【高等数学学习记录】无穷小的比较
1 知识点
定义:
如果 lim β α = 0 \lim \frac{\beta}{\alpha}=0 limαβ=0,就说 β \beta β 是比 α \alpha α 高阶的无穷小,记作 β = o ( α ) \beta=o(\alpha ) β=o(α)
如果 lim β α = ∞ \lim \frac{\beta}{\alpha}=\infty limαβ=∞,就说 β \beta β 是比 α \alpha α 低阶的无穷小。
如果 lim β α = c ≠ 0 \lim \frac{\beta}{\alpha}=c\neq 0 limαβ=c=0,就说 β \beta β 与 α \alpha α 是同阶无穷小。
如果 lim β α = c ≠ 0 \lim \frac{\beta}{\alpha }=c\neq 0 limαβ=c=0, k > 0 k>0 k>0,就说 β \beta β 是关于 α \alpha α 的 k k k 阶无穷小。
如果 lim β α = 1 \lim \frac{\beta}{\alpha}=1 limαβ=1,就说 β \beta β 与 α \alpha α 是等价无穷小,记作 α ∼ β \alpha \sim \beta α∼β。
定理1
β \beta β 与 α \alpha α 是等价无穷小的充分必要条件为 β = α + o ( α ) \beta = \alpha + o(\alpha ) β=α+o(α)。
定理2
设 α ∼ α ′ \alpha \sim \alpha ' α∼α′, β ∼ β ′ \beta \sim \beta ' β∼β′,且 lim β ′ α ′ \lim \frac{\beta '}{\alpha '} limα′β′ 存在,则 lim β α = lim β ′ α ′ \lim \frac{\beta}{\alpha}=\lim \frac{\beta '}{\alpha '} limαβ=limα′β′。
常用等价无穷小:
(1) 1 + x n − 1 ∼ 1 n x \sqrt[n]{1+x}-1\sim \frac{1}{n}x n1+x−1∼n1x
(2) s i n x ∼ x sinx\sim x sinx∼x
(3) t a n x ∼ x tanx\sim x tanx∼x
(4) a r c s i n x ∼ x arcsinx\sim x arcsinx∼x
(5) 1 − c o s x ∼ 1 x 2 1-cosx\sim \frac{1}{x^2} 1−cosx∼x21
2 练习题
2.1 当 x → 0 x\rightarrow 0 x→0 时, 2 x − x 2 2x-x^2 2x−x2 与 x 2 − x 3 x^2-x^3 x2−x3 相比,哪一个是高阶无穷小?
- 解:
lim x → 0 2 x − x 2 x 2 − x 3 \quad \lim_{x\rightarrow 0}\frac{2x-x^2}{x^2-x^3} limx→0x2−x32x−x2
= lim x → 0 2 − x x − x 2 =\lim_{x\rightarrow 0}\frac{2-x}{x-x^2} =limx→0x−x22−x
= lim x → 0 2 x − 1 1 − x =\lim_{x\rightarrow 0}\frac{\frac{2}{x}-1}{1-x} =limx→01−xx2−1
= ∞ =\infty =∞
∴ lim x → 0 x 2 − x 3 2 x − x 2 = 0 \therefore \lim_{x\rightarrow 0}\frac{x^2-x^3}{2x-x^2}=0 ∴limx→02x−x2x2−x3=0
∴ x 2 − x 3 \therefore x^2-x^3 ∴x2−x3 是比 2 x − x 2 2x-x^2 2x−x2 的高阶无穷小。
2.2 当 x → 1 x\rightarrow 1 x→1 时,无穷小 1 − x 1-x 1−x 和 (1) 1 − x 3 1-x^3 1−x3,(2) 1 2 ( 1 − x 2 ) \frac{1}{2}(1-x^2) 21(1−x2) 是否同阶?是否等价?
