第一篇：北大版高等數(shù)學(xué)第一章 函數(shù)及極限答案 習(xí)題1.2（范文）

習(xí)題1.2 1.求下列函數(shù)的定義域:(1)y?ln(x2?4);(2)y?ln1?x5x?x211?x;(3)y?ln4;(4)y?2x2?5x?3.解(1)x2?4?0,|x|2?4,|x|?2,D?(??,?2)?(2,??).(2)1?x1?x?0.??1?x?0或?1?x?0?1?x?0??1?x?0.?1?x?1,D?(?1,1).(3)5x?x24?1,x2?5x?4?0.x2?5x?4?0,(x?1)(x?4)?0,x1?1,x2?4.D?(1,4).(4)2x2?5x?3?0.(2x?1)(x?3)?0,x1??3,x2?1/2.D?(??,?3)?(1/2,??).2.求下列函數(shù)的值域f(X),其中X為題中指定的定義域.(1)f(x)?x2?1,X?(0,3).f(X)?(1,10).(2)f(x)?ln(1?sinx),X?(??/2,?],f(X)?(??,ln2].(3)f(x)?3?2x?x2,X?[?1,3],3?2x?x2?0,x2?2x?3?0,(x?1)(x?3)?0,x1??1,x2?3,f(X)?[0,f(1)]?[0,4].(4)f(x)?sinx?cosx,X?(??,??).f(x)?2(sinxcos(?/4)?cosxsin(?/3))?2sin(x??/4),f(X)?[?2,2].3.求函數(shù)值：設(shè)f(x)?lnx2(1)ln10,求f(?1),f(?0.001),f(100);(2)設(shè)f(x)?arcsinx1?x2,求f(0),f(1),f(?1);(3)設(shè)f(x)???ln(1?x),???x?0,??x, 0?x???,求f(?3),f(0),f(5).?cosx,0?x?1,(4)設(shè)f(x)???1/2, x?1,求f(0),f(1),f(3/2),f(2).??2x, 1?x?3解(1)f(x)?logx2,f(?1)?log1?0,f(?0.001)?log(10?6)??6,f(100)?log104=4.(2)f(0)?0,f(1)?arcsin(1/2)??/6,f(?1)?arcsin(?1/2)???/6.(3)f(?3)?ln4,f(0)?0,f(5)??5.(4)f(0)?cos0?1,f(1)?1/2,f(3/2)?22,f(2)?4.4.設(shè)函數(shù)f(x)?2?x2?x,x??2,求f(?x),f(x?1),f(x)?1,f??1?1?x??,f(x).解f(?x)?2?x2?x?13?x2?x,x??2;f(x?1)?2?x?1?1?x,x?1,x??3, 2?x4?1?2?1/x2x?1?1?,x??2;f????,x?0,x??1/2,2?x2?x?x?2?1/x2x?112?x?,x??2.f(x)2?xf(x??x)?f(x)5.設(shè)f(x)?x3,求，其中?x為一個(gè)不等于零的量.?xf(x??x)?f(x)(x??x)3?x3x3?3x2?x?3x?x2??x3?x3解???3x2?3?x??x2.?x?x?x6.設(shè)f(x)?lnx,x?0,g(x)?x2,???x???,試求f(f(x)),g(g(x)),f(g(x)),g(f(x)).f(x)?1?解f(f(x))?f(lnx)?lnlnx,x?1;g(g(x))?g(x2)?x4,???x???;f(g(x))?f(x2)?lnx2,x?0;g(f(x))?g(lnx)?ln2x,x?0.?0, x?0,?x, x?0;7.設(shè)f(x)??g(x)??求f(g(x)),g(f(x)).?x,x?0;1?x,x?0,??解?x,g(x)?0,f(g(x))?0.?g(0), x?0,?0, x?0,g(f(x))????g(?x),x?0.???x,x?0.8.作下列函數(shù)的略圖:(1)y?[x],其中[x]為不超過x的最大整數(shù);(2)y?[x]?x;1(3)y?sinhx?(ex?e?x)(???x???);21(4)y?coshx?(ex?e?x)(???x???);2?x2, 0?x?0,(5)y???x?1,?1?x?0.（1）

