2 4与2 5极限的运算法则

1、2.4极限的运算法则极限的运算法则:性质设limf(x) = A, lim g(x) = B,则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结论: 如果limf(x)存在,而n是正整数,则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 求下列极限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.求下列极限- 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当n = m b0xm -1+ a+L+ axmalim 01m = 0当n m当n N时,有yn xn zn ,且lim yn= lim zn= a,则lim xn= annn111设y=+L例1,求lim ynn+ 1+ 2+ nn2n2n2n（2）函数极限的夹逼准则若在极限过程x X所允许的某一邻域

4、内, g(x) f (x) h(x),且lim g(x) = limh(x) = A,xXxX则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当x 0,有1 - x x1 1 lim x1 = 1xxx0+当x 0,有1 x1 1 - x lim x1 = 1xxx0-1lim x = 1xx02、极限存在的准则(II)准则单调有界数列必有极限.几何解释:xx2xn+1x1x3xnMAlim(1 + 1)n = e-重要极限()可证:nn 1 , x例3 : 设数列 x满足0 x= x(1 -

5、 2 x),n+1n1nn2单减且0 x 1 , n = 1,2,Ln = 1,2,L, 证: (1) xnn2(2)求lim xnn二、两个重要极限()()lim(1 + 1 ) x= ex xlim sin x = 1x 0xC利用夹逼定理可证重要极限Blim sin x = 1()xx 0xpoA设单位圆 O, 圆心角AOB = x,(0 x )2作单位圆的切线，得DACO .DDOAB的高为BD ,于是有sin x = BD,扇形OAB的圆心角为x ,x = 弧AB,tan x = AC ,sin x x tan x, 即cos x sin x 1,(对- p x 0,xX= AB1y = f(x)g(x),lim g(x) = BxX设lim f (x)g(x) xX则有x - 11sin()例7求lim()xxx四.连续复利设一笔贷款A0 (本金)，年利率为r,则一年后的本利和为 A1 = A0 (1+ r)A= A (1+ r)kk年后的本利和为k0如果一年分n期计息，年利率仍为r,则每期利率为r ,n于是一年后的本利和为rA= A (1+)n10nrA= A (1+n