- 解:
- (1)
lim x → 1 1 − x 1 − x 3 \quad \lim_{x\rightarrow 1}\frac{1-x}{1-x^3} limx→11−x31−x
= lim x → 1 1 − x ( 1 − x ) ( 1 + x + x 2 ) =\lim_{x\rightarrow 1}\frac{1-x}{(1-x)(1+x+x^2)} =limx→1(1−x)(1+x+x2)1−x
= lim x → 1 1 x 2 + x + 1 =\lim_{x\rightarrow 1}\frac{1}{x^2+x+1} =limx→1x2+x+11
= 1 3 =\frac{1}{3} =31
∴ 1 − x \therefore 1-x ∴1−x 与 1 − x 3 1-x^3 1−x3 同阶,但不等价。 - (2)
lim x → 1 1 − x 1 2 ( 1 − x 2 ) \quad \lim_{x\rightarrow 1}\frac{1-x}{\frac{1}{2}(1-x^2)} limx→121(1−x2)1−x
= lim x → 1 2 1 + x =\lim_{x\rightarrow 1}\frac{2}{1+x} =limx→11+x2
= 1 =1 =1
∴ 1 − x \therefore 1-x ∴1−x 与 1 2 ( 1 − x 2 ) \frac{1}{2}(1-x^2) 21(1−x2) 等价。
2.3 证明:当 x → 0 x\rightarrow 0 x→0 时,有(1) a r c t a n x ∼ x arctanx\sim x arctanx∼x,(2) s e c x − 1 ∼ x 2 2 secx-1\sim \frac{x^2}{2} secx−1∼2x2
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证明:
-
(1)
令 t = a r c t a n x t=arctanx t=arctanx
∵ lim x → 0 a r c t a n x = lim x → 0 x = 0 \because \lim_{x\rightarrow 0}arctanx=\lim_{x\rightarrow 0}x=0 ∵limx→0arctanx=limx→0x=0
∴ lim x → 0 a r c t a n x x \therefore \lim_{x\rightarrow 0}\frac{arctanx}{x} ∴limx→0xarctanx 可以转换为:
lim t → 0 t t a n t \quad \lim_{t\rightarrow 0}\frac{t}{tant} limt→0tantt
= lim t → 0 t c o s t s i n t =\lim_{t\rightarrow 0}\frac{tcost}{sint} =limt→0sinttcost
= lim t → 0 c o s t =\lim_{t\rightarrow 0}cost =limt→0cost
= 1 =1 =1
∴ a r c t a n x \therefore arctanx ∴arctanx 与 x x x 是等价无穷小。 -
(2)
∵ lim x → 0 s e c x − 1 x 2 2 \because \lim_{x\rightarrow 0}\frac{secx-1}{\frac{x^2}{2}} ∵limx→02x2secx−1
= lim x → 0 2 ( 1 − c o s x ) x 2 c o s x =\lim_{x\rightarrow 0}\frac{2(1-cosx)}{x^2cosx} =limx→0x2cosx2(1−cosx)
= lim x → 0 2 ( 1 − 1 + 2 s i n 2 x 2 ) x 2 c o s x =\lim_{x\rightarrow 0}\frac{2(1-1+2sin^2\frac{x}{2})}{x^2cosx} =limx→0x2cosx2(1−1+2sin22x)
= lim x → 0 s i n 2 x 2 ( x 2 ) 2 c o s x =\lim_{x\rightarrow 0}\frac{sin^2\frac{x}{2}}{(\frac{x}{2})^2cosx} =limx→0(2x)2cosxsin22x
= 1 =1 =1
∴ s e c x − 1 ∼ x 2 2 \therefore secx-1\sim \frac{x^2}{2} ∴secx−1∼2x2
2.4 利用等价无穷小的性质,求下列极限:
(1) lim x → 0 t a n 3 x 2 x \lim_{x\rightarrow 0}\frac{tan3x}{2x} limx→02xtan3x
(2) lim x → 0 s i n ( x n ) ( s i n x ) m \lim_{x\rightarrow 0}\frac{sin(x^n)}{(sinx)^m} limx→0(sinx)msin(xn)( n n n、 m m m 为正整数)
(3) lim x → 0 t a n x − s i n x s i n 3 x \lim_{x\rightarrow 0}\frac{tanx-sinx}{sin^3x} limx→0sin3xtanx−sinx
(4) lim x → 0 s i n x − t a n x ( 1 + x 2 3 − 1 ) ( 1 + s i n x − 1 ) \lim_{x\rightarrow 0}\frac{sinx-tanx}{(\sqrt[3]{1+x^2}-1)(\sqrt{1+sinx}-1)} limx→0(31+x2−1)(1+sinx−1)sinx−tanx
- 解:
- (1)
∵ t a n 3 x ∼ 3 x ( x → 0 ) \because tan3x\sim 3x(x\rightarrow 0) ∵tan3x∼3x(x→0)
∴ \therefore ∴ 原式 = lim x → 0 3 x 2 x = 3 2 =\lim_{x\rightarrow 0}\frac{3x}{2x}=\frac{3}{2} =limx→02x3x=23 - (2)
