3. 函数极限与连续函数
3.3 无穷小量与无穷大量的阶
3.3.3 应用等价量的关系求函数极限
【定理3.3.1】 u ( x ) , v ( x ) , w ( x ) u(x),v(x),w(x) u(x),v(x),w(x)在 x 0 x_{0} x0的某个去心邻域上有定义,且 lim x → x 0 v ( x ) w ( x ) = 1 \lim\limits_{x\to x_{0}}\frac{v(x)}{w(x)}=1 x→x0limw(x)v(x)=1,即 v ( x ) ∼ w ( x ) , ( x → x 0 ) v(x)\sim w(x),(x\to x_{0}) v(x)∼w(x),(x→x0),则:
(1) lim x → x 0 u ( x ) w ( x ) = A ⇔ lim x → x 0 u ( x ) v ( x ) = A \lim\limits_{x\to x_{0}}u(x)w(x)=A\Leftrightarrow\lim\limits_{x\to x_{0}}u(x)v(x)=A x→x0limu(x)w(x)=A⇔x→x0limu(x)v(x)=A;
(2) lim x → x 0 u ( x ) w ( x ) = A ⇔ lim x → x 0 u ( x ) v ( x ) = A \lim\limits_{x\to x_{0}}\frac{u(x)}{w(x)}=A\Leftrightarrow\lim\limits_{x\to x_{0}}\frac{u(x)}{v(x)}=A x→x0limw(x)u(x)=A⇔x→x0limv(x)u(x)=A
【注】用等价量代替乘积或除的因子,只限乘和除,加减不行。
【常见的等价无穷小量如下( x → 0 x\to 0 x→0), x x x可以换成 f ( x ) f(x) f(x)进行广义的替换】
(1) x ∼ sin x ∼ tan x ∼ arcsin x ∼ arctan x ∼ ln ( 1 + x ) ∼ e x − 1 ∼ ln ( 1 + 1 + x 2 ) x \sim \sin x \sim \tan x \sim \arcsin x \sim \arctan x \sim \ln (1+x) \sim e^{x}-1 \sim \ln \left(1+\sqrt{1+x^{2}}\right) x∼sinx∼tanx∼arcsinx∼arctanx∼ln(1+x)∼ex−1∼ln(1+1+x2);
(2) ( 1 + x ) α − 1 ∼ α x (1+x)^{\alpha}-1 \sim \alpha x (1+x)α−1∼αx;
(3) 1 − cos x ∼ 1 2 x 2 , 1 − cos α x ∼ α 2 x 2 1-\cos x \sim \frac{1}{2} x^{2}, \quad 1-\cos ^{\alpha} x \sim \frac{\alpha}{2} x^{2} 1−cosx∼21x2,1−cosαx∼2αx2;
(4) a x − 1 ∼ x ln a a^{x}-1 \sim x \ln a ax−1∼xlna;
(5) x − sin x ∼ x 3 6 , arcsin x − x ∼ x 3 6 x-\sin x \sim \frac{x^{3}}{6}, \quad \arcsin x-x \sim \frac{x^{3}}{6} x−sinx∼6x3,arcsinx−x∼6x3;
(6) tan x − x ∼ x 3 3 , x − arctan x ∼ x 3 3 \tan x-x \sim \frac{x^{3}}{3}, \quad x-\arctan x \sim \frac{x^{3}}{3} tanx−x∼3x3,x−arctanx∼3x3;
(7) x + sin x ∼ 2 x x + \sin x \sim 2x x+sinx∼2x;
(8) tan x − sin x ∼ x 3 2 , arcsin x − arctan x ∼ x 3 2 \tan x-\sin x \sim \frac{x^{3}}{2}, \quad \arcsin x-\arctan x \sim \frac{x^{3}}{2} tanx−sinx∼2x3,arcsinx−arctanx∼2x3;
(9) 若 x → 0 , α ( x ) → 0 , α ( x ) β ( x ) → 0 则有 ( 1 + α ( x ) ) β ( x ) − 1 ∼ α ( x ) β ( x ) \text { 若 } x \rightarrow 0, \alpha(x) \rightarrow 0, \alpha(x) \beta(x) \rightarrow 0 \text { 则有 }(1+\alpha(x))^{\beta(x)}-1 \sim \alpha(x) \beta(x) 若 x→0,α(x)→0,α(x)β(x)→0 则有 (1+α(x))β(x)−1∼α(x)β(x)
【例3.3.9】计算 lim x → 0 ln ( 1 + x 2 ) ( e 2 x − 1 ) tan x \lim\limits_{x\to 0}\frac{\ln(1+x^{2})}{(e^{2x}-1)\tan x} x→0lim(e2x−1)tanxln(1+x2)
【解】 lim x → 0 ln ( 1 + x 2 ) ( e 2 x − 1 ) tan x = lim x → 0 x 2 2 x ⋅ x = 1 2 \lim\limits_{x\to 0}\frac{\ln(1+x^{2})}{(e^{2x}-1)\tan x}=\lim\limits_{x\to 0}\frac{x^{2}}{2x\cdot x}=\frac{1}{2} x→0lim(e2x−1)tanxln(1+x2)=x→0lim2x⋅xx2=21
【例3.3.10】计算 lim x → 0 1 + x − e x 3 ln ( 1 + 2 x ) \lim\limits_{x\to 0}\frac{\sqrt{1+x}-e^{\frac{x}{3}}}{\ln(1+2x)} x→0limln(1+2x)1+x−e3x.
