广东版高数 第01章
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关于工期滞后的函关于工程严重滞后的函关于工程进度滞后的回复函关于征求同志党风廉政意见的函关于征求廉洁自律情况的复函
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:
1yx,,,2(1); 21,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,Dx(),,解:依题意有,则函数定义域( 3,2,xx,,,60,2(3); yxx,,,,ln(32)
2,,,,xx320解:依题意有,则函数定义域( Dxxx()|12,,,,,
13xx,(4); y,2
3xx,,0解:依题意有,则函数定义域( Dxxx()|x0,1,,,,,,,,,且,,
1,sin1,, x,,(5) y, x,1,
,21;, x,,
解:依题意有定义域( Dxxx()|,,,,,,,,,
1yx,,,arctan3(6). x
x,0,解:依题意有,则函数定义域( Dxxx()|3x0,,,且,,,30,,x,
2[0,1]fx()2(已知定义域为,求 fxfxfxafxafxa(), (sin), (), ()(),,,,
(a,0)的定义域(
22[0,1]01,,x[1,1],fx()解:因为定义域为,所以当时,得函数的定义域为; fx()
[2kkπ,(21),π]fx(sin)当0sin1,,x时,得函数定义域为;
fxa(),[,1],,,aa当01,,,xa时,得函数定义域为; 01,,,xa,1a,fxafxa()(),,,当时,得函数定义域为:(1)若,;xaa,,,1,,,201,,,xa,
111a,x,a,x,,(2)若,;(3)若,( 222
,,1ax,fx()1,,,faf(2),(1)a,0,3(设其中求函数值( ,,222xaaxx,,2,,
,,1ax,fx()1,,解:因为,则 ,,222xaaxx,,2,,
,,0 ,>1,a,11a,11,a,,f(1)1,,,,( fa(2)1,,,,,,,,222,,2 ,0<<1a11a,42aaa,,,,,
1||1,x,,,xfgx(())gfx(())4(设,求与,并做出函数图形( fxxgx()0||1,()2,,,,
,,,1||1.x,
x,121,10x,,,,xfgx(())021,,解:,即, fgxx(())00,,,,,x,,,1 0x,,1 21,,
,1,2||1x,,2||1x,,,0gfxx(())1||1,,gfxx(())2||1,,,即,函数图形略( ,,,,,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[()],,,1,1x,,1,()0fx,,,
fx()gx()6(下列各组函数中,与是否是同一函数,为什么,
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(
7(确定下列函数在给定区间内的单调性:
yxx,,3lnx,,,(0,)(1),;
yx,3yx,lnyyy,,x,,,(0,)解:当时,函数单调递增,也是单调递增,则在1212
(0,),,内也是递增的(
,xx,,,(,1)y, (2),( 1,x
,,,xx(1)11yx,,1x,,,(,1)解:y,,,,1,当时,函数单调递增,则1111,,,xxx
,x11y,是单调递减的,故原函数是单调递减的( y,,21,xyx,11
8. 判定下列函数的奇偶性(
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)
y,0 (2);
fxfx()0(),,,y,0解:因为,所以是偶函数(
2(3); yxxx,,,,2cossin1
2fxfxfxfx()()()(),,,,,且解:因为,,所以fxxxx()2cossin1,,,,,
2既非奇函数,又非偶函数( yxxx,,,,2cossin1
xx,aa,(4). y,2,xxxx,aa,aa,解:因为,所以函数是偶函数( ()()y,fxfx,,22
fx()[,],ll9(设是定义在上的任意函数,证明:
fxfx()(),,fxfx()(),,(1)是偶函数,是奇函数;
fx()(2)可表示成偶函数与奇函数之和的形式.
gxfxfxhxfxfx()()(),()()(),,,,,,证明:(1)令,则 gxfxfxgxhxfxfxhx()()()(),()()()(),,,,,,,,,,,fxfx()(),,,所以是偶函数,
fxfx()(),,是奇函数(
fxfxfxfx()()()(),,,,fxfx()(),,(2)任意函数,由(1)可知是偶函fx(),,222
fxfx()(),,数,是奇函数,所以命题得证( 2
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
fx()证明:(必要性)若函数在区间上有界,则存在正数,使得,都有xI,IM
,,,MfxM()fx()成立,显然,即证得函数在区间上既有上界又有下界 IfxM(),
MMfx()(充分性)设函数在区间上既有上界,又有下界,即有I21
fxMfxM()(),,且fx(),取,则有,即函数在区间上有IMMM,max{,}fxM(),1212
界(
11(下列函数是否是周期函数,对于周期函数指出其周期:
yx,|sin|(1);
周期函数,周期为( π
yx,,1sinπ(2);
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,y,1
x,1fxx()log,(,0)(1,),,,,,,,( 3x,1
axb,yadbc,,()(2); cxd,
bdx,bdy,,1fxadbc()()解:依题意,,,,则反函数( x,cxa,cya,
2yxx,,,lg1(3); ,,
11yy,,,1xxx,,fxxR,,,(1010)()(1010),解:依题意,,所以反函数( 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
xx值和的函数值: 12
u2(1); yuxxx,,,,e,1,0,2,122v(2)( yuuvxxx,,,,,,,,,1,e1,1,1,1122x,15解:(1) yfxfefe,,,,()e,(0),(2)
x,1242f(1)1,,(2),,( yfx,,,,()(e1)1fee(0)22,,,
r14(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H液面的高度为时,溶液的体积为(试把表示为的函数,并指出其定义区间( hVhV
V22Vrh,πhVrH,,,[0,π]解:依题意有,则( 2πr
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,an,,(1,2,3,)?1(设, n31n,
222(1) 求||,||,||aaa,,,的值; 110100333
2,4a,,||10(2) 求N,使当nN,时,不等式成立; n3
2||a,,,(3) 求N,使当nN,时,不等式成立( n3
232122121||||,a,,,,||||,a,,,,解:(1) 11034312331393
220121 ||||a,,,,( 10033013903
2119997,4a,,n,,||10,,(2) 要使 即 , 则只要 取N,n4933310(n+1)
29997,,,4a,,||10 故当n>1110时,不等式成立( ,1110,n,,39,,
213,,213,,,,||a,,,n,,||a,,,(3)要使成立, 取,那么当nN,时, N,nn,,3939,,,,
成立.