（2）

（3）

（4）

（5）

?x29.設(shè)f(x)??,x?0,求下列函數(shù)并且作它們的圖形?x, x?0,:(1)y?f(x2);(2)y?|f(x)|;(3)y?f(?x);(4)y?f(|x|).解(1)y?x4,???x???.(2)y?|f(x)|???x2,x?0，??x, x?0.(3)y?f(?x)???x2,?x?0,?x2,x?0,??x, ?x?0????x, x?0.(4)y?f(|x|)?x2,???x???.3

求下列函數(shù)的反函數(shù):(1)y?x2?2x(0?x???);(2)y?sinhx(???x???);(3)y?coshx(0?x???).解(1)x2?2x?y,x2?2yx?4?0,x?y?y2?4,y?x?x2?4(???x???).ex?e?x(2)?y,z?ex,z2?2yz?1?0,ex?z?y?y22?1,x?ln(y?y2?1),y?ln(x?x2?1),(???x???).(3)ex?e?x2?y,z?ex,z2?2yz?1?0,ex?z?y?y2?1,x?ln(y?y2?1),y?ln(x?x2?1),(x?1).證明cosh2x?sinh2x?1.?ex?e?x?2?ex?e?x?2(e2x證coshx?sinhx???e?2x?2)?(e2x?e?2x22?2)?2?????2???4?1.下列函數(shù)在指定區(qū)間內(nèi)是否是有界函數(shù)?(1)y?ex2,x?(??,??);否(2)y?ex2x?(0,1010);是(3)y?lnx,x?(0,1);否(4)y?lnx,x?(r,1),其中r?0.是2(5)y?e?x2?sinx?cos(2x),x?(??,??);是|y|?12?1?1?2.4 10.11.12.(6)y?x2sinx,x?(??,??);否.(7)y?x2cosx,x?(?1010,1010).是

13.證明函數(shù)y?1?x?x在(1,??)內(nèi)是有界函數(shù).證y?1?x?x?(1?x?x)(1?x?x)1?x?x?11?x?x?12?1(x?1).13.研究函數(shù)y?x6?x4?x21?x6在(??,??)內(nèi)是否有界.|x|?1時(shí),x6?x4?x2x6?x4?x23x6解1?x6?3,|x|?1時(shí),1?x6?x6?3,|y|?y?3,x?(??,??).5

第二篇：北大版高等數(shù)學(xué)第一章 函數(shù)及極限答案習(xí)題1.6

習(xí)題1.6

1.證明:任一奇數(shù)次實(shí)系數(shù)多項(xiàng)式至少有一實(shí)根.證設(shè)P(x)是一奇數(shù)次實(shí)系數(shù)多項(xiàng)式,不妨設(shè)首項(xiàng)系數(shù)是正數(shù),則limP(x)???,x???

limP(x)???,存在A,B,A?B,P(A)?0,P(B)?0,P在[A,B]連續(xù),根據(jù)連續(xù)函數(shù)

x???的中間值定理,存在x0?(A,B),使得P(x0)?0.2.設(shè)0???1,證明對(duì)于任意一個(gè)y0?R,方程y0?x??sinx有解,且解是唯一的.證令f(x)?x??sinx,f(?|y0|?1)??|y0|?1????|y0|?y0,f(|y0|?1)?|y0|?1???|y0|?y0,f在[?|y0|?1,|y0|?1]連續(xù),由中間值定理,存在x0?[?|y0|?1,|y0|?1],f(x0)?y0.設(shè)x2?x1,f(x2)?f(x1)?x2?x1??(sinx2?sinx1)?x2?x1??|x2?x1|?0,故解唯一.3.設(shè)f(x)在(a,b)連續(xù),又設(shè)x1,x2?(a,b),m1?0,m2?0,證明存在??(a,b)使得f(?)?

