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目录
写在前面
关于无穷小量的计算、常用等价无穷小,可以看我的另一篇笔记 → 链接
无穷小等价替换定理
设函数 f , g , h f,g,h f,g,h 在 U ˚ ( x 0 ) \mathring{U}(x_{0}) U˚(x0) 内有定义,且有 f ( x ) ∼ g ( x ) f(x)\sim g(x) f(x)∼g(x)
(1) 若 lim x → x 0 f ( x ) h ( x ) = A \lim\limits_{x\to x_0}f(x)h(x)=A x→x0limf(x)h(x)=A,则 lim x → x 0 g ( x ) h ( x ) = A \lim\limits_{x\to x_0}g(x)h(x)=A x→x0limg(x)h(x)=A;
(2) 若 lim x → x 0 h ( x ) f ( x ) = B \lim\limits_{x\to x_0}\displaystyle\frac{h(x)}{f(x)}=B x→x0limf(x)h(x)=B,则 lim x → x 0 h ( x ) g ( x ) = B \lim\limits_{x\to x_0}\displaystyle\frac{h(x)}{g(x)}=B x→x0limg(x)h(x)=B。
注意: 等价无穷小一般只能在乘除中替换,在加减中替换有时会出错(加减时可以整体代换,不一定能随意单独代换或分别代换)。
下面做详细解释
1. 问题的提出
2. 等价无穷小代换的原理
将 tan x 和 sin x 都展开到 x 3 时: tan x = x + 1 3 x 3 + ο ( x 3 ) , sin x = x − x 3 3 ! + ο ( x 3 ) . \begin{aligned} \text{将}&\tan x\text{和}\sin x\text{都展开到 }x^3\text{时:}\\ &\tan x = x+\frac{1}{3}x^{3}+\omicron(x^{3}) , \\ &\sin x =x-\frac{x^{3}}{3!}+\omicron(x^{3}). \\ \end{aligned} 将tanx和sinx都展开到 x3时:tanx=x+31x3+ο(x3),sinx=x−3!x3+ο(x3).
则 lim x → 0 tan x − sin x x 3 = lim x → 0 x + 1 3 x 3 + o ( x 3 ) − [ x − x 3 3 ! o ( x 3 ) ] x 3 = lim x → 0 x + 1 2 x 3 + o ( x 3 ) x 3 = lim x → 0 ( 1 2 + o ( x 3 ) x 3 ) = 1 2 . \begin{aligned} \text{则 } &\lim\limits_{x\to0} \frac{\tan x-\sin x}{x^{3}} \\ =&\lim\limits_{x\to0}\frac{x+\frac{1}{3}x^{3}+o(x^{3})-\left[x-\frac{x^{3}}{3!}o(x^{3})\right]}{x^{3}} \\ =&\lim\limits_{x\to0}\frac{x+\frac{1}{2}x^{3}+o(x^{3})}{x^{3}} \\ =&\lim\limits_{x\to0}\Bigl({\frac{1}{2}}+{\frac{o(x^{3})}{x^{3}}}\Bigr)={\frac{1}{2}}. \end{aligned} 则 ===x→0limx3tanx−sinxx→0limx3x+31x3+o(x3)−[x−3!x3o(x3)]x→0limx3x+21x3+o(x3)x→0lim(21+x3o(x3))=21.
3. 等价无穷小代换条件
3.1. 商(积)式代换
定理 1 在自变量同一变化过程中,设 α ∼ α ′ , β ∼ β ′ \alpha\sim\alpha^{\prime},\beta\sim\beta^{\prime} α∼α′,β∼β′,且 lim β ′ α ′ \lim\displaystyle\frac{\beta^{\prime}}{\alpha^{\prime}} limα′β′存在,则
lim β α = lim β ′ α ′ \lim\frac\beta\alpha=\lim\frac{\beta^{\prime}}{\alpha^{\prime}} limαβ=limα′β′
【例 3】求: ( 1 ) lim x → 1 = ln ( 1 + x − 1 3 ) arcsin x − 1 3 ( 2 ) lim x → 0 = ln ( sin 2 x + e x ) − x ln ( x 2 − e 2 x ) − 2 x . \begin{aligned} \textbf{【例 3】求:} &(1)\ \lim\limits_{x\to1}=\frac{\ln(1+\sqrt[3]{x-1})}{\arcsin\sqrt[3]{x-1}} \\ &(2)\ \lim\limits_{x\to0}=\frac{\ln(\sin^2x+\mathrm{e}^x)-x}{\ln(x^{2}-\mathrm{e}^{2x})-2x}. \end{aligned} 【例 3】求:(1) x→1lim=arcsin3x−1ln(1+3x−1)(2) x→0lim=ln(x2−e2x)−2xln(sin2x+ex)−x.
