高等数学
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
与详解 1 广东
第一章 函数与极限习题详解
第一章 函数与极限
习 题 1-1
1(求下列函数的自然定义域:
1(1); yx,,,221,x
2,10,,x解:依题意有,则函数定义域( Dxxx()|2x1,,,,,且,,,x,,20,
21x,arccos3(2); y,2xx,,6
,,21x,1,解:依题意有,则函数定义域( Dx(),,3,2,xx,,,60,2(3); yxx,,,,ln(32)
2解:依题意有,则函数定义域( Dxxx()|12,,,,,,,xx320,,
13xx,(4); y,2
3解:依题意有,则函数定义域( Dxxx()|x0,1,,,,,,,,,且xx,,0,,
1,sin1,, x,,(5) y, x,1,
,21;, x,,
解:依题意有定义域( Dxxx()|,,,,,,,,,
1(6). yx,,,arctan3x
x,0,解:依题意有,则函数定义域( Dxxx()|3x0,,,且,,,30,,x,
22(已知定义域为,求 fx()[0,1]fxfxfxafxafxa(), (sin), (), ()(),,,,
()的定义域( a,0
22解:因为定义域为,所以当时,得函数的定义域为; fx()[0,1][1,1],fx()01,,x
当时,得函数定义域为; fx(sin)[2kkπ,(21),π]0sin1,,x
当时,得函数定义域为; fxa(),[,1],,,aa01,,,xa
01,,,xa,1111当时,得函数定义域为:(1)若,;(2)若,;(3)若,( fxafxa()(),,,xaa,,,1x,,a,a,x,a,,,,01,,,xa2222,
,,1ax,3(设其中求函数值( faf(2),(1)a,0,fx()1,,,,,222xaaxx,,2,,
,,1ax,解:因为,则 fx()1,,,,222xaaxx,,2,,
,,0 ,>1,a,11,a11a,,,,( fa(2)1,,,f(1)1,,,,,,,,222,,2 ,0<<1a42aaa11a,,,,,,
1||1,x,,,x4(设,求与,并做出函数图形( fgx(())gfx(())fxxgx()0||1,()2,,,,
,,,1||1.x,
x,,12110x,,,,x解:,即, fgx,,(())021fgxx(())00,,,,,,x,,1 0x,,1 21,,
,1,2||1x,,2||1x,,,0,即,函数图形略( gfxx(())2||1,,gfxx(())1||1,,,,
,,,112||1x,,, ||1x,,2
1,0,,,xx2,1,,,,xx,,5(设试证: fx(),ffx[()],,,1,0,x,1,1.x,,,,
1(),()0,,fxfx2,1,,,,xx,,证明:,即,得证( ffx[()],ffx[()],,,1,()0fx,1,1x,,,,6(下列各组函数中,与是否是同一函数,为什么, fx()gx()
22(1) ; fxxxgxx()ln3,()ln33,,,,,,,,,,,
不是,因为定义域和对应法则都不相同(
33532(2); fxxxgxxx()2,()2,,,,
是(
22(3); fxgxxx()2,()sectan,,,
不是,因为对应法则不同(
2(4); fxxgxx()2lg,()lg,,
不是,因为定义域不同(
1
第一章 函数与极限习题详解 7(确定下列函数在给定区间内的单调性:
(1),; yxx,,3lnx,,,(0,)
解:当时,函数单调递增,也是单调递增,则在内也是递增的( x,,,(0,)yx,3yx,lnyyy,,(0,),,1212
,x (2),( y,x,,,(,1)1,x
,x11,,,xx(1)11解:,当时,函数单调递增,则是单调递减的,故原函数是单调递y,x,,,(,1)yx,,1y,,y,,,,1211,xyx,1111,,,xxx1
减的(
. 判定下列函数的奇偶性( 8
2(1); yxx,,,lg(1)
2212,解:因为, fxxxxxxxfx()lg(1)lg(1)lg(1)(),,,,,,,,,,,,,,
2所以是奇函数( yxx,,,lg(1)
(2); y,0
解:因为,所以是偶函数( fxfx()0(),,,y,0
2(3); yxxx,,,,2cossin1
22解:因为,,所以既非奇函数,又非偶函数( fxfxfxfx()()()(),,,,,且fxxxx()2cossin1,,,,,yxxx,,,,2cossin1
xx,aa,(4). y,2,xxxx,aa,aa,解:因为,所以函数是偶函数( ()()fxfx,,y,22
9(设是定义在上的任意函数,证明: fx()[,],ll
(1)是偶函数,是奇函数; fxfx()(),,fxfx()(),,
(2)可表示成偶函数与奇函数之和的形式. fx()
证明:(1)令,则 gxfxfxhxfxfx()()(),()()(),,,,,,
,所以是偶函数,是奇函数( gxfxfxgxhxfxfxhx()()()(),()()()(),,,,,,,,,,,fxfx()(),,fxfx()(),,
fxfxfxfx()()()(),,,,fxfx()(),,fxfx()(),,(2)任意函数,由(1)可知是偶函数,是奇函数,所以命题得证( fx(),,2222
10(证明:函数在区间上有界的充分与必要条件是:函数在上既有上界又有下界. II证明:(必要性)若函数在区间上有界,则存在正数,使得,都有成立,显然,即证得函数在IMfxM(),fx(),,,MfxM()fx()xI,
区间上既有上界又有下界 I
(充分性)设函数在区间上既有上界,又有下界,即有,取,则有,即函数Ifx()MMM,max{,}fxM(),fx()fxMfxM()(),,且MM122112
在区间上有界( I
11(下列函数是否是周期函数,对于周期函数指出其周期: (1); yx,|sin|
周期函数,周期为( π
(2); yx,,1sinπ
周期函数,周期为2(
(3); yxx,tan
不是周期函数(
2(4). yx,cos
周期函数,周期为( π
12(求下列函数的反函数: x3(1); y,x,31
yyx解:依题意,,则,所以反函数为 3x,log,31y,1y,
x,1( fxx()log,(,0)(1,),,,,,,,3x,1
axb,(2); yadbc,,()cxd,
bdxbdy,,,1解:依题意,,则反函数( fxadbc()()x,,,cya,cxa,
2(3); yxx,,,lg1,,
11yy,,,1xx解:依题意,,所以反函数( x,,fxxR,,,(1010)()(1010),22
ππ,,(4)( yxx,,,,3cos2,,,44,,
yxarccosarccos,133解:依题意,,所以反函数( ,,fxx(),[0,3]x,22
