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習(xí)題
1.按定義證明下列極限:
(1) limx???6x?5=6 ;(2) lim(x2-6x+10)=2; x?2x
x2?5?1 ;(4) lim?(3) lim2x???x?1x?2
(5) limcos x = cos x0 x?x04?x2=0;
2.根據(jù)定義2敘述limf (x) ≠ A.x?x0
3.設(shè)limf (x) = A.,證明limf (x0+h) = A.x?x0h?0
4.證明:若limf (x) = A,則lim| f (x)| = |A|.當(dāng)且僅當(dāng)A為何值時(shí)反之也成立? x?x0x?x0
5.證明定理3.1
6.討論下列函數(shù)在x0→0 時(shí)的極限或左、右極限: (1)f(x)=x
x;(2) f(x) = [x]
?2x;x?0.?(3) f (x)=?0;x?0.
?1?x2,x?0.?
7.設(shè) limf (x) = A,證明limf (x???x?x01) = A x
8.證明:對(duì)黎曼函數(shù)R(x)有l(wèi)imR (x) = 0 , x0∈[0,1](當(dāng)x0=0或1時(shí),考慮單側(cè)極限).x?x0
習(xí)題
1. 求下列極限:
x2?1 (1)lim2(sinx-cosx-x);(2)lim; ?x?02x2?x?1x?22
x2?1?x?1???1?3x?;
lim(3) lim;(4)
x?12x2?x?1x?0x2?2x3
xn?1(5) limm(n,m 為正整數(shù));(6)lim
x?1xx?4?1
(7)lim
x?0
?2x?3x?2
70
;
20
a2?x?a?3x?6??8x?5?.
(a>0);(8) lim
x???x5x?190
2. 利用斂性求極限: (1) lim
x???
x?cosxxsinx
;(2) lim2
x?0xx?4
x?x0
3. 設(shè) limf(x)=A, limg(x)=B.證明:
x?x0
(1)lim[f(x)±g(x)]=A±B;
x?x0
(2)lim[f(x)g(x)]=AB;
x?x0
(3)lim
x?x0
f(x)A
=(當(dāng)B≠0時(shí)) g(x)B
4. 設(shè)
a0xm?a1xm?1???am?1x?am
f(x)=,a0≠0,b0≠0,m≤n,nn?1
b0x?b1x???bn?1x?bn
試求 limf(x)
x???
5. 設(shè)f(x)>0, limf(x)=A.證明
x?x0
x?x0
lim
f(x)=A,
其中n≥2為正整數(shù).6.證明limax=1(0\n
x?0
7.設(shè)limf(x)=A, limg(x)=B.
x?x0
x?x0
(1)若在某∪(x0)內(nèi)有f(x)
(2)證明:若A>B,則在某∪(x0)內(nèi)有f(x) > g(x).8.求下列極限(其中n皆為正整數(shù)): (1) lim ?
x?0
x
x11
lim;(2);nn?x?0x1?xx1?x
x?x2???xn?n
(3) lim ;(4) lim
x?0x?0x?1
?x?1
x
(5) lim
x??
?x?(提示:參照例1)
x
x?0
x?0
x?0
9.(1)證明:若limf (x3)存在,則limf (x)= lim f (x3)(2)若limf (x2)存在,試問(wèn)是否成立limf (x) =limf (x2) ?
x?0
x?0
x?0
習(xí)題
1.敘述函數(shù)極限limf(x)的歸結(jié)原則,并應(yīng)用它證明limcos x不存在.
n???
n???
2.設(shè)f 為定義在[a,+?)上的增(減)函數(shù).證明: lim= f(x)存在的充要條件是f在
n???
[a,+?)上有上(下)界.
3.(1)敘述極限limf (x)的柯西準(zhǔn)則;
n???
(2)根據(jù)柯西準(zhǔn)則敘述limf (x)不存在的充要條件,并應(yīng)用它證明limsin x不存在.
n???
n???
4.設(shè)f在∪0(x0)內(nèi)有定義.證明:若對(duì)任何數(shù)列{xn}?∪0(x0)且limxn=x0,極限limf(xn)都
n??
n??
存在,則所有這極限都相等.
提示: 參見(jiàn)定理3.11充分性的證明.
5設(shè)f為∪0(x0)上的遞減函數(shù).證明:f(x0-0)和f(x0+0)都存在,且f(x0-0) =supf(x),f(x0+0)=
0x?u?
?x0?
0x?un(x0)
inff (x)
6.設(shè) D(x)為狄利克雷函數(shù),x0∈R證明limD(x)不存在.
x?x0
7.證明:若f為周期函數(shù),且limf(x)=0,則f(x)=0
x???
8.證明定理3.9
習(xí)題
1.求下列極限
sin2xsinx3
(1) lim;(2) lim
x?0x?0sinx2x
(3) lim
x?
cosxx?
?
tanx?sinxarctanx
lim(5) lim;(6) ; 3x?0x?0xx
sin2x?sin2a1
(7) limxsin ;(8) lim;
x???x?axx?a
;(4) lim
x?0
tanx
; x
?cosx2
(9) lim;(10) lim
x?0x?01?cosxx?1?1
sin4x
2.求下列極限
12?x
(1) lim(1?);(2) lim?1?ax?x(a為給定實(shí)數(shù));
n??x?0x
x
(3) lim?1?tanx?
x?0
cotx
;(4) lim?
?1?x?