m1f(x1)?m2f(x2)

m1?m2

.證如果f(x1)?f(x2),取??x1即可.設(shè)f(x1)?f(x2),則f(x1)?

m1f(x1)?m2f(x1)

m1?m2

?

m1f(x1)?m2f(x2)

m1?m2

?

m1f(x2)?m2f(x2)

m1?m2

?f(x2),在[x1,x2]上利用連續(xù)函數(shù)的中間值定理即可.4.設(shè)y?f(x)在[0,1]上連續(xù)且0?f(x)?1,?x?[0,1].證明在存在一點(diǎn)t?[0,1]使得f(t)?t.證g(t)?f(t)?t,g(0)?f(0)?0,g(1)?f(1)?1?0.如果有一個(gè)等號(hào)成立,取t為0或1.如果等號(hào)都不成立,則由連續(xù)函數(shù)的中間值定理,存在t?(0,1),使得g(t)?0,即f(t)?t.5.設(shè)y?f(x)在[0,2]上連續(xù),且f(0)?f(2).證明在[0,2]存在兩點(diǎn)x1與x2,使得|x1?x2|?1,且f(x1)?f(x2).證令g(x)?f(x?1)?f(x),x?[0,1].g(0)?f(1)?f(0),g(1)?f(2)?f(1)?f(0)?f(1)??g(0).如果g(0)?0,則f(1)?f(0),取x1?0,x2?1.如果g(0)?0,則g(0),g(1)異號(hào),由連續(xù)函數(shù)的中間值定理,存在??(0,1)使得g(?)?f(??1)?f(?)?0,取x1??,x2???1.

第三篇：北大版高等數(shù)學(xué)第一章 函數(shù)及極限答案習(xí)題1.4

習(xí)題1.4

1.直接用?-?說法證明下列各極限等式:(1)limx?ax?a(a?0);(2)limx?a;(3)lime?e;(4)limcosx?cosa.x?ax?ax?a22xa證(1)???0,要使||x?a|x?a|?|x-a|x?a??,由于|x-a|x?a?|x-a|ax?,a|??,故lim只需??,|x?a|?a?.取??a?,則當(dāng)|x?a|??時(shí),|x?a.ax?a(2)???0,不妨設(shè)|x?a|?1.要使|x2?a2|?|x?a||x?a|??,由于|x?a|?|x?a|?|2a|?1?|2a|,只需(1?|2a|)|x?a|??,|x?a|??當(dāng)1?|2a|.取??min{?1?|2a|,1},則|x?a|??時(shí),|x2?a2|??,故limx2?a2.x?a(3)???0,設(shè)x?a.要使|ex?ea|?ea(ex?a?1)??,即0?(ex?a?1)??ea,1?ex?a?1??ea,0?x?a?ln????min{??1?,1},則當(dāng)0?x?a??時(shí),|ex?eaa?,取?|??e?|2a|?,?1故limex?ea.類似證limex?ea.故limex?ea.x?a?x?a?x?a(4)???0,要使|cosx?cosa|?2sinx?aa2sinx?a2?2sinx?a2sinx?2?|x?a|,取???,則當(dāng)|x?a|??時(shí),|cosx?cosa|??,故limcosx?cosa.x?a2.設(shè)limf(x)?l,證明存在a的一個(gè)空心鄰域(a??,a)?(a,a??),使得函數(shù)u?f(x)在x?a該鄰域內(nèi)使有界函數(shù).證對(duì)于??1,存在??0,使得當(dāng) 0?|x-a|??時(shí),|f(x)?l|?1,從而|f(x)|?|f(x)?l?l|?|f(x)?l|?|l|?1?|l|?M.3.求下列極限:2(1)lim(1?x)2?1?lim2x?x?lim(1?x1.x?02xx?02xx?02)?22sin2?x???(2)lim1?cosx?2???1?sin?x?????1x?0x2?limx?0x22lim??2????12?1.x?0?x??22?2??(3)limx?a?axx?lim?1(a?0).x?0x?0x(x?a?a)2a(4)limx2?x?2x?12x2?2x?3??2?3.x2(5)lim?x?2?2x?02x2?2x?3??3.1