13(在下列各题中,求由所给函数构成的复合函数,并求这函数分别对应于给定自变量值和的函数值: xx12u2(1); yuxxx,,,,e,1,0,2,122v(2)( yuuvxxx,,,,,,,,,1,e1,1,1,1122x,15解:(1) yfxfefe,,,,()e,(0),(2)
x,1242(2),,f(1)1,,( yfx,,,,()(e1)1fee(0)22,,,
rH14(在一圆柱形容器内倒进某种溶液,该容器的底半径为,高为(当倒进溶液后液面的高度为时,溶液的体积为(试把表示为的hVhV
函数,并指出其定义区间(
V22解:依题意有,则( hVrH,,,[0,π]Vrh,π2πr
2
第一章 函数与极限习题详解
15(某城市的行政管理部门,在保证居民正常用水需要的前提下,为了节约用水,制定了如下收费方法:每户居民每月用水量不超过4.5吨
时,水费按0.64元,吨计算(超过部分每吨以5倍价格收费(试建立每月用水费用与用水数量之间的函数关系(并计算用水量分别为3.5吨、4.5
吨、5.5吨的用水费用(
0.64,04.5xx,,,解:依题意有,所以 fx(),,4.50.64(4.5)3.2,4.5,,,,,xx,
( fff(3.5)2.24(4.5)2.88(5.5)6.08,,,元,元,元
习 题 1-2
21n,1(设, an,,(1,2,3,)n31n,
222(1) 求的值; ||,||,||aaa,,,110100333
2,4(2) 求,使当时,不等式成立; a,,||10NnN,n3
2(3) 求,使当时,不等式成立( a,,,NnN,||n3
232122121解:(1) ||||,a,,,,||||,a,,,,11034312331393
220121 ( ||||a,,,,10033013903
211999799972,,,4,4(2) 要使 即 , 则只要 取N, 故当n>1110时,不等式成立( a,,a,,n,,||10,1110,||10,,nn4,,99333310(n+1),,
213,,13,,2,,(3)要使成立, 取,那么当时, a,,,a,,,n,,nN,||N,||nn,,9933,,,,
成立.
2(根据数列极限的定义证明:
21n,3(1); (2)( ,lim0lim1,n,,n,,n!n
1111111,,,,解:(1), 要使, 只要取, 所以,对任意,存在,当时,总有,则. ,,,,,,,,,,,,0,,,0,nN,m0ilNN|0||0|,,,,n,,,,n!n!nnn!!,,,,
2,,,,333n,332 (2) ,要使, 即,只要取,所以,对任意的>0,存在, 当, 总有,,,,,0nN,,nN,N,|1|,,,,,,,,22,,,222nn2nnn(3),,,,,,
22n,3n,3, 则. ,|1|,,lim1,n,,nn
3(若证明(并举例说明:如果数列有极限,但数列未必有极限( ||xxlimxa,,lim||||xa,,,,,nnnn,,,,nn
证明: 因为, 所以, , 当时, 有.不妨假设a>0, 由收敛数列的保号性可知:, 当时, 有,NnN,||xa,,,nN,,N,,,0limxa,11n22n,,n
, 取, 则对, , 当时, 有.故. 同理可证时, NNN,max,x,0||||||||xaxa,,,,,,,,0,NnN,lim||||xa,a,0lim||||xa,,,12nnnnn,,,,nn
成立. n反之,如果数列有极限, 但数列未必有极限.如:数列, , 显然, 但不存在( ||x||xx,,1||1x,lim||1x,limx,,,,,,nnnnnn,,,,nn
4(设数列有界,又(证明:( xlim0y,lim0xy,,,nnnn,,,,nn
证明: 依题意,存在M>0, 对一切n都有, 又, 对, 存在, ||xM,lim0y,,,,0Nnn,,n
当时, , 因为对上述, 当时, ,由的任意性, 则( |0|y,,,|0|||||xyxyMyM,,,,,,nN,NnN,lim0xy,nnnnnnnn,,n
1(3)n,π5(设数列的一般项,求( xx,coslimx,,nnn,,n2n
1(3)n,π1(3)n,π解: 因为, , 所以 . ,lim0limcos0,|cos|1,x,,x,,22nn
6(对于数列,若,,证明:( xxAk,,,()xAk,,,()xAn,,,(),,n21k,2kn
证明: 由于, 所以, , , 当时,有, 同理, ,, 当时, 有(取,,N0||xA,,,,,N0kN,||xA,,,kN>,,,0,,,0limxA,1121k,222k21k,k,,
=maxNN,, , 当时, 成立, 故( ||xA,,,xAn,,,()N,,,0nN,,,12nn
习 题 1-3
21(当时,(问等于多少,使当时,, |1|x,,,|4|0.01y,,yx,,,34x,1,
135解:令 ,则,要使 |1|x,,,,,|1|x222
522, |4||34||1||1||1||1|0.01yxxxxx,,,,,,,,,,,,2只要|1|0.004x,,,所以取,使当 |1|x,,, 时,|4|0.01y,,成立( ,,0.004
221x,X2(当时,(问等于多少,使当时,, ||xX,|2|0.001y,,x,,y,,22x,32217x,22解:要使<0.001, 只要, 即. 因此,只要就可以了,所以取( |3|7000x,,||7003x,x,,37000X,7003|2||2|y,,,,22xx,,3|3|
3(根据函数极限的定义证明:
35x,(1); (2); lim(21)5x,,lim3,x,,x,3x,1
3
第一章 函数与极限习题详解
2sinxx,4(3); (4). ,lim0lim4,,x,,,x,,2x,2x
,,证明:(1) 由于, 任给,要使,只要(因此取,则当时, 总有,|(21)5|2|3|xx,,,,|(21)5|x,,,,|3|x,,0|3|,,,x,|(21)5|x,,,,,,,0,22
故( lim(21)5x,,x,3
358x,835x,8888 (2) 由于,任给, 要使,只要,即或, 因为,所以, 取,,,,,,,,,,,,0x1x1,,0|3|,,|3|,,|1||1|,,,,,|1|xxx,,1|1|x,1
835x,35x,,则当时, 对,总有,故有( ,,||xM,,,,,0lim3,M|1||3|,,x,,,x,1x,122x,4x,4 (3)由于,任给,,要使,只要,因此取,则当时,总有|2|x,,,0|(2)|,,,,x,,,,0,,,|(4)|,,,|(4)||2|,,,,xx,2x,222x,4x,4,故. ,lim4,,|(4)|,,,x,,2x,2x,2
sin|sin|1xxsinx111sinx (4) 由于,任给,要使,只要,即,因此取,则当x>M时,总有,故,,,,,,,,,,Mx,,0|0|,,,|0||0|22,,xxxxxx
sinx. ,lim0x,,,x
4(用或语言,写出下列各函数极限的定义: ,,X,,,
(1); (2); lim()1fx,lim()fxa,x,,,x,,