?;
x?01?x??
(5) lim(
x???
3x?22x?1?
);(6) lim(1?)?x(?,?為給定實(shí)數(shù))
n???3x?1x
3.證明:lim?lim?cosxcoxcos4.利用歸結(jié)原則計(jì)算下列極限: (1) limnsin
n??
?
x?0n??
??
?
x2
xx???cos?1 2n??22??
?
n
;(2)
習(xí)題
1. 證明下列各式
(1) 2x-x2=O(x) (x→0);(2)x sinx?O(x)(x→0);
+
(3)?x?1?o(1) (x→0);
(4) (1+x)n= 1+ nx+o (x) (x→0)(n 為正整數(shù)) (5) 2x3 + x2=O(x3)(x→∞) ;
(6) o (g(x))±o(g(x)) =o(g(x))(x→x0)
(7) o(g1(x))·0(g2(x))=o(g1(x)g2(x))(x→x0) 2. 應(yīng)用定理3.12求下列極限:
?x2?1x(1) lim(2)lim x?01?cosxx??x?cosx
x3. 證明定理3.13
4. 求下列函數(shù)所表示曲線的漸近線:
13x3?4
(1) y = ;(2) y = arctan x ;(3)y = 2
xx?2x
5. 試確定a的值,使下列函數(shù)與xa當(dāng)x→0時(shí)為同階無(wú)窮小量:
(1) sin2x-2sinx ;(2)
- (1-x); 1?x
(3)?tanx??sinx;(4)
x2?4x3
6. 試確定a的值,使下列函數(shù)與xa當(dāng)x→∞時(shí)為同階無(wú)窮大量:
(1)
x2?x5;(2)x+x2 (2+sinx);
(3) (1+x)(1+x2)…(1+xn).
7. 證明:若S為無(wú)上界數(shù)集,則存在一遞增數(shù)列{xn}?s,使得xn→+∞(n→∞)
8. 證明:若f為x→r時(shí)的無(wú)窮大量,而函數(shù)g在某U0(r)上滿足g(x)≥K>0,則fg為x→r
時(shí)的無(wú)窮大量。
9. 設(shè) f(x)~g(x) (x→x0),證明:
f ( x )-g ( x ) = o ( f ( x ) )或 f ( x )-g ( x ) = o ( g ( x ) )
總 練 習(xí) 題
1. 求下列極限:
?1
(x?[x])lim([x]?1)(1) lim;(2)??
x?3
x?1
(3) lim(
x???
a?xb?x?a?xb?x)
xx?a
(4) lim
x???
(5)lim
xx?a
x???
(6) lim
?x??x?x??x
x?0
(7) lim?
n??m
,m,n 為正整數(shù) ?n?x?11?xm1?x??
2. 分別求出滿足下述條件的常數(shù)a與b:
?x2?1?
(1) lim??ax?b???0 x????x?1??
x(3) limx
(2) lim
x???x???x?2
??x?1?ax?b??0
?x?1?ax?b?0
x?2
3. 試分別舉出符合下列要求的函數(shù)f:
(1) limf(x)?f(2);(2) limf(x)不存在。
4. 試給出函數(shù)f的例子,使f(x)>0恒成立,而在某一點(diǎn)x0處有l(wèi)imf(x)?0。這同極限的
x?x0
局部保號(hào)性有矛盾嗎?
5. 設(shè)limf(x)?A,limg(u)?B,在何種條件下能由此推出
x?a
g?A
limg(f(x))?B?
x?a
6. 設(shè)f (x)=x cos x。試作數(shù)列
(1){xn} 使得 xn→∞(n→∞), f(xn)→0 (n→∞); (2){yn} 使得 yn→∞(n→∞), f(yn)→0 (n→∞); (3){zn} 使得 zn→∞(n→∞), f(zn)→0 (n→∞).
7. 證明:若數(shù)列{an}滿足下列條件之一,則{an}是無(wú)窮大數(shù)列:
(1) liman?r?1
n??
(2) lim
an?1
?s?1(an≠0,n=1,2,…)
n??an
n2
n2
8. 利用上題(1)的結(jié)論求極限:
(1) lim?1?
?n??
?1??1??(2) lim?1??
n??n??n?
9. 設(shè)liman???,證明
n??
(1) lim
(a1?a2???an)??? n??n
n??
(2)若an > 0(n=1,2,…),則lima1a2?an??? 10.利用上題結(jié)果求極限:
(1)limn!(2) lim
n??
In(n!)
n??n
11.設(shè)f為U-0(x0)內(nèi)的遞增函數(shù)。證明:若存在數(shù)列{xn}?U-0(x0)且xn→x0(n→∞),使得
limf(xn)?A,則有
n??
f (x0-0) =
supf(x)?A
0x?U?(x0)
12.設(shè)函數(shù)f在(0,+∞)上滿足方程f(2x)=f(x),且limf(x)?A。證明:f(x)?A,x∈(0,+∞)
x???
13.設(shè)函數(shù)f在(0,+∞)此上滿足方程f (x2) = f (x),且
f(x)=limf(x)?f(1)lim?
x?0
x???
證明:f(x)?f(1),x∈(0,+∞)
14.設(shè)函數(shù)f定義在(a,+∞)上,f在每一個(gè)有限區(qū)間內(nèi)(a,b)有界,并滿足
x???
lim(f(x?1)?f(1))?A證明
x???
lim
f(x)
?A x