2(根据数列极限的定义证明:
21n,3,lim0(1); (2)( lim1,n,,n,,n!n
1111,,,,,,,|0|解:(1),,,0, 要使, 只要取, 所以,对任意,,0,N,,,nnn!!,,,
111,,,,,,存在,当时,总有|0|,则lim0. nN,N,,,n,,nn!!,,,
23n,332 (2) ,要使, 即,只要取,,,0,,n|1|,,,,22,2nn2nnn(3),,
2,,,,33n,3N,N,,所以,对任意的>0,存在, 当, 总有, 则nN,,|1|,,,,,,,2,2,n,,,,
2n,3. lim1,n,,n
3(若limxa,,证明lim||||xa,(并举例说明:如果数列有极限,但数列||xx,,,,nnnn,,,,nn
未必有极限(
,NnN,||xa,,,证明: 因为limxa,, 所以, , 当时, 有.不妨假设a>0, ,,,011nn,,n
nN,x,0,N由收敛数列的保号性可知:, 当时, 有, 取, 则对NNN,max,,,22n12
||||||||xaxa,,,,,lim||||xa,,,,0, ,N, 当nN,时, 有.故. 同理可证a,0nnn,,n
时, lim||||xa,成立. n,,nn||1x,反之,如果数列有极限, 但数列未必有极限.如:数列, , x,,1||x||x,,,,,,nnnn
limx显然lim||1x,, 但不存在( nn,,,,nn
lim0y,lim0xy,4(设数列有界,又(证明:( x,,nnnn,,,,nn
||xM,lim0y,证明: 依题意,存在M>0, 对一切n都有, 又, 对,,,0, 存在N, nn,,n
|0|y,,,|0|||||xyxyMyM,,,,,当时, , 因为对上述, 当时, ,由nN,NnN,,nnnnnn
lim0xy,的任意性, 则( nn,,n
1(3)n,πlimx5(设数列的一般项,求( xx,cos,,nnn,,n2n
(3)n,π11(3)n,π|cos|1,解: 因为, , 所以 . ,lim0limcos0,x,,x,,22nn
xAk,,,()xAk,,,()xAn,,,()6(对于数列,若,,证明:( x,,21k,2knn
,,N0||xA,,,kN>limxA,证明: 由于, 所以, ,,,0, , 当时,有, 同理, 1121k,21k,k,,
,,N0kN,||xA,,,,,,0,, 当时, 有(取N=max, ,,,0, 当nN,时, NN,,,222k12
||xA,,,xAn,,,()成立, 故( nn
习 题 1-3
2|1|x,,,|4|0.01y,,1(当x,1时,(问等于多少,使当时,, ,yx,,,34
135|1|x,,,,,|1|x解:令 ,则,要使 222
522|4||34||1||1||1||1|0.01yxxxxx,,,,,,,,,,,,, 2
|1|x,,,|4|0.01y,,|1|0.004x,,,,0.004只要,所以取,使当 时,成立(
221x,y,,2||xX,|2|0.001y,,2(当时,(问等于多少,使当时,, Xx,,2x,32217x,22x,,37000解:要使<0.001, 只要, 即. 因|3|7000x,,|2||2|y,,,,22xx,,3|3|
X,7003此,只要就可以了,所以取( ||7003x,
3(根据函数极限的定义证明:
35x,lim3,(1)lim(21)5x,,; (2); x,,x,3x,12sinxx,4(3); (4). ,lim0lim4,,x,,,x,,2x,2x
|(21)5|2|3|xx,,,,|(21)5|x,,,,证明:(1) 由于, 任给,要使,只要,,0,,0|3|,,,x,|(21)5|x,,,,|3|x,,,(因此取,则当时, 总有,故lim(21)5x,,( ,x,322
35x,358x,8, (2) 由于,任给, 要使,只要,即,,0|3|,,,,|3|,,x,1,xx,,1|1||1|x88888,,,,,,,,,||xM,x1x1|1||1|M|1|或, 因为,所以, 取,则当时, 对,,0,,,,,
35x,35x,,lim3,,,,0,总有|3|,,,故有( x,,x,1x,122x,4x,4,|(4)|,,,|2|x,,, (3)由于,任给,,要使,只要,因,,0|(4)||2|,,,,xx,2x,222x,4x,4,|(4)|,,,0|(2)|,,,,x,此取,,,,则当时,总有,故. lim4,,x,,2x,2x,2
1sin|sin|1xxsinx1x, (4) 由于,任给,要使,只要,即,,,0,,,,,|0|,,,|0|2,xxxxx
1sinxsinxM,因此取,则当x>M时,总有,故. ,,,,|0|lim02x,,,,xx
4(用,,X或,,,语言,写出下列各函数极限的定义: lim()1fx,lim()fxa,(1); (2); x,,,x,,
lim()fxb,lim()8fx,,(3); (4)( ,,xa,x,3
|()1|fx,,,,,,0,解: (1) ,,M0, 当x<-M时, 总有;
,,,0,||xM,|()|fxa,,,(2) ,,M0, 当, 总有;
,,,0,|()|fxb,,,(3) ,,,0, 当axa,,,,时, 总有;
|()8|fx,,,,,,0,(4) ,,,0 当33,,,,x时, 总有(
lim||0x,5(证明:. x,0
lim||0x,lim||lim0xx,,lim||lim()0xx,,,证明: 由于, ,所以. ,,,,x,0xx,,00xx,,00
fx()m()ilfxA,6(证明:若及时,函数的极限都存在且都等于,则( Ax,,,x,,,x,,
,,M0xM,|()|fxA,,,lim()fxA,证明: 由于,则对,,,0,,当时,有(又11x,,,
,,M0xM,,|()|fxA,,,lim()fxA,,则,当,有.取那么对,,,0,当MMM,max,,,2212x,,,
|()|fxA,,,||xM,lim()fxA,时,总有,故有. x,,
习 题 1-4
1(根据定义证明:
2x,1(1)y,为当x,1时的无穷小; x,1
1yx,sin(2)为当时的无穷小; x,,x
13,xy,(3)为当时的无穷大( x,0x
