第三章 函数极限(下载后可解决看不到公式的问题)4 两个重要的极限一、证明：limx→0sin x x=1.证：∵sinx<x<tanx(0<x<π2)，∴1<xsin x <1cos x (0<x<π2)，∴cosx<sin x x<1 (0<x<π2)，又cos-x=cosx ，sin−x −x=sin x x，∴对0<|x|<π2，有cosx<sin x x<1.由lim x→0cosx=1，根据极限的迫敛性，lim x→0sin x x=1.例1：求limx→πsin x π−x.解：令t=π-x ，则sinx=sin(π-t)=sint ，且当x →π时，t →0， ∴lim x→πsin x π−x=lim t→0sin t t=1.例2：求lim x→01−cos x x 2.解：lim x→01−cos x x 2=lim x 2→012(sinx 2x 2)2=12，二、证明lim x→∞(1+1x )x=e.证：设f(x)= (1+1n+1)n, g(x)= (1+1n )n+1, n ≤x<n+1, n=1,2,…,则f(x)递增且有上界，g(x)递减且有下界，∴lim x→+∞f (x )与lim x→+∞g (x )都存在， 取{x n }={n}，由归结原则得lim x→+∞f (x )=lim n→+∞(1+1n+1)n=e ，lim x→+∞g (x )=lim n→+∞(1+1n )n+1=e ， 又1+1n+1<1+1x ≤1+1n ，则(1+1n+1)n<(1+1x )x<(1+1n )n+1，根据迫敛性定理得lim x→+∞(1+1x )x= e.设x=-y ，则(1+1x )x =(1−1y )−y =(1+1y−1)y，且当x →-∞，y →+∞，从而有lim x→−∞(1+1x )x=lim y→+∞(1+1y−1)y−1·(1+1y−1)=e.∴lim x→∞(1+1x )x=e.注：e 的另一种形式：lim a→0(1+a )1a =e.证：令a=1x ，则当a →0时，1x →∞，∴lim a→0(1+a )1a=lim 1x→∞(1+ 1x )x=e.例3：求lim x→0(1+2x )1x.解：lim x→0(1+2x )1x =lim 12x→∞[(1+2x )12x]2=e 2.例4：求lim x→0(1−x )1x.解：lim x→0(1−x )1x=lim −1x→∞1[1+(−x)]−1x=1e .例5：求lim n→∞(1+1n −1n 2)n.解：(1+1n −1n 2)n <(1+1n )n→e(n →∞)，又当n>1时有 (1+1n −1n 2)n=(1+n−1n 2)n 2n−1−nn−1≥(1+n−1n 2)n 2n−1−2→e(n →∞，即n−1n 2→0).由迫敛性定理得：lim n→∞(1+1n −1n 2)n=e.习题1、求下列极限： (1)lim x→0sin2x x；(2)lim x→0sin x 3(sin x)2；(3)lim x→π2cos xx−π2；(4)limx→0tan x x；(5)limx→0tan x−sin xx 3；(6)limx→0arctan xx ；(7)lim x→+∞x sin 1x ；(8)lim x→asin 2 x−sin 2 ax−a；(9)limx→0√x+1−1；(10)limx→0√1−cos x 21−cos x.解：(1)limx→0sin2x x=lim2x→02sin2x 2x=2；(2)lim x→0sin x 3(sin x)2=lim x→0x 3sin x 3x 3(sin x)2=lim x 3→0sin x 3x 3·lim x 2→0(xsin x )2·lim x→0x =0； (3)lim x→π2cos x x−π2=lim x−π2→0−sin (x−π2)x−π2= -1；(4)limx→0tan x x=limx→0sin x x·lim x→01cos x =1；(5)lim x→0tan x−sin xx 3=limx→0sin x (1cos x −1)x 3=limx→0sin x·1−cos x cos x x3=limx→02sin x 2cos x 2·2(sin x 2 )2cos xx 3=limx→04(sin x 2 )3·cos x2cos x x3=limx→0(sin