201030(6)lim(2x?3)(2x?2)x??(2x?1)30?2230?1.(7)lim1?x?1?x?lim2x?1.x?0xx?0x(1?x?1?x)(8)lim?13?x2?x?1?3x2?x?2x??1???x?1x3?1??lim?x??1(x?1)(x2?x?1)?limx??1(x?1)(x2?x?1)?lim(x?1)(x?2)(x?2)??3x??1(x?1)(x2?x?1)?limx??1(x2?x?1)3??1.(9)lim1?2x?3?lim(1?2x?3)(x?2)(1?2x?3)x?4x?2x?4(x?2)(x?2)(1?2x?3)?lim(2x?8)(x?2)?2?4x?4(x?4)(1?2x?3)6?43.n(n?1)2nlimxn?1n(10)?1ny?2y???yx?1?lim(1?y)x?1y?0y?lim?n.y?0y(11)limx2?1?x2?1??lim2?0.x???x??x2?1?x2?1mm?1(12)lima0x?a1x???amamx?0bnn?10x?b???b(bn?0)?1xnb.n?1?a0/b0,m?n(13)lima0xm?a1xm???amx??bnbn?1???b(a?b?00?0)??0, n?m0x?1xn???, m?n.x4?81?8/x4(14)limx?11?1/x2?1.x??2?limx??31?3x?3(15)lim1?2xx?0x?x2(3221?3x?333?lim1?2x)(1?3x?1?3x?31?2x?31?2x)x?0x?x2)(321?3x?31?3x?31?2x?32(1?2x)?lim5xx?0x(1?x)(321?3x?321?3x?31?2x?31?2x)

?lim522?5x?0(1?x)(31?3x?31?3x?31?2x?31?2x)3.(16)a?0,limx?a?x?a?lim?x?a1?x?a?0x2?a2x?a?0???x2?a2?x?a????lim?(x?a)(x?a)?1?x?a?0???x?ax?a(x?a)x?a???2

?lim?(x?a)?1?x?a?0??x?ax?a(x?a)x?a??

?lim?x?a?1?1.x?a?0???x?a(x?a)x?a????2ax4.利用limsinx?1及l(fā)im?1x?xx???1???x??e求下列極限:?(1)limsin?xsin?x?x?0tan?x?limx?0sin?xlimcos?x?x?0?.sin(2x2)sin(2x2(2)lim)2x2x?3x?lim1?0?0x?02x2?limx?03x?(3)limtan3x?sin2x?limtan3xsin2x21x?0sin5xx?0sin5x?limx?0sin5x?35?5?5.(4)limx?limxx?0?1?cosxx?0?2sinx?2.2cosx?aa(5)limsinx?sina2sinx?2?cosa.x?ax?a?limx?ax?a2?k?k??xx(?k)?x(6)limlimk???k?k??e?k.??1?x??x??x???1??k???x???limx???1?????x????5(7)lim(1?5y)1/y??1/(5y)??5y?0??lim(1?5y)??e.y?0?x?100x10(8)lim?1?10?lim?1?1?e.x?????x??x??????1???x???lim?x???1??x????5.給出limf(x)???及l(fā)imf(x)???的嚴(yán)格定義.x?ax???limf(x)???:對(duì)于任意給定的A?0,存在??0,使得當(dāng)0?|x-a|??時(shí)f(x)?A.x?alimf(x)???:對(duì)于任意給定的A?0,存在??0,使得當(dāng)x???時(shí)f(x)??A.x???3

第四篇：北大版高等數(shù)學(xué)第一章 函數(shù)及極限答案習(xí)題1.3

習(xí)題1.3

1.設(shè)xn?

nn?2

(n?1,2,?),證明limxn?1,即對(duì)于任意??0,求出正整數(shù)N,使得

n??