【解】 ( 1 ) 原式 = lim x → 1 = x − 1 3 x − 1 3 = 1. ( 2 ) 原式 = lim x → 0 ln e x ( 1 + sin 2 x e x ) − x ln e 2 x ( 1 + x 2 e 2 x ) − 2 x = lim x → 0 ln ( 1 + sin 2 x e x ) ln ( 1 + x 2 e 2 x ) [ ln ( 1 + x ) ∼ x ] = lim x → 0 sin 2 x e x x 2 e 2 x = lim x → 0 sin 2 x x 2 e x = 1. \begin{aligned} \textbf{【解】 }&(1)\ 原式 = \lim\limits_{x\to 1}=\frac{\sqrt[3]{x-1}}{\sqrt[3] {x-1}}=1. \\ &\begin{aligned} (2)\ \text{原式}&=\lim\limits_{x\to0}\frac{\ln \mathrm{e}^{x}\left(1+\displaystyle\frac{\sin^{2}x}{\mathrm{e}^{x}}\right)-x}{\ln \mathrm{e}^{2x}\left(1+\displaystyle\frac{x^{2}}{\mathrm{e}^{2x}}\right)-2x} \\ &=\lim\limits_{x\to0}\frac{\ln\left(1+\displaystyle\frac{\sin^{2}x}{\mathrm{e}^{x}}\right)}{\ln\left(1+\displaystyle\frac{x^{2}}{\mathrm{e}^{2x}}\right)} \uad \Big[\ln (1+x)\sim x\Big]\\ &=\lim\limits_{x\to0}\frac{\displaystyle\frac{\sin^{2}x}{\mathrm{e}^{x}}}{\displaystyle\frac{x^{2}}{\mathrm{e}^{2x}}} =\lim\limits_{x\to0}\frac{\sin^{2}x}{x^{2}}\mathrm{e}^{x}=1. \end{aligned} \end{aligned} 【解】 (1) 原式=x→1lim=3x−13x−1=1.(2) 原式=x→0limlne2x(1+e2xx2)−2xlnex(1+exsin2x)−x=x→0limln(1+e2xx2)ln(1+exsin2x)[ln(1+x)∼x]=x→0lime2xx2exsin2x=x→0limx2sin2xex=1.
3.2. 幂式代换
定理 2 设 α ∼ α ′ , β ∼ β ′ \alpha\sim\alpha^{\prime},\beta\sim\beta^{\prime} α∼α′,β∼β′
(1) 若 α > 0 \alpha>0 α>0,则 1 ln α ∼ 1 ln α ′ \displaystyle\frac1{\ln\alpha}\sim\frac1{\ln\alpha’} lnα1∼lnα′1
(2) 若 α , β > 0 \alpha,\beta>0 α,β>0,且 lim β ′ ln α ′ \lim\beta’\ln\alpha’ limβ′lnα′存在,则
lim α β = lim α ′ β ′ \displaystyle\lim\alpha^{\beta }= \lim{\alpha’} ^{\beta’} limαβ=limα′β′
(3) 设 α ∼ α ′ , β ∼ β ′ \alpha\sim\alpha^{\prime},\beta\sim\beta^{\prime} α∼α′,β∼β′,且 lim ( 1 + α ′ ) 1 β ′ = A \lim (1+\alpha’)^{\frac{1}{\beta’}}= A lim(1+α′)β′1=A,则
lim ( 1 + α ) 1 β = lim ( 1 + α ′ ) 1 β ′ = A \lim(1+\alpha)^{\frac{1}{\beta}}=\lim(1+\alpha’)^{\frac{1}{\beta’}}=A lim(1+α)β1=lim(1+α′)β′1=A
【例 4】 求: ( 1 ) lim x → 0 ( 1 − 2 x ) 1 x ; ( 2 ) lim x → ∞ ( cos 1 x ) x 2 . \begin{aligned} \textbf{【例 4】 求:}&(1)\ \lim\limits_{x\to0}(1-2x)^{\frac{1}{x}}; \\ &(2)\ \lim\limits_{x\to\infty}\left(\cos\frac{1}{x}\right)^{x^{2}}. \\ \end{aligned} 【例 4】 求:(1) x→0lim(1−2x)x1;(2) x→∞lim(cosx1)x2.
