时，确定f(x)=∣x∣(x2−1)tanπx的间断点并判断其类型1.{limx→0+x(x−1)tanπx=−πlimx→0−−x(x−1)tanπx=π⟹x=0为跳跃间断点2.x→1limx(x+1)(x−1)tanπx=21x→1limx−1tanπx=2π⟹x=1为可去间断点3.x→21limx(x2−1)tanπx=∞⟹x=21为无穷间断点