(3); (4)( lim()fxb,lim()8fx,,,,xa,x,3
解: (1) , 当x<-M时, 总有; |()1|fx,,,,,,0,,,M0
(2) , 当, 总有; ||xM,|()|fxa,,,,,,0,,,M0
(3) , 当时, 总有; |()|fxb,,,,,,0,,,,0axa,,,,
(4) 当时, 总有( |()8|fx,,,,,,0,,,,033,,,,x
5(证明:. lim||0x,x,0
证明: 由于, ,所以. lim||lim0xx,,lim||lim()0xx,,,lim||0x,,,,,x,0xx,,00xx,,00
6(证明:若及时,函数的极限都存在且都等于A,则( fx()x,,,x,,,lim()fxA,x,,证明: 由于,则对,,当时,有(又,则,当,有.取,,M0|()|fxA,,,|()|fxA,,,xM,,,M0xM,,,,,0lim()fxA,lim()fxA,1122x,,,x,,,
那么对,当时,总有,故有. MMM,max,||xM,|()|fxA,,,,,,0lim()fxA,,,12x,,
习 题 1-4
1(根据定义证明:
2x,1(1)为当时的无穷小; x,1y,x,1
1(2)为当时的无穷小; x,,yx,sinx
13,x(3)为当时的无穷大( y,x,0x
证明: 2x,1(1) ,因为,取,则当时, 总有,故 0|1|,,,x,,,,0,,,x,0|0||1|,,,xx,12x,1( lim0,x,1x,1
1111(2) ,因为,取, 则当时, 总有 ||xM,,M,,,0|sin0||sin|xx,,,,xxx||||
1|sin|1x1, 故. ,,,,,x,limsin0|sin0|xx,,xxxx||||
11311,x(3) , ,当时,总有,所以 0||,,x,,,,,,M0|||3|3,,,,,M,M3xxx||13,x. lim,,x,0x
2(函数yxx,sin在(0,),,内是否有界,该函数是否为时的无穷大, x,,,
解答: 取,则,因此当n,,时, yx,,,,0故函数 xn,2πy,0xn,2π,,,,nnnnn
yxx,sin 当时,不是无穷大量( x,,,
π下证该函数在0,,,内是无界的. , 且xn,,,,,, ,,,xn,,M02π,,,,nn2
πππππ,,,,,取NM,,1, ,有,所以是无界的. ynnn,,,,,yNM,,,yxx,sin2π,,,,,,xN2π(0,)2πsin2π2π,,n000n0,,,,22222,,,,
11,3(证明:函数在区间(0,1]上无界,但这函数不是时的无穷大( x,0y,cosxx
1证明: 令,类似第2题可得( ,tx
习 题 1-5
4
第一章 函数与极限习题详解
1(求下列极限:
2,,11131nn,,(1); (2); lim,,,lim,,32n,,n,,nn1223(1),,,nn,,41,,nn12n32,,,(3); (4); ,,,limlim,,22211nn,,n,,n,,32,nnn,,23x,1x,1(5); (6); limlim22x,x,12xx,,54xx,,53221x,22(7); (8); lim1xxx,,,lim,,2x,,x,,,xx,,53332()xhx,,31x,(9); (10); limlim2h,x,01hxx,,41231xx,,,(11); (12); lim,lim,,33x,1x,,xx531xx,,,,11,,
311,,,xxx(13); (14); limlim33x,1x,,,21x11,,,xx32xxx,,,3273(15); (16)( lim(236)xx,,limx,x,,3x,3
解:
311,,22331nn,,nnn(1) = ( lim,lim032n,,n,,41nn,,41,,13nn
,,111111111,,(2) = lim,,,,,,,,,lim()()(),,,,n,,n,,nnnn1223(1),,,12231,,,,,
1 = ( ,,lim(1)1n,,n,1
1nn,(1)12n1,,2(3) =( lim,,,,lim,,2222n,,n,,n2nnn,,
2n,1()nn32,13(4) =( lim,lim11nn,,n,,n,,232,3n,,32()32x,12x,1(1)(1)xx,,(5) ==( limlimlim,,2x,1x,x,11(1)(4)xx,,xx,,54x,43
33x,121,(6) =( lim,,322x,2xx,,532523,,,
2222xxxxxx,,,,,,11,,,,22(7) = lim1xxx,,,lim,,x,,,x,,,22xxx,,,1
1,1x,11x==( lim,lim22x,,,x,,,211xxx,,,1,,,112xx
1,22221x,x(8) =( lim,lim22x,,x,,53xx,,53,,12xx3332233()xhx,,(33)xxhxhhx,,,,322(9) ==( lim(33)3xxhhx,,,limlimh,h,h,000hh2,,3(1),,,xx(1)(2),,xx31,,(10) == limlim,lim,,,,233x,x,x,111xx1,x(1)(1),,,xxx,,11,,,,
2,x=( lim1,2x,11,,xx
11,22xx,xx(11) =( lim,lim03x,,x,,31531xx,,,,523xx
11,,,xx(12) lim33x,111,,,xx
22333(11)(11)((1)(1)(1)(1)),,,,,,,,,,,,xxxxxxxx= limx,22133333(11)((1)(1)(1)(1))(11),,,,,,,,,,,,xxxxxxxx
223332((1)(1)(1)(1))xxxxx,,,,,,6==( 2limx,12(11)xxx,,,
32xx(13) =lim,,,( limx,,x,,1,21x2,x
3633(14) =x,,,,( lim(236)xx,,lim(2)23x,,x,,xx
5
第一章 函数与极限习题详解
321xxx,,,32732(15) =( xxx,,,,,,lim(327)limlimx,xx,,333x,x,33
x,ex,0,,(设 问当为何值时,极限存在( 2fx(),alim()fx,x,02,0.xax,,,x解:因为,所以,当,即时,存在( lim()lim1,lim()lim(2)fxefxxaa,,,,,lim()lim()fxfx,a,1lim()fx,,,,,,x,0,,,,0000xx,,00xxxx
12x,1x,13(求当时,函数的极限( x,1ex,1112x,1xx,,11解:因为 limlim(1)0,exe,,,,,xx,,11x,1112x,1xx,,11 limlim(1),exe,,,,,,,xx,,11x,112x,1x,1 所以不存在。 limex,1x,1