证明: 2x,10|1|,,,x,(1) ,因为,取,则当时, 总有,故 ,,,0,,,x,0|0||1|,,,xx,12x,1( lim0,x,1x,1
1111M,||xM,(2) ,因为,取, 则当时, 总有 ,,,0|sin0||sin|xx,,,,xxx||||
11|sin|1xx,, 故limsin0. ,,,,,|sin0|xx,,xxxx||||
11311,x,,0||,,x,,(3) , ,当时,总有,所以 ,,M0|||3|3,,,,,M,M3xxx||13,xlim,,. x,0x
(0,),,yxx,sin2(函数在内是否有界,该函数是否为时的无穷大, x,,,
xn,2πy,0xn,2π解答: 取,则,因此当时, 故函数 n,,yx,,,,0,,,,nnnnn
yxx,sin 当时,不是无穷大量( x,,,
π,,,xn下证该函数在内是无界的. ,2π 且, ,,M00,,,xn,,,,,,,,,nn2
ππππ,,,,,取, ,,,,,,xN2π(0,),有ynnn,,,,,NM,,12πsin2π2π,,00n0,,,,2222,,,,
πyNM,,,2πyxx,sin,所以是无界的. n02
11,x,0y,cos(0,1]3(证明:函数在区间上无界,但这函数不是时的无穷大( xx
1,t证明: 令,类似第2题可得( x
习 题 1-5
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,,nnn32,,,23x,1x,1(5); (6); limlim22x,x,21xx,,53xx,,54221x,22(7)lim1xxx,,,; (8); lim,,2x,,,x,,xx,,5323331x,()xhx,,lim(9); (10); lim21x,h,0xx,,41h
231xx,,,(11); (12); ,limlim,,33x,1x,,xx11,,531xx,,,,
311,,,xxx(13); (14); limlim33x,1x,,,21x11,,,xx
32xxx,,,3273(15)lim(236)xx,,; (16)( limx,,x,3x,3
解:
311,,22331nn,,nnnlim0,(1) = ( lim32n,,n,,41nn,,411,,3nn
,,111111111,,(2) = lim,,,?,,,,,,?lim()()(),,,,n,,n,,nnnn1223(1),,,12231,,,,,
1lim(1)1,, = ( n,,n,1
1nn,(1)12n1,,2(3) =( ,lim,,,?lim,,2222n,,n,,n2nnn,,
2n1(),nn132,3lim,(4) =( lim11nn,,n,,2n,,3n32,,,32()32x,12(1)(1)xx,,x,1(5) ==( lim,,limlim2x,1x,1x,1x,43(1)(4)xx,,xx,,54
33x,121,(6) =( lim,,322x,2xx,,532523,,,
2222xxxxxx,,,,,,11,,,,22(7) = lim1xxx,,,lim,,x,,,22x,,,xxx,,,1
1,1x,11x==( ,limlim22x,,,x,,,211xxx,,,1,,,112xx
12,2221x,xlim2,(8) =( lim2x,,x,,53xx,,53,,12xx3332233()xhx,,(33)xxhxhhx,,,,322lim(33)3xxhhx,,,(9) ==( limlimh,0h,h,00hh2,,3(1),,,xx31(1)(2),,xx,,(10) =lim= lim,lim,,,,332x,1x,1x,1xx1,x,,11(1)(1),,,xxx,,,,
2,xlim1,=( 2x,11,,xx
11,22xx,xxlim0,(11) =( lim3x,,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,,,,,,62lim==( x,12(11)xxx,,,
23xx(13) =( lim,,,limx,,x,,1,21x2,x
3633x,,,,lim(236)xx,,(14) =lim(2)( 23x,,x,,xx321xxx,,,32732xxx,,,,,,(15) =lim(327)lim( limxx,,33x,3x,3x,3
x,ex,0,,lim()fx2(设 问当为何值时,极限存在( fx(),a,x,02,0.xax,,,
xlim()lim1,lim()lim(2)fxefxxaa,,,,,解:因为,所以,当lim()lim()fxfx,,,,,,,,,,,,0000xx,,00xxxx
lim()fx即a,1时,存在( x,0
12x,1x,13(求当时,函数的极限( ,1xex,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
2abc,,4(已知,其中为常数,求和b的值( lim(5)1xaxbxc,,,,ax,,,
22(5)(5)xaxbxcxaxbxc,,,,,,2 lim(5) limxaxbxc,,,,解:因为 2xx,,,,,,5xaxbxc,,,
c250,,a,(25),,,axb2a,25,(25),,,axbxc,x,所以,则( = limlim1,,b,,2xx,,,,,,,1b,10bc,5,,,xaxbxc,5,,,a5,a,2xx
5(计算下列极限:
1sin1xx,,,,(1)limsin0; (2)limlimsin0x; x,0xx,,,,xxx11arctan1xlimsin0,,,limlimarctan0x(3); (4)( x,,xx,,,,xxxx
1,5sin,0,,,xx,x,,10,0,x,6(试问函数fx(),在处的左、右极限是否存在,当时,x,0x,0,,,25,0.,,xx,,
fx()的极限是否存在,
12,,fxx,,,lim()lim(5)5fxx,,,解:,lim()lim(5sin)5,因为,所以ff(0)(0),,,,,xx,,00xx,,00x
lim()5fx,( x,0
习 题 1-6
1( 计算下列极限:
