數(shù)學(xué)分析上冊練習(xí)題及答案第四,五章

1、一元函數(shù)的連續(xù)性一元函數(shù)的連續(xù)性第四章第四章函數(shù)的連續(xù)性函數(shù)的連續(xù)性1連續(xù)性概念連續(xù)性概念1.按定義證明下列函數(shù)在其定義域內(nèi)連續(xù):(1)(2).xxf1)(?xxf?)(證明(1)的定義域是且取由函數(shù)極限四則運(yùn)算可知xxf1)(?Rx?0?x00?x所以在連續(xù).由在定義域內(nèi)的任意性知)(11lim)(lim0000xfxxxfxxxx?????)(xf0x0x在其定義域內(nèi)連續(xù).)(xf(2)的定義域是任取由于所以對任給的取)(xfRx?

2、Rx?000xxxx???0??使得當(dāng)時(shí)有.??????0xx???????000)()(xxxxxfxf按函數(shù)在一點(diǎn)連續(xù)的定義在連續(xù)由在中的任意性知在定義域???)(xf0x0xR)(xf內(nèi)連續(xù).R2.指出下列函數(shù)的間斷點(diǎn)并說明其類型:(1)(2)(3)(4)xxxf1)(??xxxfsin)(?]cos[)(xxf?xxfsgn)(?(5)(6))sgn(cos)(xxf??????)(為無理數(shù)為有理數(shù)xxxxxf(7)??????

3、?????????????????.111sin)1(17771)(xxxxxxxxf解(1)因僅在處無定義故為函數(shù)的間斷點(diǎn)又因)(xf0?x0?x?????)(lim0xfx所以為第二類間斷點(diǎn).?????)(lim0xfx0?x(2)因僅在處無定義故為函數(shù)的間斷點(diǎn)又因)(xf0?x0?x所以是的1sinlimsinlim)(lim1sinlim)(lim00000??????????????????xxxxxfxxxfxxxxx0?x

4、)(xf第一類間斷點(diǎn)且為跳躍間斷點(diǎn).為的可去間斷點(diǎn)令則為在上的延拓2?x)(xf2122)()(??????xxxfxF)(xF)(xfRx?且在上連續(xù).)(????(2)在時(shí)無定義)(xf0?x而所以為該函數(shù)的可2122sin21lim2sin2limcos1lim)(lim0220200??????????????????????xxxxxxxfxxxx0?x去間斷點(diǎn).令則為在上的延拓且在上連0210)()(????????xxxf

5、xF)(xF)(xfRx?)(????續(xù).(3)在時(shí)無定義)(xf0?x而所以為該函數(shù)的可去間斷點(diǎn).01coslim)(lim00?????xxxfxx0?x令則為在上的延拓.000)()(??????xxxfxF)(xF)(xfRx?4.證明:若在點(diǎn)連續(xù)則與也在點(diǎn)連續(xù).又問:若或在上連續(xù)那么f0xf2f0xf2fI在上是否必連續(xù)fI分析將和與的不等式關(guān)系找出從而利用)()(0xfxf?)()(022xfxf?)()(0xfxf?極限定

6、義求證其連續(xù)即運(yùn)用極限理論討論可得結(jié)論.證明(1)因?yàn)樵邳c(diǎn)連續(xù)所以則根據(jù)極限的定義對任給的)(xf0x)()(lim00xfxfxx?????存在使得當(dāng)時(shí)有.0??0?????0xx???)()(0xfxf又因所以當(dāng)時(shí)也有.所以即)()()()(00xfxfxfxf??????0xx)()(lim00xfxfxx??可知在點(diǎn)連續(xù).f0x(2)因在連續(xù)即所以由函數(shù)極限的局部有界性知存在)(xf0x)()(lim00xfxfxx??0?M使