24(已知,其中为常数,求和的值( abc,,alim(5)1xaxbxc,,,,bx,,,
22(5)(5)xaxbxcxaxbxc,,,,,,2解:因为 lim(5) limxaxbxc,,,,xx,,,,,,25xaxbxc,,,
c250,,a,(25),,,axb2a,25,(25),,,axbxc,x,所以,则( = limlim1,,b,,xx,,,,,,2,1b,10bc,5,,,xaxbxc,5,,,a5,a,2xx
5(计算下列极限:
1sin1x(1); (2); x,,,,limsin0limlimsin0xx,0xx,,,,xxx
11arctan1x(3); (4)( ,,,limsin0limlimarctan0xx,,xx,,,,xxxx
1,5sin,0,,,xx,x,,10,0,x,fx(),6(试问函数在处的左、右极限是否存在,当时,的极限是否存在, fx()x,0x,0,,,25,0.,,xx,,
1,,2解:,,因为,所以( lim()lim(5)5fxx,,,fxx,,,ff(0)(0),lim()5fx,lim()lim(5sin)5,,,,x,0xx,,00xx,,00x
习 题 1-6
1( 计算下列极限:
,12xx2x,,,,(1); (2); lim1,lim1,,,,,,,xx,0x2,,,,
1xx,2x,5x,,,,(3); (4)( limlim,,,,,,xx,22x,5,,,,
1211,x,,,,(4)(),,222224x,xx222解:(1)((2)( ,,,,e,,,,elim(1)lim(1)lim(1)lim(1)xx,,000xx,,,xxxx,
1211,xx,2xx,,2222(3)( lim()lim(1),,,exx,,2222
x,5,10,,x51010,x510(4) lim()lim(1)(1),,,,,,xx,,,,xxx,,,555,,
x,5,10101051010( ,,,,,elim(1)lim(1)xx,,,,xx,,55
2(计算下列极限:
sin2x(1); (2) ; limlimcotxxx,0x,03x
coscos3xx,cos1x,(3); (4); limlim3,x,0x,05x2x
1xn(5)x,; (6)lim2sin(为不等于零的常数)( xlimsinnx,,,,n2x
解:
6
第一章 函数与极限习题详解
xxcossin22sincos2xxx(( ,,,,(2)limlim(1)limcotlim1xxxx,,00xx,,00333xxsinx2xx,,22sinsin,,,coscos32sin2sinxxxx,,cos1xx,,22 ( (3)limlim0,,(4)limlimlim0,,,,,33,xx,,00xx,,00x,0x255xx,,22xx2,,
1xsinsinn1xnx2(( ,,,,x(6)lim2sinlim(5)limsinlim1xxnxx,,,,,,,,nn1xx2nx2
3(利用极限存在准则证明:
(1)数列,,,的极限存在; 333,,333,
证明:先用数学归纳法证明数列单调递增。由于。假设成立,则,所xxx,,0xxxx,,,,,33xx,,,,,3330,,nnn,121nnnn,,11
以数列单调递增( x,,n
下证有界性
,假设,则 x,,,313x,,131n
,故,即数列有界 xxx,,,,,,,,,,33(13)312313013,,,x,,nnn,1n
113,113,113,根据单调有界准则知存在(不妨设,则有,解得,(舍去),即有( xAA,,3limxlimxA,A,lim,A,12nnn,,,,n,,nn222
3(2); ,,lim11n,,n
3333,, 证明:因为 ,又,所以( ,,lim1lim11,,,lim11111,,,,,,nn,,,,n,,nnnn,,
22,,121n(3) ; lim,,,,,,n,,66623nnnnnn2,,,,,
nn22kk,,2212n,,11kk证明:因为, ,,,,,......6266626,,,,,2nnnnnnnnnn
nn22kk,,1,,11kk 又,所以原式成立( ,,limlim626,,,,nn3,,nnnn
1,,(4) ( x,lim1,,,x,0x,,
111,, 证明:对任一,有,则当时,有(于是 xxx,,,1xR,x,0,,,1,,,,xxx,,
1111,,,,(1)当时,,由夹逼准则得( xx,0,xxx(1),,,lim1,,,,,x,0xxxx,,,,
1111,,,,(2)当时,,同样有( xx,0,xxx,,,(1)lim1,,,,,x,0xxxx,,,,
习 题 1-7
2231( 当时,与相比,哪一个是高阶无穷小, xx,2x,032xx,2332xx,232解:因为,所以是比高阶无穷小( xx,232xx,lim0,2x,0xx,2
2x2( 证明:当时,( sec1x,x,02
1,122xxsec1x,xx,,sec11cos1xcos证明:因为,又,则,故( (1cos),xsec1x,lim1,,,limlimlim2222x,0xxx,,,0002x2xxxxcos
2222
3( 利用等价无穷小的性质,求下列极限:
2tannx121,,,xx(1)为正整数); (2); lim(,nmlimx,0x,0sin3xsinmxsinx23ln(123),,xxe,1(3); (4); limlimx,,x00arctanx4x
1cos,x21cos,,x(5)lim; (6); lim2,x,0x,0sin3xxx(1cos),
232357xxx,,xxx,,sintan3(7); (8); limlim32x,x,0042tanxx,sin52xx,3xxx,,ab,lim(9),其中ab,,0,0,均为常数. ,,,x02,,
7
第一章 函数与极限习题详解
tan()nxnxn解:( ,,(1)limlimxx,,00sinmxmxm
12xx,(2)2,,,xx12112( ,,(2)limlimxx,,00xxsin33622ln(123)231,,,xxxx( (3)limlim,,xx,,00442xx
sinxsinx3,e113( ,,(4)limlim,,xx00arctan3xx
12x,,xx1cos1cos112( ,,,(5)limlimlimlim,,,,xxxx,,,,000012,,,,xxxxxx(1cos)(1cos)(1cos)1cosxx2
12x,,,,112xx21cos2(1cos)12 ( ,,,,,,(6)limlimlimlim223xxxx,,,,00001872xxsin3(3)22,,,,xxxsin3(21cos)21cos