,12xx2x,,,,(1); (2)lim1,; lim1,,,,,,,xx,0x2,,,,
1xx,2x,5x,,,,(3); (4)lim( lim,,,,,,xx,2x,52,,,,
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,,x,51010x510lim()lim(1)(1),,,,(4) ,,xx,,,,xxx,,,555,,
x,5,10101051010( ,,,,,elim(1)lim(1)xx,,,,xx,,55
2(计算下列极限:
sin2xlimcotxxlim(1); (2) ; x,0x,03x
cos1x,coscos3xx,limlim(3); (4); 3,x,0x,05x2x
1xnlimsinx,lim2sin(x(5); (6)为不等于零的常数)( nx,,,,nx2解:
xxcossin22sincos2xxx,,,,(1)limcotlim1xx((2)limlim( xx,,00xx,,00sinx333xx
coscos32sin2sinxxxx,,(3)limlim0,, xx,,0055xx
2xx,,22sinsin,,,cos1xx,,22( (4)limlimlim0,,,,,33,xx,,00x,x02,,22xx2,,
1xsinsinn1xnx2x,,,,(5)limsinlim1(6)lim2sinlimxx(( nnnxx,,,,,,,,1x2xn2x
3(利用极限存在准则证明:
3333,,?(1)数列,,,的极限存在; 33,
证明:先用数学归纳法证明数列单调递增。由于xx,,,,,3330。假设x,,21n
xx,,0成立,则,所以数列单调递增( xxxx,,,,,33x,,nn,1nnnn,,11n下证有界性
,假设,则 x,,,313x,,131n
xx,,,,,,,,,,33(13)312313,故,即数列有界 013,,,xx,,nn,1nn
113,AA,,3limxlimxA,根据单调有界准则知存在(不妨设,则有,解得,A,nn1,,,,nn2
113,113,(舍去),即有( xA,lim,2n,,n22
3(2); ,,lim11n,,n
3333,, 证明:因为 ,又,所以( ,,,,,lim11lim1lim11111,,,,,,nn,,,,n,,nnnn,,
22,,121nlim,,,,?(3) ; ,,6662n,,3nnnnnn,,,2,,
nn22kk,,2212n,,kk11,,,,,......证明:因为, 6266626,,,,,2nnnnnnnnnn
nn22kk,,1,,kk11,,limlim 又,所以原式成立( 626,,,,nn3,,nnnn
1,,(4) ( xlim1,,,,x,0x,,
111,, 证明:对任一xR,,有,则当x,0时,有(于是 xxx,,,1,,,1,,,,xxx,,
1111,,,,x,0(1)当时,,由夹逼准则得x( lim1,xxx(1),,, ,,,,,x,0xxxx,,,,
1111,,,,x,0(2)当时,,同样有x( lim1,xxx ,,,(1),,,,,x,0xxxx,,,,
习 题 1-7
223xx,232xx,1( 当时,与相比,哪一个是高阶无穷小, x,02332xx,232xx,232xx,解:因为,所以是比高阶无穷小( lim0,2x,0xx,2
2x2( 证明:当时,( x,0sec1x, 2
1,12xx,,sec11cos1xxcoslimlimlim,, ,又,则证明:因为(1cos),x 222xxx,,,000xxxxcos2
2222sec1x,x,故( lim1,sec1x, 2x,0x2
2
3( 利用等价无穷小的性质,求下列极限:
2tannx121,,,xxlim(,nm(1)为正整数); (2); limx,0x,0sinmxsin3x
sinx23ln(123),,xxe,1(3); (4); limlim,x,x00arctanx4x
1cos,x21cos,,x(5); (6); limlim2,x,0x,0sin3xxx(1cos),
232357xxx,,xxx,,sintan3(7); (8); limlim32x,x,0042tanxx,sin52xx,3xxx,,ab,ab,,0,0(9),其中,均为常数. lim,,,x02,,
tan()nxnxn,,解:(1)limlim( xx,,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)1cos xx2
12x,,,,xx21cos2(1cos)12 ,,,(6)limlimlimlim223xxxx,,,,0000xxsin3(3),,,,xxxsin3(21cos)21cos
112( ,,,18722223357333xxxxx,, ( (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,,,0x0x,,00xx2xxxx2 (9)lim(),,ee,,eex,0233ab,(lnln)22 ( ,,eab()
124,(当时,若与是等价无穷小,试求( x,0xxsin(1)1,,axa
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:
1,,xQ,xx,||1,,,, (1) (2) fx(),fx(),,,c1,||1;x,0,.xQ,,,解答:(1)在和内连续,x,,1为跳跃间断点; ,,,,1,,,1,,,,,
fx() (2)在上处处不连续。 R
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(
1fx(),(1); 1x1,e
fx()解答:在和内连续,x,0为跳跃间断点( ,,,00,,,,,,,
1,xxsin,0,,,(2) fx(),x,
,0,0;x,,
fx()解:在上是连续的( R
2x,1(3); fx(),2xx,,32
,,fx()f(1)2,,解:在(,1),(1,2)和(2,)内连续,x=1为可去向断点,若令,,,fx()则在x=1连续;x=2为第二类向断点(
1fx()sin,(4); x
