大一上高数b1老师版

2/ ln(1+ x→0 −1)tan√

1+x−e3ln(1+ etanx−esin x→0(x+x2)ln(1+x)arcsin 3/若limf(x)=0,x→x

=o(1)(x→x0)..4/设x→x0时,u(x)v(x)为无穷小limu(x)0,u(x)是v(x)的高阶无穷小,vx→x0vu(x)

o(

limx→0

=0,x2=o(x)limx→0

=

x3=o(x)5/limu(x)=1,······等价无穷小,u(x)∼v(x)(xx),x→x0

u(x)=v(x)+o(v(x))(x→x0).例如

=1,sinx∼x(x→0),sinx=x+o(x)(x→0).6/limu(x)=C̸=0,······同阶无穷小,x→x0u(x)=O(v(x))(x→x0), 或u(x)∼Cv(x)(x→x0).例如

1−cosx

1,1−cosx

1

(x0),

o =1 2 ()(→0).当x→x0,u(x)→0时,常取v(x)=(x− 7/定义若limf(x)=∞,x→x

时,f(x)是无穷大量设x→x0时,u(x)v(x)为无穷大量limu(x)=∞,u(x)是v(x)的高阶无穷大x→x0limu(x)=1,······等价无穷大,记为u(x)∼v(x)(x→x limu(x)=C̸=0,u(x)是v(x)的同阶无穷大x→x0,设a>1,p>0 ,

=+∞,

=+∞x→+∞ x→+∞ln 8/例.设f(xx2x5.x→∞,f(x)∼1例.设g(xx√x.1

x→0,f(x)∼1x→+∞,g(x)∼x21结论:设u(xu1(x

x→0,g(x)∼x4u(x为无穷大量ui(x等价u(x为无穷小量,uj(x等价9/等价量:等价无穷小量或等价无穷大量.当x→0时,常用的等价无穷小:sinx∼x,ln(1+x)∼x,ex−1∼x,(1+x)α−1∼1−cosx

1x2,tanx∼x,arcsinx∼x,arctanx∼210/定理6.4(积与商的代换u(x)v(x)w(x)在x

的某个去心邻域上有定义,且limv(x)=x→x0即v(x)∼w(x)(x→x0),limu(x)v(x)=lim limv(x)=limw(x)x→x0 limu(x)=limu(x)x→x0 x→x0等价代换的目的:为了化简将积或商中表达式复杂的因子换11/例如:(1)

ln(1+x→0(e2x−1)sinx(2)

31+tan2x−

sin(x12/问6.5和与差的等价代换问题u(x)−求

.假设u(x)∼u1(x),v(x)∼

1(x)(x→x

limu(x)− =

u1(x)−

(∗) √

例如:(3)

1+x−eln(1+ (4)limtanx−sinx 13/limlimu(x)−??=limu1(x)−在以下过程中,情形1. limu−v

=lim

ww–limww等.. =limw−lim=limu1 w等式成立的条件:limu1,limv1都存在例如:

√ 31+x 3le(1 14/limlimu(x)−=limu1(x)−

u−v∼u1−v,1即 u−v

=u1−limu−v−(u1−v1)=u1−=u−u1−v−v1=u/u1−1−v/v1−1u1− u1− 1− u1/v1−分子中,u/u1−1,v/v1−1→0.欲使整个式子极限为零,极限非零,即limu1=:设u∼u,v∼v.当muv

1时,u−v∼u1−v15/定理6.6和与差中的等价代换求limu(x−v(x)时,设下面两个条件之一成立 (1)limu(x) limv(x)都存在 (2)

u(x)̸=1.x→x0 x→x0 x→x0u∼u1,v∼v1,limu(x)−v(x)=limu1(x)−v1(x).√

例如:

1+x le(1

,(1)(2)都满足limtanx−sinx,(1)(2都不满足 16/tanx

sinx(1−cos

x·1 lim =

=lim =

x3cos

.tanx∼x,sinx∼将得到错误的结果.tanx=x+o(x),sinx=x+o(x),limtanx−sinx=limo(x)=? tanx∼x,sinx∼x,tanx−sinxx的高阶无穷小.但具体是2阶,还是3阶,我们并不清楚.17/作业18/定理6.7(无穷小量的运算当x→0时,用o(x)表示比x高阶的无穷小量 o(xm)+o(xn)=o(xm)(mo(xn)±o(xn)=xm·o(xn)=o(xm)·o(xn)=k为常数时,k·o(xo(xφ(x有界时,φ(x·o(xn 19/

etanx−esin.2.x→0(x+x)ln(1+x)arcsin分ߤ.当x→0时,x+x2=x(1+x)∼x,ln(1+x)∼x,arcs