23357333xxxxx,, ( (7)limlimlim,,,3xxx,,,00042tan2tan22xxxx,2xxxx,,sintan34422 (( (sintan34,sin525)因为xxxxxxx,,,(8)limlim,,2xx,,00sin5255xxx,xxab,,23xx33ln(1),ab,,(2)xxxx3ab,ab,,,3(1)(1)xx22xlimlimln(),limlimab,,,,00,0x0xxxxx2xx2 (9)lim(),,ee,,eex,0233ab,(lnln)22 ( ,,eab()
124,(当时,若与是等价无穷小,试求( (1)1,,axax,0xxsin
124(1)1,,ax解:依题意有, 因为 lim1,x,0xxsin11412224,,,,则 sinxx(1)11()1(),,,,,,,,axaxax,,4
112,,ax()24,,axa(1)14,故( a,,4,,,,limlim12xx,,00xxxsin4
习 题 1-8
1(研究下列函数的连续性:
xx,||1,,1,,xQ,,, (1) (2) fx(),fx(),,,c1,||1;x,0,.xQ,,,
解答:(1)在和内连续,为跳跃间断点; ,,,,1,,,1,x,,1,,,,
R (2)在上处处不连续。 fx()
2(讨论下列函数的间断点,并指出其类型(如果是可去间断点,则补充或改变函数的定义使其连续(
1(1); fx,()1x,e1
解答:在和内连续,为跳跃间断点( ,,,00,,,fx()x,0,,,,
1,xxsin,0,,,(2) fx(),x,
,0,0;x,,
R解:在上是连续的( fx()
2x,1(3); fx(),2xx,,32
,,,,解:fx()在(,1),(1,2)和(2,)内连续,x=1为可去向断点,若令f(1)2,,,则fx()在x=1连续;x=2为第二类向断点(
1(4); fx()sin,x
,,解:在(,0)和内连续,x=0为第二类向断点; fx()(0,),,
2xx,(5); fx(),2||(1)xx,
1,1fx()解:在(,1),,,,(,1,0),(0,1)和(1,),,内连续;x=是第二类间断点;x=0是跳跃间断点;x=1是可去间断点,若令,f(1),2
则fx()在x=1处连续(
8
第一章 函数与极限习题详解
xx,,1,3,,(6) fx(),,4,3.,,xx,
解:在和内连续,x,3为跳跃间断点( fx()(,3),,[3,),,
3(讨论下列函数的连续性,若有间断点,判别其类型(
1(1); fxx,,()lim(0)n,,n,x1
,1,01,,,x,1,解: 为跳跃间断点; x,1fxx()1,,,,2,,
,0,1.x,,
2n(1),xx(2)( fx()lim,2n,,n1,x
xx,||1,,,解: 为跳跃间断点( xx,,,11和fxx()0,||1,,,
,,,xx,||1,
sin2x,,0,x,,4(设函数试确定的值,使函数在处连续( fx()ax,0fx(),x,2,xax,,,0.,
sin2x2解:因为所以,依题意有=2. fa(0),,,,,,,alim()lim2,lim()lim(),fxfxxaa,,,,xxxx,,,,0000x
ln(13),x,,0,x,,5(设函数在点处连续,求和的值( ax,0bfx(),sinax,
,bxx,,1, 0,
ln(13)3,x解:因为,依题意有为任意实数( f(0)1,,ab,3,lim()lim(1)1,lim()limfxbxfx,,,,,,,,,xxxx,,,,0000sinaxa
6(试分别举出具有以下性质的函数的例子: fx()
11π(1) 是的所有间断点,且它们都是无穷间断点,例如: ; fx()xn,,,,,,0,1,2,,,,,fxx()cot(,,π)cotx2n
1,,xQ,,(2) 在R上处处不连续,但在R上处处连续;例如: fx()fx()fx(),,c,,1,;xQ,
xxQ,,,,(3) 在R上处处有定义,但仅在一点连续,例如: fx()fx(),,c,,xxQ,.,
习 题 1-9
1(研究下列函数的连续性:
2x(1); fxxx()cose,,
2x解答:因为在,,,,,上是初等函数,所以在,,,,,上连续( fx()fxxxe()cos,,,,,,
x,3(2); fx(),3x,27
x,31x,3解答:显然当时,fx()无意义,但,则是函数的可去间断点( x,3x,3lim,fx(),33x,3x,2727x,27
2(3); fxxx()12,,,,
2解答:当时,即x,,4,3时,连续( fx(),,,,xx120,,
,(求下列极限:
,,x1,(1); limsinπ,,,x,1,53x,,,
x,1,解: limsin()sin1;,,,x,1532x,
2(2); limarcsinxxx,,,,x,,,
22()()xxxxxx,,,,2解: limarcsin()xxx,,,limarcsinx,,,x,,,2xxx,,
9
第一章 函数与极限习题详解
x1π, ; ,,limarcsinarcsin2x,,,26,,xxx
1,,xln(2)2(3); limx,1πx,3arctan4
11,,,,xln(2)ln(21)122解: ,,lim;x,1πππx,,3arctan3arctan144
1,xx(4); lim14,x,,x,011,x1,x,,,,(4)x,44xxx解:; lim[1(4)],,x,elim14,,x,,x,0x,0
2x(5); lim[1ln1],,x,,,x012ln(1),x2,2ln(1),xxx解: ; ,lim[1ln(1)]lim[1ln(1)],,,,,xxexx,,00
1x2,x1cos(6); lim(1e),x,x021x1x,,ex2222xx,1cosx,1cosxxe解: lim(1)lim(1);,,,,xexee,,xx00
1tan1sin,,,xx(7); lim2x,0xxx1sin,,
1tan1sin,,,xx解: lim2x,0xxx1sin,,
(1tan1sin)(1tan1sin),,,,,,xxxx21sin1,,xlim ,22,x,0()(,)xxx1sin11sin1,,,1tan1sin,,,xx
,2(tansin)1sin1xxx,,, ,,limlim2xx,,00xx,sin1tan1sin,,,xx2x,xtan(1cos)xx,12 ; lim,lim,,22x,0x,02,xxxx,sin
2cotx(8); lim(cos)x,x01cos1x,1,,22cotxcos1x,tanx2解: ; lim(cos)x,,,,lim(1cos1)xe,x0x,0
(9)( lim[lnln(2)]nnn,,n,,
n2解: ,,,lim[lnln(2)]nnn,,limlnlimln(1)nnnn,,,,n,,,,nn22nn,,22,,,,,2,2n,2,,22n,, ; ,,elimln(1),,,,,limln(1)ln2,,nn,,n,n2,2
3(设函数与在点连续,证明函数 fx()gx()x0
,()max(),()xfxgx,, ,()min(),()xfxgx,,,,,
在点也连续( x0
证明:略(
