2 4與2 5極限的運(yùn)算法則

1、2.4極限的運(yùn)算法則極限的運(yùn)算法則:性質(zhì)設(shè)limf(x) = A, lim g(x) = B,則xXxXlimf(x) g(x) = limf(x) lim g(x) = A B;(1)(2)(3)xXxXxXlimcf(x) = climf(x)xXxXlimf(x) g(x) = limf(x) lim g(x) = A B;xXxXlimf(x)xXf (x)A=其中B 0.xXlim g(x)xX(4)lim,xX g(x)B結(jié)論: 如果limf(x)存在,而n是正整數(shù),則xXlimf(x)n = limf(x)n .xXxX定理：如果j(x) y (x),而lim j(x) = a,

2、 limy (x) = b,x Xx X那么a b.例1 求下列極限x3- 1(1) lim.- 3x + 52x2 x= lim x 2 - lim 3x + lim 5Qlim( x 2 - 3x + 5)解x2x2x2x2= (lim x)2 - 3 lim x + lim 5x2x2x2= 22 - 3 2 + 5= 3 0,lim x 3 - lim 1x 3- 1- 1237lim= x 2x 2=.- 3 x + 5x 2lim( x 2 - 3 x + 5)33x 2x 2(2)lim xe xx2例2.求下列極限- 1xn(1)limx - 1x1lim (1 + x -(2

3、)x )x+ 13-(3)lim()1 - x1 - x2x1+ x - 2x2(4)limx- 12x1+ 6x + 38x2(5)lim- 4x + 7- 2 x - 122xx3 x2(6)lim- x+ 5322 xx- x2 + 52x3(7)lim- 2 x - 123 xx a0當(dāng)n = m b0xm -1+ a+L+ axmalim 01m = 0當(dāng)n m當(dāng)n N時(shí),有yn xn zn ,且lim yn= lim zn= a,則lim xn= annn111設(shè)y=+L例1,求lim ynn+ 1+ 2+ nn2n2n2n（2）函數(shù)極限的夾逼準(zhǔn)則若在極限過程x X所允許的某一鄰域

4、內(nèi), g(x) f (x) h(x),且lim g(x) = limh(x) = A,xXxX則limf(x) = AxX例2求lim x 1 x x0解: Q 1 - 1 1 1(如: 2.5 = 2,2.5 - 1 2.5 2.5,)xxx當(dāng)x 0,有1 - x x1 1 lim x1 = 1xxx0+當(dāng)x 0,有1 x1 1 - x lim x1 = 1xxx0-1lim x = 1xx02、極限存在的準(zhǔn)則(II)準(zhǔn)則單調(diào)有界數(shù)列必有極限.幾何解釋:xx2xn+1x1x3xnMAlim(1 + 1)n = e-重要極限()可證:nn 1 , x例3 : 設(shè)數(shù)列 x滿足0 x= x(1 -

5、 2 x),n+1n1nn2單減且0 x 1 , n = 1,2,Ln = 1,2,L, 證: (1) xnn2(2)求lim xnn二、兩個(gè)重要極限()()lim(1 + 1 ) x= ex xlim sin x = 1x 0xC利用夾逼定理可證重要極限Blim sin x = 1()xx 0xpoA設(shè)單位圓 O, 圓心角AOB = x,(0 x )2作單位圓的切線，得DACO .DDOAB的高為BD ,于是有sin x = BD,扇形OAB的圓心角為x ,x = 弧AB,tan x = AC ,sin x x tan x, 即cos x sin x 1,(對- p x 0,xX= AB1y = f(x)g(x),lim g(x) = BxX設(shè)lim f (x)g(x) xX則有x - 11sin()例7求lim()xxx四.連續(xù)復(fù)利設(shè)一筆貸款A(yù)0 (本金)，年利率為r,則一年后的本利和為 A1 = A0 (1+ r)A= A (1+ r)kk年后的本利和為k0如果一年分n期計(jì)息，年利率仍為r,則每期利率為r ,n于是一年后的本利和為rA= A (1+)n10nrA= A (1+n