【解】 ( 1 ) 原式 = lim x → 0 e 1 x ln ( 1 − 2 x ) = lim x → 0 e 1 x ( − 2 x ) = e − 2 ( 2 ) 原式 = lim x → ∞ [ 1 − 2 sin 2 ( 1 2 x ) ] x 2 ( cos 2 φ = 1 − 2 sin 2 φ ) = lim x → ∞ ( 1 − 1 2 x 2 ) x 2 = lim x → ∞ [ ( 1 − 1 2 x 2 ) 2 x 2 ] 1 2 [ lim x → ∞ ( 1 − 1 x ) x = e − 1 ] = e − 1 2 . \begin{aligned} \textbf{【解】 }&(1)\ 原式=\lim\limits_{x\to0}\mathrm{e}^{\frac{1}{x}\ln(1-2x)}=\lim\limits_{x\to0}\mathrm{e}^{\frac{1}{x}(-2x)}=\mathrm{e}^{-2} \\ &\begin{aligned} (2)\ 原式&=\lim\limits_{x\to\infty}\left[1-2\sin^{2}\left(\frac{1}{2x}\right)\right]^{x^{2}} \uad(\cos2\varphi=1-2\sin^{2}\varphi) \\ &=\lim\limits_{x\to\infty}\left(1-\frac{1}{2x^{2}}\right)^{x^{2}} \\ &=\lim\limits_{x\to\infty}\left[\left(1-\frac{1}{2x^{2}}\right)^{2x^{2}}\right]^{\frac{1}{2}} \uad\quad \Big[\lim \limits_{x\to \infty} (1-\frac{1}{x})^x=e^{-1}\Big] \\ &=\mathrm{e}^{-\frac{1}{2}}. \end{aligned} \end{aligned} 【解】 (1) 原式=x→0limex1ln(1−2x)=x→0limex1(−2x)=e−2(2) 原式=x→∞lim[1−2sin2(2x1)]x2(cos2φ=1−2sin2φ)=x→∞lim(1−2x21)x2=x→∞lim[(1−2x21)2x2]21[x→∞lim(1−x1)x=e−1]=e−21.
3.3. 和差代换
定理 3 设 α ∼ α ′ , β ∼ β ′ \alpha\sim\alpha^{\prime},\beta\sim\beta^{\prime} α∼α′,β∼β′,则
(1) 若 α \alpha α 与 β \beta β 不等价,则 α − β ∼ α ′ − β ′ . \alpha-\beta\sim\alpha^\prime-\beta^{\prime}. α−β∼α′−β′.
(2) 若 α \alpha α 与 β \beta β 等价,则 α − β \alpha-\beta α−β 与 α ′ − β ′ \alpha^\prime-\beta^{\prime} α′−β′ 未必等价.
推论 设 α , β , γ \alpha,\beta,\gamma α,β,γ 是自变量同一变化过程中的无穷小量,且 α ∼ α ′ , β ∼ β ′ \alpha\sim\alpha^{\prime},\beta\sim\beta^{\prime} α∼α′,β∼β′,则
(1) 当 α \alpha α 与 β \beta β 不等价时,则 lim α − β γ = lim α ′ − β ′ γ . \lim\displaystyle\frac{\alpha-\beta}{\gamma}=\lim\frac{\alpha^{\prime}-\beta^{\prime}}{\gamma}. limγα−β=limγα′−β′.
(2) 当 α \alpha α 与 β \beta β 等价时,上式极限未必成立.