2,abxx,,,0,,4(若函数在内连续,则和的关系是( )( (,),,,,abfx(),,sinbx, 0x,,x,
A(( B(( C(( D (不能确定( ab,ab,ab,
sinbx2解答:因为依题意有( ,,,,,fa(0),,ab,lim()lim(),lim()lim,fxabxafxb,,,,xxxx,,,,0000x
xxa,2,,5(设且,求常数的值( lim8,aa,0,,,,xxa,,,
xaax,3,xaaa,2333axxa33axa,解:因为,则,所以( e,8a,ln2lim()lim(1)lim(1),,,,,exxx,,,,,,xaxaxa,,,
习 题 1-10
10
第一章 函数与极限习题详解
1. 证明方程在内至少有一实根( (1,)exxln2,
证明:令,则在上连续,又,根据零点定理, fx()fxxx()ln2,,[1,e]f(1)20,,,fee()20,,,,fxxx()ln2,,
在开区间内至少有一点使,即在内至少有一实根( (1,)e(1,)e,f()0,,xxln2,
52(证明方程有正实根( xx,,1
5证明:令,则在内连续,又,, fx()(,),,,,f(0)10,,,f(1)10,,fxxx()1,,,
55根据零点定理,在内至少有一点,使,即有正实根( (0,1),f()0,,fxxx()1,,,xx,,1
3(设函数对于闭区间上的任意两点、,恒有,其中为正常数,fxfyLxy()(),,,Lfx()[,]abyx
且(证明:至少有一点,使得( fafb()()0,,,,(,)abf()0,,
证明:任取,取,使,依题意有,则,xab,(,)xxab,,,(,)0()(),,,,,,fxxfxLx,xlim()()0fxxfx,,,,,,x0即,由的任意性,可知在内连续,同理可证在点右连续,点左连续,fx()(,)abfx()xablim()()fxxfx,,,,,x0
那么,在上连续。而且,根据零点定理,至少有一点,使得( fx()[,]abfafb()()0,,,,(,)abf()0,,
fxfxfx()()(),,,12n4(若在上连续,,则在内至少有一点,使( fx()[,]abaxxxb,,,,,(,)xx,,f(),12n1nn
证明:因为在上连续,所以在上有最小值,最大值,使得Mfx()[,]abfx()[,]abm
fxfxfx()()...(),,,12n,,,,因此, mfxM,,()mfxM,,()...mfxM,,()mM,,12nn
fxfxfx()()(),,,12n由介值定理得,在内至少有一点,使. ,(,)xx,f(),1nn
n
,(若在上连续,,且(试证至少 fx()[,]abxabtin,,,[,],0(1,2,3,,)t,1,iii,0i
存在一点使得( ,,(,)abftfxtfxtfx()()()(),,,,,1122nn
证明:因为在上连续,所以在上有最小值,最大值,使得Mfx()[,]abfx()[,]abm
,,,,那么,,,即mfxM,,()mfxM,,()mfxM,,()tt()tmfxM,,tt()tmfxM,,...tt()tmfxM,,...12n11112222nnnnnnnnn
,又,故,由介值定理可知,至少存在一点使得 ,,(,)abmttfiMt,,()t,1mtfiM,,(),,,,,iiiii,,,111,1,1iiiii
( ftfxtfxtfx()()()(),,,,,1122nn
6(证明:若在内连续,且存在,则必在内有界( fx()(,),,,,fx()(,),,,,lim()fxx,,
证明:因为存在,则必有,使得当时,对任意的,有,因此,在xX,fxa(),,,fx(),X,0lim()fxx,,
区间及区间上有界,即当,存在,有fxM(),,同时,在(,),,,X(,)X,,xXX,,,,,,(,)(,)M,0fx()[,],XX11上连续,有由有界性定理知,存在,当,取,则当时,总有,xXXfxM,,,[,],()fxM(),M,0max{,}MMM,x,,,,,(,)2212即在内有界( fx()(,),,,,
复习题A
x11.设, , 求及其定义域. gx(),fgxgfx(),,,fx(),,,,,2,,1,x2,x21,x,,12解: , 其定义域为且, xx,,,420x,1fgx(),,,,22xx,,42x,,2,,,1,x,,
Dxxx,,,,,,|221且即;. ,,
121222,x,其定义域为 203-0,,,,xx且gfx,,,,,,2,,13,x1,22,x
Dxxx,,,,,|23且. ,,
22.求函数的反函数. fxxx()(1)sgn,,
11
第一章 函数与极限习题详解
2,,,,,(1),0xx(1),1xx,,,,,,,,1解: 因, 所以, fxx()0,0,,fxx()0,0,,,,2,,1,0,,xxxx,,1,1,,
3(单项选择题
(1)下列各式中正确的是( )
x11,,xA(; B(; lim1e,,x,,lim1e,,,,,,,0,xx0x,,
x,x11,,,,C(; D(( lim1e,,,lim1e,,,,,,,,xx,,,xx,,,,(2)当时,下列四个无穷小量中,哪一个是比其它三个更高阶的无穷小( )( x,0
2A(; B(; x1cos,x
2C(; D(( 11,,xxx,tan
12x,1x,1(3)极限为( ) limex,1x,1
A(; B(; C(,; D(不存在但不为,( 10
(4) 若当时,和都是无穷小,则当时,下列表达式中哪一个不一定是无穷小( )( xx,,()x,()xxx,00
22A(; B(和; ,,()()xx,,()x,()x
2,()xC(; D( ( ln1()(),,,,xx,,()x,
2,xxb,,2,1,x,,(5) 设适合,则以下结果正确的是( )( lim()fxA,fx(),x,1,x,1,ax,1,,
A(; B(可取任意实数; abA,,,,4,3,4aAb,,4,4,C(可取任意实数; D(都可取任意实数( bAa,,,3,4,abA,,解答:
(1) A; (2) D; (3) D: (4) D; (5) C(
4.求下列极限:
πsinn,1πn3(1) ; ,,lim3sinlim3π3πn,1nn,,,,π3n,13
1,1n,11188(2) ; ,,,,,,lim(1)limnnn,,,,1887,18
22()()xxxxxx,,,,2(3) ; limarccos()limarccosxxx,,,nn,,,,,,2xxx,,
x1π=,,; limarccosarccos2n,,,23,,xxx21cos22sin2,xx(4) ; limlim,,xx,,00xxxxsin333
1122xxsincoscos1xx(5) ,,,xlimlimlimcos0;xxx,,,000xxx
16121612,,,,,,xxxx,,,,1612,,,xx(6) limlim,22xx,,00xx,4xxxx,,,,41612,,,,