當(dāng)n?N時(shí)有 |xn-1|??,并填下表:

n

?1|?

2n?2

??,只需n?

2?2,取

證???0,不妨設(shè)??1,要使|xn-1|?|N?

n?2?

?2?

?2,則當(dāng)n?N時(shí),就有|xn-1|??.?????

n??

n??

2.設(shè)liman?l,證明lim|an|?|l|.證???0,?N,使得當(dāng)n?N時(shí),|an?l|??,此時(shí)||an|?|l||?|an?l|??,故lim|an|?|l|.n??

3.設(shè){an}有極限l,證明

(1)存在一個(gè)自然數(shù)N,n?N|an|?|l|?1;

(2){an}是一個(gè)有界數(shù)列,即存在一個(gè)常數(shù)M,使得|an|?M(n?12,?).證(1)對(duì)于??1,?N,使得當(dāng)n?N時(shí),|an?l|?1,此時(shí)|an|?|an?l?l|?|an?l|?|l|?|l|?1.(2)令M?max{|l|?1,|a1|,?,|aN|},則|an|?M(n?12,?).4.用?-N說法證明下列各極限式:

(1)lim

n??

3n?12n?3

?

;(2)lim

n??

n?1

?0;

(3)limnq?0(|q|?1);(4)lim

n??

n??

2n

n!n

n

?0;

?1?11(5)lim???????1;n??1?22?3(n?1)?n????11(6)lim?????0.3/23/2?n??(n?1)(2n)??證(1)??>0,不妨設(shè)?<1,要使

3n?12n?3

?32?

112(2n?3)

??,只需n?

112?

?3,取N?

3n?133n?13?11?

?3,當(dāng)n?N時(shí),???,故lim?.?2??n??2n?32n?322??

(2)??>0,要使

??,由于

?

只需

??,n?

?

3,?1

取N?

??3?(3)|q|?|nq|?

n

?,當(dāng)

n?N時(shí)??1

??.1??n

(??0).n?4

?

1?n???124n?

n

n(n?1)

(1??)6n

n

??

n(n?1)(n?2)

?????

?}.??

3n

?

(n?1)(n?2)?n!n

n

??,n??1????.??

??,N?max{4,?24???3?

(4)?

1n

??,n?

?,N?

?1?11(5)???????1

(n?1)?n??1?22?3

??11??11??11??11?1

?????????????????1???,n?,N?

?n????(n?1)n????12??23?

?

.??

(6)

1(n?1)

n??

3/2

???

1(2n)

3/2

?

n(n?1)

3/2

?

??,n?

?,N?

?1??2??.??

5.設(shè)liman?0,{bn}是有界數(shù)列,即存在常數(shù)M,使得|bn|?M(n?1,2,?),證

明limanbn?0.n??

證???0,?正整數(shù) N,使得

|an|?故limanbn?0.n??

?

M,|anbn|?|an||bn|?

?

M

?M??,6.證明lim

n??

?1.證???0,要使1|n(1??)

n

1??,只需

n(1??)

n

?1.4n?

而?

1?n??

nn(n?1)

?

?

(n?1)?

?

4n?,只需?1,n?

?,N?

?4

??2??.??

7.求下列各極限的值:(1)limn??

?lim

n??

?0.22

(2)lim

n??

n?3n?1004n?n?2(2n?10)n?n

?lim

n??

1?3/n?100/n4?1/n?2/n

?

.(3)lim

n??

?lim

n??

(2?10/n)1?1/n

n

?16.?2

1??

(4)lim?1??

n??n??

?2n

?1???

??lim?1???

n??n??????