x 2 )3·cos x2cos x2(x 2)3=lim x2→0(sinx2x 2)3·lim x2→0cos x 22lim x→0cos x =12；(6)令arctan x=y ，则x=tany ，且x →0时，y →0， ∴limx→0arctan xx=lim y→0y tan y =limy→0cos ysin y y=1；(7)lim x→+∞x sin 1x =lim 1x→0sin1x1x =1；(8)lim x→asin 2 x−sin 2 ax−a =limx→a (sin x−sin a )(sin x+sin a)x−a=limx→a2cosx+a 2 sin x−a2x−a·2sin a=limx−a 2→0sinx−a2x−a 2·cos a ·2sin a= sin2a ； (9)lim√x+1−1=limx→0(√x+1+1)sin 4xx=8lim4x→0sin 4x 4x=8；(10)lim x→0√1−cos x 21−cos x=limx→0√2sin x 222(sin x 2)2=√2limx→0sinx 22 x 22 (sinx 2x 2)2=√2.2、求下列极限：(1)lim x→∞(1−2x )−x；(2)lim x→0(1+ax )1x(a 为给定实数)；(3)lim x→0(1+tan x )cot x ；(4)lim x→0(1+x 1−x )1x ；(5)lim x→+∞(3x+23x−1)2x−1；(6)lim x→+∞(1+αx )βx(α,β为给定实数)解：(1)lim x→∞(1−2x )−x=lim −x2→∞[(1+1−x 2)−x 2]2=e 2；(2)lim x→0(1+ax )1x=lim ax→0[(1+ax )1ax]a=e a ；(3)lim x→0(1+tan x)cot x =lim tan x→0(1+tan x )1tan x=e ；(4)lim x→0(1+x1−x )1x=limx→0(1+x )1x(1−x )1x=lim x→0(1+x )1xlim−x→0{[1+(−x )]1−x }−1=e 2；(5)lim x→+∞(3x+23x−1)2x−1=lim x→+∞(1+33x−1)6x−33=lim33x−1→0+(1+33x−1)2(3x−1)−13=lim33x−1→0+(1+33x−1)2(3x−1)3lim 33x−1→0+(1+33x−1)13=e 2；(6)lim x→+∞(1+αx )βx=lim x→+∞(1+αx )αβxα=lim αx→0+[(1+αx )x α]αβ=e αβ.3、证明：lim x→0{lim n→∞[cos xcos x2cos x22…cos x2n ]}=1.证：∵cos xcos x2cos x 22…cos x2n =2n+1cos xcos x 2cos x 22…cos x 2n sin x2n2n+1sin x2n=sin 2x2n+1sin x2n=sin 2x 2x sin x 2nx 2n=x 2nsin x 2n·sin 2x 2x；∴当x ≠0时，lim n→∞[cos xcos x2cos x22…cos x2n ]=lim x2n →0x 2nsin x 2n·sin 2x 2x=sin 2x 2x；lim x→0{lim n→∞[cos xcos x 2cos x 22…cos x 2n ]}=lim2x→0sin 2x 2x=1.当x=0时，cos xcos x2cos x22…cos x2n =1， ∴lim x→0{lim n→∞[cos xcos x2cos x22…cos x2n ]}=1.4、利用归结原则计算下列极限：(1)limn→∞√n sinπn；(2)limn→∞(1+1n+1n2)n.解：(1)∵limx→∞√x sinπx=limx→∞sinπxπx·√x=limπx→0sinπxπx·lim√x=0根据归结原则，limn→∞√n sinπn=0.(2)∵当x>0时，(1+1x +1x2)x>(1+1x)x→e(x→+∞)，又(1+1x +1x2)x=(1+x+1x2)x2x+1+xx+1<(1+x+1x2)x2x+1→e(x→+∞，即x+1x2→0)，∴limx→+∞(1+1x+1x2)x=e根据归结原则，limn→∞(1+1n+1n2)n=e.。