【解】 lim x → 0 1 + x − e x 3 ln ( 1 + 2 x ) = lim x → 0 ( 1 + x − 1 ) + 1 − e x 3 2 x = lim x → 0 ( 1 + x − 1 ) − ( e x 3 − 1 ) 2 x \lim\limits_{x\to 0}\frac{\sqrt{1+x}-e^{\frac{x}{3}}}{\ln(1+2x)}=\lim\limits_{x\to 0}\frac{(\sqrt{1+x}-1)+1-e^{\frac{x}{3}}}{2x}=\lim\limits_{x\to 0}\frac{(\sqrt{1+x}-1)-(e^{\frac{x}{3}}-1)}{2x} x→0limln(1+2x)1+x−e3x=x→0lim2x(1+x−1)+1−e3x=x→0lim2x(1+x−1)−(e3x−1)
由于 1 + x − 1 ∼ x 2 ⇔ 1 + x − 1 = x 2 + o ( x ) \sqrt{1+x}-1\sim\frac{x}{2}\Leftrightarrow\sqrt{1+x}-1=\frac{x}{2}+o(x) 1+x−1∼2x⇔1+x−1=2x+o(x)
e x 3 − 1 ∼ x 3 ⇔ e x 3 − 1 = x 3 + o ( x ) e^{\frac{x}{3}}-1\sim\frac{x}{3}\Leftrightarrow e^{\frac{x}{3}}-1=\frac{x}{3}+o(x) e3x−1∼3x⇔e3x−1=3x+o(x)
所以 lim x → 0 ( 1 + x − 1 ) − ( e x 3 − 1 ) 2 x = lim x → 0 ( x 2 + o ( x ) ) − ( x 3 + o ( x ) ) 2 x = lim x → 0 1 6 x + o ( x ) 2 x = 1 12 \lim\limits_{x\to 0}\frac{(\sqrt{1+x}-1)-(e^{\frac{x}{3}}-1)}{2x}=\lim\limits_{x\to 0}\frac{(\frac{x}{2}+o(x))-(\frac{x}{3}+o(x))}{2x}=\lim\limits_{x\to 0}\frac{\frac{1}{6}x+o(x)}{2x}=\frac{1}{12} x→0lim2x(1+x−1)−(e3x−1)=x→0lim2x(2x+o(x))−(3x+o(x))=x→0lim2x61x+o(x)=121
【注】 o ( f ( x ) ) o(f(x)) o(f(x))代表函数类,里面的函数都是关于某一个函数的低价无穷小或低阶无穷大,函数类加减不改变原来的性质。 o o o和 O O O不表示函数,而是表示函数渐进性质的特征。所以 o ( x ) ± o ( x ) = o ( x ) o(x)\pm o(x)=o(x) o(x)±o(x)=o(x), o ( x ) o(x) o(x)是 x x x的高阶无穷小量,除上 2 x , ( x → 0 ) 2x,(x\to 0) 2x,(x→0)的极限是0.
【例3.3.11】计算 lim x → ∞ x ( x 3 + x 3 − x 3 − x 3 ) \lim\limits_{x\to \infty}x(\sqrt[3]{x^{3}+x}-\sqrt[3]{x^{3}-x}) x→∞limx(3x3+x−3x3−x).