?e.?2

1?1?

(5)lim?1???limn?1

n??n??n??1??1??

1?1?????

n?1??n?1???

1??

lim?1??n??n?1??1??

(6)lim?1??

n??n??

n

n

n

n?1

?

1??

lim?1??n??n?1??

n

n

1e

.??1??11??

?lim??1???,取q?(,1),?N,當(dāng)n?N時(shí),?1???qn??n??en??????

??1??1??

1??0,即lim1???????n??nn????????

n

n

n

n

n

??1??nn

0???1????q,limq?0,lim

n??n??n??????

n

n

n

?0.1?1?1?1???

(7)lim?1?2??lim?1??lim?1???e?1.n??n??n?n?n???n?e??

8.利用單調(diào)有界序列有極限證明下列序列極限的存在性:(1)xn?xn?1?(2)xn?

11?11?212?1

???

1n,xn?1?xn??2?

12?1

n

1(n?1)

?xn,???

1(n?1)n1

1n

?2.xn單調(diào)增加有上界，故有極限.,xn?1?xn?

n?1

?

2?1

???

?1

?xn,1?n

1111?111?1?1.xn??2???n??1??2???n?1??2222?222?21?1

2xn單調(diào)增加有上界，故有極限.(3)xn?

1n?1

?

1n?2

???

1n?n

.xn?1?xn?

12n?2

?

1n?1

??

12n?2

?0,xn?1?xn,xn?0,xn單調(diào)減少有下界，故有極限.(4)xn?1?1?

12!???

1n!

.xn?1?xn?

1(n?1)!

?0,1??11?1?1??1

xn?2??1?????????????3??3.2??23?n??n?1n?xn單調(diào)增加有上界，故有極限.11??

9.證明e=lim?1?1?????.n??2!n!??

1?1n(n?1)1n(n?1)?(n?k?1)1?

證?1???1?n?2????k

nn2!nk!n????

n(n?1)?(n?n?1)1

n!

n

n

n

?2?

1?1?1?1??k?1?1?1??n?1?1??1??1??1??1???????????2!?n?k!?n??n?n!?n??n?1

n

1?11????1?1????.e?lim?1???lim?1?1?????.n??n??2!n!n?2!n!???對(duì)于固定的正整數(shù)k,由上式,當(dāng)n?k時(shí),1?1?1?1?1??k?1??1??2?1??1??1?????????,n?2!?n?k!?n??n??

11??

令n??得e??1?1?????,2!k!??

11?11???

e?lim?1?1?????lim1?1?????n????.k??2!k!2!n!????

10.設(shè)滿足下列條件:|xn?1|?k|xn|,n?1,2,?,其中是小于1的正數(shù).證明limxn?0.n??

n

n?1

證由|xn?1|?k|xn|?k|xn?1|??k|x1|?0(n??),得limxn?0.n??