88x=,,; limlim12xx,,00,,,,,,,,xxxxxxx4161241612,,,,,,,,
sinsinxx222sin()sin(),1cos(sin)1x22(7) ,,,limlimlim;222xxx,,,000,2(1)24Inxxx
12
第一章 函数与极限习题详解
1sin11111,,x(8) xxxx,,,,,x,,,limsinsinlimsinlimsinlimlimsin101;,,xxx,,,,,,xx,,,,1xxxxx,,
xxaax,2xaa,2xa22axa,(9) lim()lim(1);,,,exx,,,,xaxa,,
11cos1x,1,,xxxcos12x,,,,,(10) ( limcoslim1(cos1)xx,,,,,,,lim1(cos1)xe,,,,,,,,,xxx,000
5(设当时,是比高阶的无穷小(证明:当时,与是等价无穷小( xx,fx()gx()xx,fxgx()(),gx()00
fxgxfx()()(),fx()证明: 由已知得则即是等价无穷小( ,fxgxgx()()(),与lim0,limlim(1)1,,,,xx,xxxx,,000gx()gxgx()()
3xaxb,,6(已知,求常数与的值( lim8a,bx,22x,
3xaxb,,3解:因为,所以,则lim8lim()820xaxbab,,,,,,,x,x,222x,
32xaxbxxxa,,,,,,(2)[2(4)]2,( 解得ab,,,4,0,,,,,,,lim[2(4)]128xxaaba,,,,82,limlim那么xx,,x,222xx,,2(2)
7.设并求此极限( xxxnx,,,,10,6(1).{}证明数列极限存在,11nnn,
证明:用数学归纳法证明此数列的单调性(因为 及,可知假设则 ,所xx,,xx,,xxxx,,,,,66x,10xx,,,6412112nn,1nnnn,,,112
以单调递减,又显然,即有下界,由单调有界准则知存在极限,设,两边取极限,有A=,x,0对xx,,6{}x{}x{}xlimxA,6,Annnnnn,1n,,n解之得A=3或A=,2(舍去) ,即得=3( limxn,,n
sin6x,x,,0,,2x,8.确定常数a与b的值,使得函数=处处连续( fx()axx,,3,0,,
,1x,,,bxx(1),0.,
1sin6xsin6xx解:当时,=和当时,=,显然它们都是连续的,又f(x)==3, f(x)fx()fx()(1),bxx,0x,0limlimlim,,,x,0x,0x,02x2x
1bbbxbx===,当时,=a,要使f(x)在x=0点也连续,则=3=a,即a=3,b=ln3( fx()(1),bx(1),bxex,0elimlim,,x,0x,0
9.求下列函数的间断点,并判断其类型(
1(1)=; fx()arctanx
1,1,解:因为==,==,又时,连续,所以只有x=0为间断点,x=0为跳跃间断点( fx()fx()fx()arctan,arctanlimlimlimlimx,0,,,,x,0x,0x,0x,022xx
x(2)=; fx()tanx
xx,解:当tanx=0时,有x=0或x=n(n=,1,,2,…)因为=1,所以x=0为可去间断点(又=(n=,1,,2,…),lim()lim,,limfxxx,,00xn,,tanxtanx
,,x,,xn,,xn,,所以(n=1,2,…)为无穷间断点(当(n=1,2,…)时,=0,所以(n=1,2,…)是可去间断点( ,,lim,,xn,,,,,22tanxxn,,,2
nxxe,(3); ()limfx,nx,,x1,e
1,0,x,,
nx,xe,1,解:=,因f(x)=0, fx()=1,所以x=0为跳跃间断点( ()limlimlimfx,,0x,,nx,,,,xx,0x,01,e2,xx,0,,,,
复习题B
1(单项选择题
13
第一章 函数与极限习题详解
(1)当时,下列无穷小量中与不等价的是( )( xx,0
2ln(1),x2x2423A(( B(( C(( D(( sin(6sin)xx,e251,,,xxxxx,,3x
(2)下列极限不存在的是( )(
12xx,,311ln(1)1,xsinxxA(( B(( C(( D(( xlim,limsinlim(arctan),lim(2),x,0x,,,,,xx,0xxxxx
(3)极限( )等于( e
1111x,1xxxA(( B(( C(( D(( ,,,lim(1),xlim(1)lim(1)lim(1),,,,0,,xxxx,,,xxx
(4)设,数列,如果,则的值为( )( ,n|()|()fngn,lim()3gn,lim()fnn,,n,,
A(( B(( C(( D(( lim()3fn,,,,,3lim()3fnlim()3fn,,,,3lim()3fnn,,n,,n,,n,,
22x(5)已知,其中与为常数(则( )( a,,,blim()1axbx,,,x1
A(,( B(,( C(,( D(,( a,2a,,2b,3a,2b,,3a,,2b,3b,,3
3,xx(6)设函数,则( )( fx(),sin,x
A(有无穷多个第一类间断点( B(只有个可去间断点( 1
C(有个跳跃间断点( D(有个可去间断点( 23
解答:
(1)D;(2)C;(3)B;(4)B;(5)C;(6)D(提示:x=0,1为可去间断点) ,,
2(填空题
x,1(1)设函数的定义域是,则的定义域是_________( fx()[0,1]f()x,1
211,,xxlimln(2)计算=_________( 20x,xxx1,,
1222axbx,(3)设,则=_________, ( ab,lim(122)e,,,xxx,0
2,xxxcos,(4)设时,与是同阶无穷小,则 ( x,0ee,x,,
fx()fx()fx()(5)设,则 , ( ,,,,lim3limlim32x,x,0x,00xxx
有界是数列收敛的 条件;函数的极(6) 在“充分”、“必要”和“充分必要”三者中选择一个正确的填入空格内:数列fx()xx,,,,nn
限存在是在的某一去心邻域内有界的 条件;函数在的某一去心邻域内无界是的 条件;fx()fx()lim()fxxxlim()fx,,00xx,xx,00
函数在左连续且右连续是在连续的 条件( fx()fx()xx00
9 答案:(1);(2)1;(3)a=1;b为任意实数;(4);(5)0,0;(6)必要,充分,必要,充要( 1,,,,,2
22211,,xxln(1)ln(1),,,,,xxxx(2)题解答过程:= lnlimlim2x,0x,0xxx1,,2x
2222,,,,ln(1+(),,xxln1),,(xx11xx,,,xx,,,,====1( lim,,()limlim,,x,x,0x,0022x222xx2x
(3)题解答过程:
21122xx,2,222222axbx,a22xxaxbx,,因为=,所以,a=1,b为任意实数( (122),,xxee,lim(122),,,xxlimx,0x,0