,,(0,),,解:fx()在(,0)和内连续,x=0为第二类向断点;
2xx,(5); fx(),2||(1)xx,
fx()(,1),,,(1,),,解:在,(,1,0),(0,1)和内连续;x=是第二类间断点;,1
1f(1),fx()x=0是跳跃间断点;x=1是可去间断点,若令,则在x=1处连续( 2
xx,,1,3,,(6) fx(),,4,3.,,xx,
(,3),,[3,),,fx()解:在和内连续,x,3为跳跃间断点(
(讨论下列函数的连续性,若有间断点,判别其类型( 3
1fxx,,(1)()lim(0); n,,n,x1
,1,01,,,x,1,: 为跳跃间断点; 解x,1fxx()1,,,,2,,
,0,1.x,,
2n(1),xx(2)( fx()lim,2n,,n1,x
xx,||1,,,xx,,,11和解: 为跳跃间断点( fxx()0,||1,,,
,,,xx,||1,
sin2x,,0,x,,fx()4(设函数试确定的值,使函数在x,0处连续( fx(),ax,2,xax,,,0.,
sin2x2lim()lim2,lim()lim(),fxfxxaa,,,,,fa(0),,解:因为所以,依题意有,,,,xxxx,,,,0000x
=2. a
ln(13),x,,0,x,,5(设函数在点x,0处连续,求和b的值( fx(),asinax,
,bxx,,1, 0,
ln(13)3,xf(0)1,,lim()lim(1)1,lim()limfxbxfx,,,,,解:因为,依题意有,,,,xxxx,,,,0000sinaxa
ab,3,为任意实数(
fx()6(试分别举出具有以下性质的函数的例子:
11fx()(1) xn,,,,,,0,1,2,,,,,??是的所有间断点,且它们都是无穷间断点,2n
πfxx()cot(,,π)cot例如: ; x
1,,xQ,,fx()(2) 在上处处不连续,但在上处处连续;例如: fx(),RRfx(),c,,1,;xQ,
xxQ,,,,fx()(3) 在上处处有定义,但仅在一点连续,例如: fx(),R,c,,xxQ,.,
习 题 1-9
1(研究下列函数的连续性:
2x(1); fxxx()cose,,
2xfx()解答:因为在上是初等函数,所以在上连续( ,,,,,,,,,,fxxxe()cos,,,,,,
x,3(2)fx(),; 3x,27
x,31x,3fx()lim,解答:显然当时,无意义,但,则是函数fx(),x,3x,333x,3x,2727x,27
的可去间断点(
2(3); fxxx()12,,,,
2,,,,xx120fx()解答:当时,即时,连续( x,,4,3,,
,(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,limarcsin解: limarcsin()xxx,,2x,,,x,,,xxx,,
x1π, ; ,,limarcsinarcsin2x,,,26,,xxx
1,,xln(2)2lim(3); x,1π3arctanx,4
11,,,,xln(2)ln(21)122lim;,,解: x,1πππ3arctan3arctan1x,,44
1,xx(4); lim14,x,,x,011,x1,x,,,,(4)x,44xxxlim[1(4)],,x,e解:lim14,,x; ,,x,0x,0
2xlim[1ln1],,x(5); ,,,x012ln(1),x2,2ln(1),xxxe,解: ; lim[1ln(1)]lim[1ln(1)],,,,,xxxx,,00
1x2,x1coslim(1e),x(6); x,021x1x,,ex2222xx,1cosx,1cosxxe解: lim(1)lim(1);,,,,xexee,,xx00
1tan1sin,,,xx(7); lim2x,0xxx1sin,,
1tan1sin,,,xx解: lim2x,0xxx1sin,,
(1tan1sin)(1tan1sin),,,,,,xxxx21sin1,,xlim,, 22x,0()(,)xxx1sin11sin1,,,1tan1sin,,,xx
,
2(tansin)1sin1xxx,,,,,limlim 2xx,,00xx,sin1tan1sin,,,xx
2xx,1tan(1cos)xx,2,,lim,lim ; 22x,0x,0xx,2xx,sin
2cotxlim(cos)x(8); x,0
1cos1x,1,,22cotxcos1x,tanx2lim(cos)x解: ; ,,,,lim(1cos1)xe,x0x,0
lim[lnln(2)]nnn,,(9)( n,,
n2,,,lim[lnln(2)]nnn,,limlnlimln(1)nn解: nn,,,,n,,,,nn22nn,,22,,,,,2,2n,2,,22n,,,,limln(1) ; e,,,,,limln(1)ln2,,nn,,n,2n,2
xfx()gx()3(设函数与在点连续,证明函数 0
,()max(),()xfxgx,,()min(),()xfxgx,, ,,,,
x在点也连续( 0
证明:略(
2,abxx,,,0,,fx(),(,),,,,4(若函数在内连续,则和的关系是( )( ba,sinbx, 0x,,x,
A(( B(( C(( D (不能确定( ab,ab,ab,
sinbx2,,,,,fa(0),,解答:因为依题意有lim()lim(),lim()lim,fxabxafxb,,,,xxxx,,,,0000x( ab,
xxa,2,,5(设且,求常数的值( lim8,a,0a,,,,xxa,,,
xaax,3,xaaa,2333axxa33axa,e,8解:因为,则,所以lim()lim(1)lim(1),,,,,exxx,,,,,,xaxaxa,,,
( a,ln2
习 题 1-10
(1,)e1. 证明方程xxln2,在内至少有一实根(
[1,e]f(1)20,,,fee()20,,,,fxxx()ln2,,fx()证明:令,则在上连续,又,根
f()0,,fxxx()ln2,,(1,)e,据零点定理, 在开区间内至少有一点使,即xxln2,在
(1,)e内至少有一实根(
5xx,,12(证明方程有正实根(