∵ s i n x n ∼ x n ( x → 0 ) , s i n x ∼ x ( x → 0 ) \because sinx^n\sim x^n(x\rightarrow 0),sinx\sim x(x\rightarrow 0) ∵sinxn∼xn(x→0),sinx∼x(x→0)
∴ \therefore ∴ 原式 = lim x → 0 x n x m =\lim_{x\rightarrow 0}\frac{x^n}{x^m} =limx→0xmxn
= { 0 , n > m 1 , n = m ∞ , n < m =\begin{cases}0,&n>m\\1,&n=m\\\infty,&n<m\end{cases} =⎩ ⎨ ⎧0,1,∞,n>mn=mn<m - (3)
原式 = lim x → 0 1 − c o s x c o s x ⋅ s i n 2 x =\lim_{x\rightarrow 0}\frac{1-cosx}{cosx\cdot sin^2x} =limx→0cosx⋅sin2x1−cosx
= lim x → 0 1 c o s x ⋅ lim x → 0 2 s i n 2 x 2 s i n 2 x =\lim_{x\rightarrow 0}\frac{1}{cosx}\cdot \lim_{x\rightarrow 0}\frac{2sin^2\frac{x}{2}}{sin^2x} =limx→0cosx1⋅limx→0sin2x2sin22x
= 2 ⋅ lim x → 0 s i n 2 x 2 s i n 2 x =2\cdot \lim_{x\rightarrow 0}\frac{sin^2\frac{x}{2}}{sin^2x} =2⋅limx→0sin2xsin22x
= 2 ⋅ lim x → 0 ( x 2 ) 2 x 2 =2\cdot \lim_{x\rightarrow 0}\frac{(\frac{x}{2})^2}{x^2} =2⋅limx→0x2(2x)2
= 1 2 =\frac{1}{2} =21 - (4)
原式 = lim x → 0 s i n x ( c o s x − 1 ) c o s x ( 1 + x 2 3 ) ( 1 + s i n x − 1 ) =\lim_{x\rightarrow 0}\frac{sinx (cosx-1)}{cosx(\sqrt[3]{1+x^2})(\sqrt{1+sinx}-1)} =limx→0cosx(31+x2)(1+sinx−1)sinx(cosx−1)
∵ c o s x − 1 ∼ − 1 2 x 2 , t a n x ∼ x , 1 + x 2 3 − 1 ∼ 1 3 x 2 , 1 + s i n x − 1 ∼ 1 2 x \because cosx-1\sim -\frac{1}{2}x^2,tanx\sim x,\sqrt[3]{1+x^2}-1\sim \frac{1}{3}x^2,\sqrt{1+sinx}-1\sim \frac{1}{2}x ∵cosx−1∼−21x2,tanx∼x,31+x2−1∼31x2,1+sinx−1∼21x
∴ \therefore ∴ 原式 = lim x → 0 − 1 2 x 2 ⋅ x 1 3 x 2 ⋅ 1 2 x = − 3 =\lim_{x\rightarrow 0}\frac{-\frac{1}{2}x^2\cdot x}{\frac{1}{3}x^2\cdot \frac{1}{2}x}=-3 =limx→031x2⋅21x−21x2⋅x=−3
2.5 证明无穷小的等价关系具有下列性质:
(1) α ∼ α \alpha \sim \alpha α∼α(自反性)
(2) 若 α ∼ β \alpha \sim \beta α∼β,则 β ∼ α \beta \sim \alpha β∼α (对称性)
(3) 若 α ∼ β \alpha \sim \beta α∼β, β ∼ γ \beta \sim \gamma β∼γ,则 α ∼ γ \alpha \sim \gamma α∼γ (传递性)
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证明:
-
(1)
∵ lim α α = 1 \because \lim \frac{\alpha}{\alpha}=1 ∵limαα=1
∴ α ∼ α \therefore \alpha\sim\alpha ∴α∼α -
(2)
∵ α ∼ β \because \alpha \sim \beta ∵α∼β
∴ lim α β = 1 \therefore \lim \frac{\alpha}{\beta}=1 ∴limβα=1
∴ lim β α = lim 1 α β = 1 \therefore \lim \frac{\beta}{\alpha}=\lim \frac{1}{\frac{\alpha}{\beta}}=1 ∴limαβ=limβα1=1
∴ β ∼ α \therefore \beta \sim \alpha ∴β∼α -
(3)
∵ α ∼ β \because \alpha \sim \beta ∵α∼β
∴ lim α β = 1 \therefore \lim \frac{\alpha}{\beta}=1 ∴limβα=1
∵ β ∼ γ \because \beta \sim \gamma ∵β∼γ
∴ lim β γ = 1 \therefore \lim \frac{\beta}{\gamma}=1 ∴limγβ=1
∴ lim α γ = lim α β ⋅ β γ = 1 \therefore \lim \frac{\alpha}{\gamma}=\lim \frac{\alpha}{\beta}\cdot \frac{\beta}{\gamma}=1 ∴limγα=limβα⋅γβ=1
∴ α ∼ γ \therefore \alpha \sim \gamma ∴α∼γ
- 学习资料
1.《高等数学(第六版)》 上册,同济大学数学系 编