第五篇：北大版高等數(shù)學(xué)第一章 函數(shù)及極限答案習(xí)題1.5

習(xí)題1.5 1.試用???說法證明(1)1?x在x?0連續(xù)(2)sin5x在任意一點(diǎn)x?a連續(xù).證(1)???0,要使|x??,|x|?221?x?21?0|?2x22??.由于22x22?x,只需221?x?11?x?11?0|??,故1?x在x?0連續(xù).5(x?a)2|??.?,取???,則當(dāng)|x|??時(shí)有|1?x?5x?5a2||sin(2)(1)???0,要使|sin5x?sin5a|?2|cos由于2|cos取??5x?5a2||sin5(x?a)2|?5|x?a|,只需5|x?a|??,|x?a|??5,?5,則當(dāng)|x?a|??時(shí)有|sin5x?sin5a|??,故sin5x在任意一點(diǎn)x?a連續(xù).2.設(shè)y?f(x)在x0處連續(xù)且f(x0)?0,證明存在??0使得當(dāng)|x?x0|??時(shí)f(x)?0.證由于f(x)在x0處連續(xù),對(duì)于??f(x0)/2,存在存在??0使得當(dāng)|x?x0|??時(shí)f(x)?f(x0)|?f(x0)/2, 于是f(x)?f(x0)?f(x0)/2?f(x0)/2?0.3.設(shè)f(x)在(a,b)上連續(xù),證明|f(x)|在(a,b)上也連續(xù),并且問其逆命題是否成立?證任取 x0?(a,b),f在x0連續(xù).任給??0,存在??0使得當(dāng)|x?x0|??時(shí)|f(x)?f(x0)|??,此時(shí)||f(x)|?|f(x0)||?|f(x)?f(x0)|??,故|f|在x0連續(xù).其逆命題?1,x是有理數(shù)不真,例如f(x)??處處不連續(xù),但是|f(x)|?1處處連續(xù).??1,x是無理數(shù)4.適當(dāng)?shù)剡x取a,使下列函數(shù)處處連續(xù): 2??ln(1?x), x?1,?1?x,x?0,(1)f(x)??(2)f(x)???aarccos?x,x?1.??a?x x?0;解(1)limf(x)?limx?0?x?0?x?1?x?1?1?x2?1?f(0),limf(x)?f(0)?a?1.x?0?x?1?x?1?(2)limf(x)?limln(1?x)?ln2?f(1),limf(x)?limaarccos?x??a?f(1)?ln2,a??ln2.5.利用初等函數(shù)的連續(xù)性及定理3求下列極限:(1)limcosx???1?x?x?22x?coslimx???1?x?xx?cos0?1.(2)limxx?2x.sin2xsin3x2sin2x(3)limex?0sin3x?elimx?0?e3.?arctanlimx??(4)limarctanx??x?8x?124x?8x?124?arctan1??4.1(5)limx??(x?1?3|x|x?1?22x?2)|x|????2x?2?x?x02lim?(x???x?1?22x?2)|x|????lim?x???x?x0??3lim???22x???1?1/x?1?2/x?g(x)32.6.設(shè)limf(x)?a?0,limg(x)?b,證明lim)f(x)x?x0lim[(lnf(x))g(x)]?a.?a.bb證lim)f(x)x?x0g(x)?lim)ex?x0(lnf(x))g(x)?ex?x0?eblna7.指出下列函數(shù)的間斷點(diǎn)及其類型,若是可去間斷點(diǎn),請(qǐng)修改函數(shù)在該點(diǎn)的函數(shù)值,使之稱為連續(xù)函數(shù):(1)f(x)?cos?(x?[x]),間斷點(diǎn)n?Z,第一類間斷點(diǎn).(2)f(x)?sgn(sinx),間斷點(diǎn)n?,n?Z,第一類間斷點(diǎn).?x,x?1,(3)f(x)??間斷點(diǎn)x?1,第一類間斷點(diǎn).?1/2,x?1.?x?1,0?x?1?(4)f(x)??間斷點(diǎn)x?1,第二類間斷點(diǎn).?,1?x?2,?sinx?1??1,0?x?1,?2?x?(5)f(x)??x,1?x?2,間斷點(diǎn)x?2,第一類間斷點(diǎn).?1?,2?x?3.?1?x22

8.設(shè)y?f(x)在R上是連續(xù)函數(shù),而y?g(x)在R上有定義,但在一點(diǎn)x0處間斷.問函數(shù)h(x)?f(x)?g(x)及?(x)?f(x)g(x)在x0點(diǎn)是否一定間斷?解h(x)?f(x)?g(x)在x0點(diǎn)一定間斷.因?yàn)槿绻趚0點(diǎn)連續(xù),g(x)?(f(x)?g(x))?f(x)將在x0點(diǎn)連續(xù),矛盾.而?(x)?f(x)g(x)在x0點(diǎn)未必間斷.例如f(x)?0,g(x)?D(x).