5f(1)10,,fx()(,),,,,f(0)10,,,证明:令,则在内连续,又,, fxxx()1,,,
55f()0,,xx,,1(0,1),根据零点定理,在内至少有一点,使,即有正fxxx()1,,,
实根(
fx()[,]ab3(设函数对于闭区间上的任意两点、,恒有,其yfxfyLxy()(),,,x
fafb()()0,,f()0,,,,(,)ab中为正常数,且(证明:至少有一点,使得( L
xxab,,,(,)xab,(,)证明:任取,取,x,使,依题意有,0()(),,,,,,fxxfxLx
(,)abfx()lim()()0fxxfx,,,,lim()()fxxfx,,,则,即,由的任意性,可知在内连x,,x0,,x0
fx()fx()[,]ab续,同理可证在点右连续,点b左连续,那么,在上连续。而且a
fafb()()0,,f()0,,,,(,)ab,根据零点定理,至少有一点,使得(
axxxb,,,,,?(,)xxfx()[,]ab,4(若在上连续,,则在内至少有一点,使12n1nfxfxfx()()(),,,?12n,f(),( n
fx()[,]abfx()[,]ab证明:因为在上连续,所以在上有最小值,最大值,使得Mm
fxfxfx()()...(),,,12nmfxM,,()mfxM,,()mfxM,,()...mM,,,,,,因此, 12nn
fxfxfx()()(),,,?12n(,)xx,,f(),由介值定理得,在内至少有一点,使. 1nn
n
xabtin,,,[,],0(1,2,3,,)?fx()[,]ab,(若在上连续,,且(试证至少 t,1ii,i,0i
ftfxtfxtfx()()()(),,,,,?,,(,)ab存在一点使得( 1122nn
证明:因为fx()在[,]ab上连续,所以fx()在[,]ab上有最小值,最大值,使得MmmfxM,,()mfxM,,()mfxM,,()tt()tmfxM,,,,,,那么,...12n1111
nnnn
tt()tmfxM,,...tt()tmfxM,,,,即,又,故mttfiMt,,()t,12222nnnn,,,,iiii,,,111,1iiiin
,,(,)ab,由介值定理可知,至少存在一点使得 mtfiM,,(),i,1i
ftfxtfxtfx()()()(),,,,,?( 1122nn
(,),,,,(,),,,,fx()lim()fxfx()6(证明:若在内连续,且存在,则必在内有界( x,,
lim()fx证明:因为存在,则必有,使得当时,对任意的,有,X,0xX,fxa(),,,,x,,
xXX,,,,,,(,)(,):fx()(,),,,X(,)X,,因此,在区间及区间上有界,即当,存在
M,0M,0fx()[,],XX,有,同时,在上连续,有由有界性定理知,存在,fxM(),121
max{,}MMM,x,,,,,(,)当,取,则当时,总有,即xXXfxM,,,[,],()fxM(),122
fx()(,),,,,在内有界(
复习题A
x1gx(),fx(),1.设, , 求及其定义域. fgxgfx(),,,,,,,2,,2,x1,x
21,x,,12xx,,,420解: , 其定义域为且x,1, fgx(),,,,22xx,,42x,,2,,,1,x,,
Dxxx,,,,,,|221且即;. ,,
121222,x203-0,,,,xx且gfx,,,,,其定义域为 ,,2,,13,x1,22,x
Dxxx,,,,,|23且. ,,
22.求函数的反函数. fxxx()(1)sgn,,
2,,,,,,,(1),1xx,,,(1),0xx,,,1fxx()0,0,,解: 因, 所以, 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,,xlim1e,,,,,,,,,0,xx0x,,
x,x11,,,,C(; D(( ,,,lim1e,,lim1e,,,,,,xx,,,xx,,,,
x,0(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( )(
2x1cos,xA(; B(;
211,,xC(; D(( xx,tan
12x,1x,1(3)极限为( ) limex,1x,1
A(; B(0; C(; D(不存在但不为( 1,,
,()x,()x 若当时,和都是无穷小,则当时,下列表达式中哪一(4)xx,xx,00
个不一定是无穷小( )(
22,,()()xx,A(; B(和; ,()x,()x
2,()xln1()(),,,,xxC(; D( ( ,,()x,
2,xxb,,2,1,x,,lim()fxA,(5) 设适合,则以下结果正确的是( )( fx(),x,1,x,1,ax,1,,
abA,,,,4,3,4aAb,,4,4,A(; B(可取任意实数;
bAa,,,3,4,abA,,C(可取任意实数; D(都可取任意实数( 解答:
(1) A; (2) D; (3) D: (4) D; (5) C(
4.求下列极限:
πsinn,1πn3,,lim3sinlim3π3π(1) ; n,1nn,,,,π3n,13
1,1n,11188,,,,,,?(2) lim(1)lim; nnn,,,,1887,18
22()()xxxxxx,,,,2limarccos()limarccosxxx,,,(3) ; 2nn,,,,,,xxx,,
x1π=; ,,limarccosarccos2n,,,23,,xxx
21cos22sin2,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
1sin11111,,x,,,,,x(8) limlimsin101;xxxx,,,limsinsinlimsinlimsin,,xx,,,,xxx,,,,,,1xxxxx,,
xxaax,2 xaa,2xa22axa,(9) lim()lim(1);,,,exx,,,,xaxa,,
11cos1x,1,,xxxcos1x,2,,,,limcoslim1(cos1)xx,,,( (10) lim1(cos1)xe,,,,,,,,,,,,,xx00x,0
fx()gx()fxgx()(),gx()5(设当时,是比高阶的无穷小(证明:当时,与xx,xx,00