【解】 lim x → ∞ x ( x 3 + x 3 − x 3 − x 3 ) = lim x → ∞ x 2 ( 1 + 1 x 2 3 − 1 − 1 x 2 3 ) = lim x → ∞ x 2 ( ( 1 + 1 x 2 3 − 1 ) − ( 1 − 1 x 2 3 − 1 ) ) = lim x → ∞ x 2 ( 1 3 x 2 + o ( 1 x 2 ) − ( − 1 3 x 2 + o ( 1 x 2 ) ) ) = lim x → ∞ 2 x 2 3 x 2 + x 2 o ( 1 x 2 ) = lim x → ∞ 2 x 2 3 x 2 + o ( 1 x 2 ) 1 x 2 = 2 3 \lim\limits_{x\to \infty}x(\sqrt[3]{x^{3}+x}-\sqrt[3]{x^{3}-x})=\lim\limits_{x\to \infty}x^{2}(\sqrt[3]{1+\frac{1}{x^{2}}}-\sqrt[3]{1-\frac{1}{x^{2}}})=\lim\limits_{x\to \infty}x^{2}(\sqrt[3]{(1+\frac{1}{x^{2}}}-1)-(\sqrt[3]{1-\frac{1}{x^{2}}}-1))=\lim\limits_{x\to \infty}x^{2}(\frac{1}{3x^{2}}+o(\frac{1}{x^{2}})-(-\frac{1}{3x^{2}}+o(\frac{1}{x^{2}})))=\lim\limits_{x\to \infty}\frac{2x^{2}}{3x^{2}}+x^{2}o(\frac{1}{x^{2}})=\lim\limits_{x\to \infty}\frac{2x^{2}}{3x^{2}}+\frac{o(\frac{1}{x^{2}})}{\frac{1}{x^{2}}}=\frac{2}{3} x→∞limx(3x3+x−3x3−x)=x→∞limx2(31+x21−31−x21)=x→∞limx2(3(1+x21−1)−(31−x21−1))=x→∞limx2(3x21+o(x21)−(−3x21+o(x21)))=x→∞lim3x22x2+x2o(x21)=x→∞lim3x22x2+x21o(x21)=32
【例3.3.12】计算 lim x → 0 ( cos x ) 1 x 2 \lim\limits_{x\to 0}(\cos x)^{\frac{1}{x^{2}}} x→0lim(cosx)x21
【解】 lim x → 0 ( cos x ) 1 x 2 = lim x → 0 e 1 x 2 ln ( cos x ) = lim x → 0 e 1 x 2 ln ( 1 − 1 + cos x ) = lim x → 0 e cos x − 1 x 2 = lim x → 0 e − x 2 2 x 2 = e − 1 2 = 1 e \lim\limits_{x\to 0}(\cos x)^{\frac{1}{x^{2}}}=\lim\limits_{x\to 0}e^{\frac{1}{x^{2}}\ln(\cos x)}=\lim\limits_{x\to 0}e^{\frac{1}{x^{2}}\ln(1-1+\cos x)}=\lim\limits_{x\to 0}e^{\frac{\cos x - 1}{x^{2}}}=\lim\limits_{x\to 0}e^{\frac{-\frac{x^2}{2}}{x^{2}}}=e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}} x→0lim(cosx)x21=x→0limex21ln(cosx)=x→0limex21ln(1−1+cosx)=x→0limex2cosx−1=x→0limex2−2x2=e−21=e1
【例】计算 lim x → 0 tan x − sin x x 3 \lim\limits_{x\to 0}\frac{\tan x - \sin x}{x^{3}} x→0limx3tanx−sinx(不能直接在加减中用等价无穷小量,这样得到的结果0是错的,因为等价无穷小实际上是等于等价量+一个高阶无穷小,此处加的是 o ( x ) o(x) o(x),但是 o ( x ) o(x) o(x)是比 x x x的高阶无穷小,分母是 x 3 x^{3} x3, lim x → 0 tan x − sin x x 3 = lim x → 0 x + o ( x ) − x − o ( x ) x 3 = lim x → 0 o ( x ) x 3 = ? \lim\limits_{x\to 0}\frac{\tan x - \sin x}{x^{3}}=\lim\limits_{x\to 0}\frac{x+o(x)-x-o(x)}{x^{3}}=\lim\limits_{x\to 0}\frac{o(x)}{x^{3}}=? x→0limx3tanx−sinx=x→0limx3x+o(x)−x−o(x)=x→0limx3o(x)=?,只说了比 x x x高阶,不知道和 x 3 x^{3} x3的阶数的大小关系,所以直接在加减中运用是错误的)
【解】实际上用刚才我给的那些等价(就是麦克劳林展开式)可以直接计算
lim x → 0 tan x − sin x x 3 = lim x → 0 x 3 2 x 3 = 1 2 \lim\limits_{x\to 0}\frac{\tan x - \sin x}{x^{3}}=\lim\limits_{x\to 0}\frac{\frac{x^3}{2}}{x^3}=\frac{1}{2} x→0limx3tanx−sinx=x→0limx32x3=21
【例】计算 lim x → 0 ( 1 + x − 1 ) − x 2 x 2 \lim\limits_{x\to 0}\frac{(\sqrt{1+x}-1)-\frac{x}{2}}{x^{2}} x→0limx2(1+x−1)−2x.(直接加减中代换是错的)
【解】 lim x → 0 ( 1 + x − 1 ) − x 2 x 2 = lim x → 0 1 + x − ( 1 + x 2 ) x 2 = lim x → 0 ( 1 + x ) − ( 1 + x 2 ) 2 x 2 ( 1 + x + 1 + x 2 ) = lim x → 0 1 + x − 1 − x − x 2 4 2 x 2 = − 1 8 \lim\limits_{x\to 0}\frac{(\sqrt{1+x}-1)-\frac{x}{2}}{x^{2}}=\lim\limits_{x\to 0}\frac{\sqrt{1+x}-(1+\frac{x}{2})}{x^{2}}=\lim\limits_{x\to 0}\frac{(1+x)-(1+\frac{x}{2})^{2}}{x^{2}(\sqrt{1+x}+1+\frac{x}{2})}=\lim\limits_{x\to 0}\frac{1+x-1-x-\frac{x^2}{4}}{2x^{2}}=-\frac{1}{8} x→0limx2(1+x−1)−2x=x→0limx21+x−(1+2x)=x→0limx2(1+x+1+2x)(1+x)−(1+2x)2=x→0lim2x21+x−1−x−4x2=−81