定理 4 设 α ∼ α ′ , β ∼ β ′ \alpha\sim\alpha^{\prime},\beta\sim\beta^{\prime} α∼α′,β∼β′,则
(1) lim α β = c ≠ − 1 或 ∞ \lim\displaystyle\frac\alpha\beta=c\neq-1\small\text{或}\infty limβα=c=−1或∞,则 α + β ∼ α ′ + β ′ \alpha + \beta \sim \alpha’+ \beta’ α+β∼α′+β′
(2) 当 γ ∼ γ ′ , δ ∼ δ ′ \gamma\sim\gamma’,\delta\sim\delta’ γ∼γ′,δ∼δ′,且 lim a α b β = e ′ ≠ − 1 \lim\displaystyle\frac{a\alpha}{b\beta}=e’\neq-1 limbβaα=e′=−1 和 lim c γ d δ = e 2 = c ≠ − 1 \lim\displaystyle\frac{c\gamma}{d\delta}=e^{2}= c\neq-1 limdδcγ=e2=c=−1(其 中 a , b , c , d a, b, c, d a,b,c,d 为常数),则
lim a α + b β c γ + d δ = lim a α ′ + b β ′ c γ ′ + d δ ′ \lim\frac{a\alpha+b\beta}{c\gamma+d\delta}=\lim\frac{a\alpha’+b\beta’}{c\gamma’+d\delta’} limcγ+dδaα+bβ=limcγ′+dδ′aα′+bβ′
(3) 如果无穷小量 α i ∼ β i , λ i \alpha_{i}\sim\beta_{i},\ \lambda_{i} αi∼βi, λi 为常数 ( i = 1 , 2 , ⋯ , n ) (i=1,2,\cdots,n) (i=1,2,⋯,n) ,且
lim λ 1 α 1 λ 2 α 2 ≠ − 1 , lim λ 1 α 1 + λ 2 α 2 λ 3 α 3 ≠ − 1 , ⋯ , lim λ 1 α 1 + λ 2 α 2 + ⋯ + λ n − 1 α n − 1 λ n α n ≠ − 1 \lim\displaystyle\frac{\lambda_{1}\alpha_{1}}{\lambda_{2}\alpha_{2}}\neq-1,\lim\frac{\lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2}}{\lambda_{3}\alpha_{3}}\neq-1,\cdots,\lim\frac{\lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2}+\cdots+\lambda_{n-1}\alpha_{n-1}}{\lambda_{n}\alpha_{n}}\neq-1 limλ2α2λ1α1=−1,limλ3α3λ1α1+λ2α2=−1,⋯,limλnαnλ1α1+λ2α2+⋯+λn−1αn−1=−1
则
λ 1 α 1 + λ 2 α 2 + ⋯ + λ n α n ∼ λ 1 β 1 + λ 2 β 2 + ⋯ + λ n β n ( n ⩾ 2 ) . \lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2}+\cdots+\lambda_{n}\alpha_{n}\sim\lambda_{1}\beta_{1}+\lambda_{2}\beta_{2}+\cdots+\lambda_{n}\beta_{n}(n\geqslant2). λ1α1+λ2α2+⋯+λnαn∼λ1β1+λ2β2+⋯+λnβn(n⩾2).
总结
可直接等价替换的类型
lim α β = lim α ′ β ′ \begin{aligned}\lim\frac{\alpha}{\beta}=\lim\frac{\alpha’}{\beta’}\end{aligned} limβα=limβ′α′
lim α β = lim α ′ β ′ \lim\alpha^{\beta}=\lim\alpha^{\prime\beta^{\prime}} limαβ=limα′β′
lim ( 1 α ) β = lim 1 α β = lim 1 α ′ β ′ = lim ( 1 α ′ ) β ′ \lim\left(\displaystyle\frac{1}{\alpha}\right)^{\beta}=\lim\displaystyle\frac{1}{\alpha^{\beta}}=\lim\frac{1}{\alpha^{\prime{\beta^{\prime}}}}=\lim\left(\frac{1}{\alpha^{\prime}}\right)^{\beta’} lim(α1)β=limαβ1=limα′β′1=lim(α′1)β′
lim ( 1 + α ) 1 β = lim ( 1 + α ′ ) 1 β ′ \lim\left(1+\alpha\right)\displaystyle^{\frac{1}{\beta}}=\lim\left(1+\alpha^{\prime}\right)\displaystyle^{\frac{1}{\beta^{\prime}}} lim(1+α)β1=lim(1+α′)β′1
以上几个性质可以用来化简一些未定式以方便运用洛必达法则.
需要满足一定条件才能替换的类型
若 lim α ′ β ′ ≠ − 1 \lim\displaystyle\frac{\alpha^{\prime}}{\beta^{\prime}}\neq-1 limβ′α′=−1,则 α + β ∼ α ′ + β ′ \alpha+\beta\sim\alpha^{\prime}+\beta^{\prime} α+β∼α′+β′
(该条性质非常重要,这是判断在加减法中能否分别等价替换的重要依据)
变上限积分函数(积分变限函数)也可以用等价无穷小进行替换。
个人笔记,如有错误,烦请指正 : )
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