是等价无穷小(
fx()fxgxfx()()(),fxgxgx()()(),与证明: 由已知得则即,lim0,limlim(1)1,,,,xx,xxxx,,000gx()gxgx()()是等价无穷小(
3xaxb,,lim8b6(已知,求常数与的值( ,ax,22x,
3xaxb,,3lim8解:因为,所以,则,lim()820xaxbab,,,,,,x,2x,22x,
32xaxbxxxa,,,,,,(2)[2(4)]ba,,,,82,limlim那么xx,,22xx,,2(2)
2解得ab,,,4,0,,,,,,,lim[2(4)]128xxaa,( x,2
7.设xxxnx,,,,10,6(1).{}证明数列极限存在,并求此极限( 11nnn,
xx,,x,10证明:用数学归纳法证明此数列的单调性(因为 及xx,,,64,可知11221
xx,,x,0{}x{}x假设则 xxxx,,,,,66,所以单调递减,又显然,即nn,1nnnnnnn,,,112
{}x有下界,由单调有界准则知存在极限,设limxA,,两边取极限,有对xx,,6nnnn,1,,n
6,AA=,解之得A=3或A=(舍去) ,即得limx=3( ,2n,,n
sin6x,,0,x,,2x,axx,,3,0,fx()8.确定常数a与b的值,使得函数=处处连续( ,,1x,(1),0.,,bxx,
1sin6xxfx()fx()解:当时,=和当时,=,显然它们都是连续的,又x,0x,0(1),bx2x
1bsin6xbxbxelimlimlimlimlimfx()f(x)==3, f(x)===,当x,0时,=a,(1),bx(1),bx,,,,,x,0x,0x,0x,0x,02x
be要使f(x)在x=0点也连续,则=3=a,即a=3,b=ln3(
9.求下列函数的间断点,并判断其类型(
1fx()arctan(1)=; x
1,1,fx(),fx()fx()limlimarctanlimlimarctan解:因为==,==,又时,x,0,,,,x,0x,0x,0x,022xx
连续,所以只有x=0为间断点,x=0为跳跃间断点(
xfx()(2)=; tanx
xlim()lim,fx解:当tanx=0时,有x=0或x=n(n=1,2,…)因为=1,所以,,,xx,,00tanx
x,limx=0为可去间断点(又=(n=1,2,…),所以(n=1,2,…)为无穷,,,,xn,,xn,,tanx
,,xxnxn,,lim,,间断点(当(n=1,2,…)时,=0,所以(n=1,2,…),,,,,,,2tanx2xn,,,2是可去间断点(
nxxe,fx()lim,(3); nxx,,1,e
1,0,x,,
,nxxe,1,()limfx,limlimfx()解:=,因f(x)=0, =1,所以x=0为跳跃间,0x,,nx,,x,,x,0x,01,e2,xx,0,,,,
断点(
复习题B
1(单项选择题
x,0(1)当时,下列无穷小量中与不等价的是( )( x
2ln(1),x23x242e251,,,xxxxx,,3A(( B(( C(( D(( sin(6sin)xx,x
(2)下列极限不存在的是( )(
12xx,,311ln(1)1,xsinxxA(( B(( C(( D(( xlimlim(arctan),limsin,lim(2),x,0x,,,,,x,0xxxxxx
(3)极限( )等于( e
1111x,1xxxlim(1),xA(( B(( C(( D(( ,,,lim(1)lim(1)lim(1),,,,,,0xxxx,,,xxx
,n|()|()fngn,lim()3gn,lim()fn(4)设,数列,如果,则的值为( )( n,,n,,lim()3fn,,,,,3lim()3fnlim()3fn,,,,3lim()3fn(( B(( C(( D(( An,,n,,n,,n,,
22xb(5)已知,其中与为常数(则( )( a,,,lim()1axbx,,,x1
a,2b,3a,,2b,3a,2b,,3a,,2b,,3A(,( B(,( C(,( D(,(
3xx,(6)设函数,则( )( fx(),,sinx
A(有无穷多个第一类间断点( B(只有个可去间断点( 1
3C(有个跳跃间断点( D(有个可去间断点( 2
解答:
2)C;(3)B;(4)B;(5)C;(6)D(提示:x=0,(1)D;(1为可去间断点) ,,
2(填空题
x,1[0,1]fx()(1)设函数的定义域是,则f()的定义域是_________( x,1
211,,xxlimln(2)计算=_________( 20x,xxx1,,
1222axbx,b,(3)设lim(122)e,,,xx,则=_________, ( ax,0
2,xxxcos,x,0ee,x(4)设时,与是同阶无穷小,则 ( ,,
fx()fx()fx()(5)设,则 , ( ,,,,lim3limlim32x,x,0x,00xxx
x(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:数列有,,n
fx()fx()x界是数列收敛的 条件;函数的极限存在是在的某一去xlim()fx,,n0xx,0
fx()心邻域内有界的 条件;函数在的某一去心邻域内无界是的 xlim()fx,,0xx,0
fx()fx()条件;函数在左连续且右连续是在连续的 条件( 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,1,,,,,2必要,充要(
22211,,xxln(1)ln(1),,,,,xxxxlimlim(2)题解答过程:ln= 2x,0x,0xxx1,,2x
2222,,,,ln(1+(),,xxln1),,(xxxx,11,,xx,,,,,,()limlim====1( ,lim,x,0x,0x,02x222x2x2x
(3)题解答过程:
21122xx,2,222222axbx,a22xxaxbx,,ee,lim因为lim(122),,,xx(122),,xx=,所以,a=1,b为任意实x,0x,0
数(
(4)题解答过程:
1x422xexx,,()xxecos,xxx(cos1),x2eee[1],exx (cos1),2因为====c(常limlimlimlimuu,uu,,,,x0,x,0x,0x0xxxx