(4)题解答过程:
1x42x2exx,,()xxx(cos1),x2xxecos,9eee[1],exx(cos1),2因为====c(常数),所以u=( limlimlimlimuuuu,,,,,,x,0x0x,0x02xxxx
fx()fx()fx()22(5)题解答过程:因为=,所以=,,3(),x (其中,()x为当x时的无穷小量),那么=,,,3()xxx,,3,0lim33x,0xxx
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第一章 函数与极限习题详解 fx()fx()fx()=,故=0,=0( ,,3()xxx,limlim22x,0x,0xxx
3(求下列极限:
3xe1,(1); (2); lim[1212(1)],,,,,,,,nnlim,n,,,0xxx1cos(1cos),,
1xxx1x,,abc,,x2,x1cos(3)(;(4); (0,0,0)abc,,,limlim(1arctan),ex,,,x0,x03,,
tanx(5); (6); lim(sin)xlim(sin1sin)xx,,πx,,,,x2
1,,1x2esinx,111,,n(7); (8); ,,,,limlim(1),4,,n,,,x0n23||x,,x1e,,,
sinx3nln(e1cos)sin,,,xx22(9); (10)( limlim(1)(1)(1),||1,,,,xxxx3,x0x,,arctan(41cos),x
解答:
nnnn(1)(1),,(1) lim[12121],,,,,,,,,nnlim[],,,n,,n,,22
nnnnnnnn(1)(1)(1)(1),,,,[][],,n2222 lim,,limn,,n,,,,(1)(1)nnnnnnnn(1)(1),,,[][],2222
21 ,,limn,,211,,11nn,22
333xxxe,1(2)( lim,lim4,lim,3,,,x,x,,000(1cos),xxxx1cos(1cos),,xx
24xxx311(1)(1)(1)abc,,,,,xxxxxxxxxabc,,abcabc,,,,,,33xxxxxabcx,,,33(3) lim(),lim(1)lim(1),,,xx,,00,x0333
lnabc33==( abce
x21arctanex1x22x2x2,1cosx,x1cosexarctan(4)=( lim(1arctan),,exelim(1arctan),ex,x,0x0
1sin(sin1)xx,,tanxsin1cosxx,(5) lim(sin)x,lim(1sin1),,xπ,x,,x22
xx2,,sin(cossin)x122,sin1x,xx22cossin,022=( lim(1sin1),,,xe,1,x,2
xxxx,,,,11(6), lim(sin1sin)xx,,,lim2sincosx,,,x,,,22
xx,,1xx,,1xx,,1,其中|2|2,即2是有界量,,故( lim(sin1sin)0xx,,,coscoslimsin0,x,,,x,,,222
1111111111nnnn(7)因为,又,所以( ,,,,,,n,,,,,lim1n,1(1....)lim(1....)1,,n,,nnn2323
11xx2sin2sin,,exex(8)因为, lim()lim()211,,,,,,44,,,,xx00||xxxx11,,ee
111xxx2sin2sin,,exex2sin,ex,所以,( lim()lim()011,,,,,,lim()1,,444,,,,,xxx000||xx||xxxx11,,ee1,e
n22n(1)(1)(1)...(1),,,,xxxx22(9) lim(1)(1)...(1)lim,,,,xxxnn,,,,1,x
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第一章 函数与极限习题详解
n,1211,x. ,,,lim(||1)xn,,11,,xx
sinsinsinxxx33ln(1cos)sinln(1cos)lnexxexe,,,,,,(10) limlim,33,,xx00arctan(41cos)41cos,,xx
331cos1cos,,xxln(1),xxsinsin111ee . ,,,,limlimlimxsin33,,,xxx00044e41cos41cos,,xx
nx4(已知函数,试确定的间断点及其类型( ()lim,fx()fx2n,,n2,x
0,||1x,,
,0,0||1,,xn,x,解:因为, fx()lim,,1,2n,,,n,1x,2,x,3,不存在,x=-1,,
所以, lim()lim()0(1)fxfxf,,,lim()lim()0,(1)fxfxf,,,不存在,,,,,xx,,11xx,,,,11
因此,均为可去间断点。 x,,1
2,axbxx,,,1,,5(设函数求,使在处连续( afx()bfxx()3,1,,, x,1,
,2,1.abxx,, ,
2解: 因为, lim()lim()fxaxbxab,,,,lim()lim(2)2fxabxab,,,,,,,,xx,,xx,,1111
abab,,,2,,要使在处连续,则,解得,. f(1)3,fx()x,1a,2b,1,ab,,3,
ππ6(求证方程在区间上至少有一个根( xx,,,1sin0(,),22
ππππ,,证明:令,显然在上连续,又, fx()fxxx()1sin,,,,,f()0,,,,,,2222,,
ππππ,,,由零点定理可知,在内至少有一个零点,即方程 fx(),,,f()20,,,,,2222,,
ππ,,在内至少有一个根. xx,,,1sin0,,,,22,,
1a7(设,任取,令(其中)(证明数列收敛(并求极限( a,0,,x,0{}xn,1,2,xx()limxnn,11nn,,n2xn
2x11an,,,,,证明:首先证明x是单调的,(1)若,则,即x单调递增有上界。(2)若,则xa,xxxx()()xa,,,,,,nnnnnnnn122xxnn
2x11an,,,,,,即 xxxx()(),nnnn122xxnn
1ax单调递减有上界,综(1)(2)知数列x有极限存在;令,则,解之得或(舍去),即. limxa,AA,,()Aa,Aa,,limxA,,,,,nnnn,,,,nn2A
8(成本—效益模型 从某工厂的污水池清除污染物的百分比与费用是由下列模型给出: pc
100c( pc(),8000,c如果费用允许无限增长,试求出可被清除污染物的百分比(实际上,可以完全清除污染吗, c
100c解:所以如果费用C允许无限增长,可被清除污染物的百分比为100%,实际上是不可能完全清除污染的. ,,lim()lim100pccc,,,,,8000c
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