9数),所以u=( 2
fx()fx(),,3(),x,()xlim(5)题解答过程:因为=,所以= (其中为当x时的,3,033x,0xx
fx()fx()fx()fx()22,,3()xxx ,limlim无穷小量),那么=,=,故=0,=0( ,,3()xxx ,22x,0x,0xxxx
3(求下列极限:
3xe1,limlim[1212(1)],,,,,,,,??nn(1); (2); ,n,,x,01cos(1cos),,xx
1xxx1x,,abc,,x2,x1cos(0,0,0)abc,,,lim(1arctan),ex(3)lim(;(4); ,,,x0,x03,,
tanx(5); (6)lim(sin1sin)xx,,; lim(sin)xπx,,,,x2
1,,1x2esin,x111,,n,lim(7); (8); ,,,,?lim(1)4,,,,0xn,n23||x,,x1e,,,
sinx3nln(e1cos)sin,,,xx22(9); (10)( lim(1)(1)(1),||1,,,,xxxx?lim3,x0x,,arctan(41cos),x
解答:
nnnn(1)(1),,(1) lim[12121],,,,,,,,,??nnlim[],,,n,,n,,22
nnnnnnnn(1)(1)(1)(1),,,,[][],,n2222 ,,limlimn,,n,,nnnn(1)(1),,,,(1)(1)nnnn[],,[]2222
12 ,,limn,,211,,11nn,22
3x33e,1xxlim,(2)( lim4,lim,,3,,,0xx,x,0(1cos),0xxx1cos(1cos),,xx
24(3)xxx31(1)(1)(1)abc,,,,,1xxxxxxxxx abcabc,,,,,,33abc,,xxxxxabcx,,,33 lim()lim(1)lim(1),,,,xx,,00,x0333
lnabc33abce==(
x21arctanex1 x2x22x2,1cosx,x1cosexarctanlim(1arctan),,exe(4)=( lim(1arctan),exx,0,x0
1sin(sin1)xx,,tanxsin1cosxx,lim(1sin1),,x(5) lim(sin)x,π,x,,x22
xx2,,sin(cossin)x122,sin1x,xx22cossin,022lim(1sin1),,,xe,1=( ,x,2
xxxx,,,,11lim(sin1sin)xx,,,(6), lim2sincosx,,,x,,,22
xx,,1xx,,1xx,,1其中|2|2,即2是有界量,,故,coscoslimsin0,x,,,222
lim(sin1sin)0xx,,,( x,,,
1111111n111nnnlim1n,(7)因为,又,所以( ,,,,,,n,,,,,lim(1....)11(1....)n,,,,nnn2323
11xx2sin2sin,,exex(8)因为, lim()lim()211,,,,,,44,,,,xx00||xxxx11,,ee
111xxx2sin2sin,,exex2sin,ex,所以,( lim()1,,lim()lim()011,,,,,,444,,,,,xxx000||x||xxxxx1,e11,,ee
n22n(1)(1)(1)...(1),,,,xxxx22(9) lim(1)(1)...(1)lim,,,,xxxnn,,,,1,xn,1211,x. ,,,lim(||1)xn,,11,,xxsinsinsinxxx33ln(1cos)sinln(1cos)lnexxexe,,,,,, (10)limlim,33,,xx00arctan(41cos)41cos,,xx
331cos1cos,,xxln(1),xxsinsin111ee . ,,,,limlimlimxsin33,,,xxx00044e41cos41cos,,xx
nxfx()(已知函数4,试确定的间断点及其类型( ()lim,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,,
lim()lim()0(1)fxfxf,,,lim()lim()0,(1)fxfxf,,,不存在,所以, ,,,,xx,,11xx,,,,11
因此,x,,1均为可去间断点。
2,axbxx,,,1,,bfx()5(设函数求,使在处连续( x,1fxx()3,1,,, a,
,2,1.abxx,, ,
2lim()lim()fxaxbxab,,,,lim()lim(2)2fxabxab,,,,解: 因为, ,,,,xx,,xx,,1111
abab,,,2,f(1)3,fx(),要使在x,1处连续,则,解得a,2,b,1. ,ab,,3,
ππ(,),6(求证方程xx,,,1sin0在区间上至少有一个根( 22
ππππ,,fxxx()1sin,,,fx()证明:令,显然在上连续,又f()0,,,,, ,,,,2222,,
ππππ,,fx(),f()20,,,,由零点定理可知,在内至少有一个零点,即方程 ,,,,2222,,
ππ,,在内至少有一个根. xx,,,1sin0,,,,22,,
1aa,0x,0{}x7(设,任取,令(其中)(证明数列收敛(并求n,1,2,?,,xx()1nnn,12xn
极限( limxn,,n
2x11an是单调的,(1)若,则,证明:首先证明,,,,,xxxx()()xa,x,,,nnnn1nn22xxnn
2x11an即单调递增有上界。(2)若,则,即 ,,,,,xxxx()()xa,x,,,nnnnnn122xxnn
1alimxA,AA,,()单调递减有上界,综(1)(2)知数列有极限存在;令,则,xx,,,,nnn,,n2A
Aa,Aa,,limxa,解之得或(舍去),即. n,,n
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:lim()lim100pc所以如果费用C允许无限增长,可被清除污染物的百分cc,,,,,8000c
比为100%,实际上是不可能完全清除污染的.