1. 解: (1)相等.因为两函数的定义域相同,都是实数集R ;x =知两函数的对应法则也相同;所以两函数相等.(2)相等.因为两函数的定义域相同,都是实数集R ,由已知函数关系式显然可得两函数的对应法则也相同,所以两函数相等.(3)不相等. 因为函数()f x 的定义域是{,1}x x x ∈≠R ,而函数()g x 的定义域是实数集R ,两函数的定义域不同,所以两函数不相等.2. 解: (1)要使函数有意义,必须400x x -≥⎧⎨≠⎩ 即40x x ≤⎧⎨≠⎩所以函数的定义域是(,0)(0,4]-∞.(2)要使函数有意义,必须30lg(1)010x x x +≥⎧⎪-≠⎨⎪->⎩ 即301x x x ≥-⎧⎪≠⎨⎪<⎩所以函数的定义域是[-3,0)∪(0,1).(3)要使函数有意义,必须210x -≠ 即 1x ≠±所以函数的定义域是(,1)(1,1)(1,)-∞--+∞.(4)要使函数有意义,必须12sin 1x -≤≤ 即 11sin 22x -≤≤即ππ2π2π66k x k -+≤≤+或5π7π2π2π66k x k +≤≤+,(k 为整数).也即ππππ66k x k -+≤≤+ (k 为整数).所以函数的定义域是ππ[π,π]66k k -++, k 为整数.3.解: 由已知显然有函数的定义域为(-∞,+∞),又当0x ≠时,1x 可以是不为零的任意实数,此时,1sinx 可以取遍[-1,1]上所有的值,所以函数的值域为[-1,1]. 4. 解: 10(0)110f -==+,1()1(),1()1x x f x x x --+-==+--1111().111x x f x x x --==++ 5.解: 1,1101,01(1).(1)1,012,13x x f x x x x x -≤-<≤<⎧⎧-==⎨⎨-+≤-≤≤≤⎩⎩6.解: ()ln (())22,g x x x f g x ==(())()ln ()2ln 2(ln 2)2,x x x g f x f x f x x ==⋅=⋅()2(())22,(())()ln ()ln ln(ln ).xf x f f xg g x g x g x x x x x ====7. 证:由321y x =-解得x =故函数3()21f x x =-的反函数是)y x =∈R ,这与()g x =数,所以3()21f x x =-和()g x =.8. 解: (1)由11x y x -=+解得11y x y -=+, 所以函数11x y x -=+的反函数为1(1)1x y x x -=≠-+.(2)由ln(2)1y x =++得1e 2y x -=-,所以,函数ln(2)1y x =++的反函数为1e 2()x y x -=-∈ R . (3)由253x y +=解得31(log 5)2x y =-所以,函数253x y +=的反函数为31(log 5)(0)2y x x =-> .(4)由31cos y x =+得cos x =又[0,π]x ∈,故x =又由1cos 1x -≤≤得301cos 2x ≤+≤,即02y ≤≤,故可得反函数的定义域为[0,2],所以,函数31cos ,[0,π]y x x =+∈的反函数为(02)y x =≤≤.9. 解: (1)()()f x f x -===()f x ∴=.(2)222222()e e sin()e e sin (e e sin )()x x x x x x f x x x x f x ----=-+-=-+=--+=-∴函数22e e sin x xy x -=-+是奇函数.10. 解: (1)函数的定义域为(-∞,+∞), 当0x ≤时,有201xx ≤+,当0x >时,有21122x x x x ≤=+,故(,),x ∀∈-∞+∞有12y ≤.即函数21x y x =+有上界. 又因为函数21x y x =+为奇函数,所以函数的图形关于原点对称,由对称性及函数有上界知,函数必有下界,因而函数21x y x =+有界.又由1212121222221212()(1)11(1)(1)x x x x x x y y x x x x ---=-=++++知,当12x x >且121x x <时,12y y >,而 当12x x >且121x x >时,12y y <.故函数21xy x =+在定义域内不单调.(2)函数的定义域为(0,+∞),10,0M x ∀>∃>且12;e 0M x M x >∃>>,使2ln x M >. 取012max{,}x x x =,则有0012ln ln 2x x x x M M +>+>>,所以函数ln y x x =+在定义域内是无界的. 又当120x x <<时,有12120,ln ln 0x x x x -<-<故1211221212(ln )(ln )()(ln ln )0y y x x x x x x x x -=+-+=-+-<. 即当120x x <<时,恒有12y y <,所以函数ln y x x =+在(0,)+∞内单调递增.11. 解: (1)124(1)y x =+是由124,1y u u x ==+复合而成. (2)2sin (12)y x =+是由2,sin ,12y u u v v x ===+复合而成.(3)512(110)x y -=+是由152,1,10,w y u u v v w x ==+==-复合而成.(4)11arcsin 2y x =+是由1,1,arcsin ,2y u u v v w w x -==+==复合而成. 12.证: (1)设()()()F x f x f x =+-,则(,)x ∀∈-∞+∞, 有()()()()F x f x f x F x -=-+= 故()()f x f x +-为偶函数.(2)设()()(),G x f x f x =--则(,)x ∀∈-∞+∞,有()()()[()()]()G x f x f x f x f x G x -=---=---=- 故()()f x f x --为奇函数.13.解: 设年销售批数为x , 则准备费为103x ;又每批有产品610x 件,库存数为6102x 件,库存费为6100.052x ⨯元.设总费用为,则63100.05102y x x ⨯=+. 14. 解: 当x 能被20整除,即[]2020x x =时,邮资0.802025x x y =⨯=; 当x 不能被20整除时,即[]2020x x ≠时,由题意知邮资0.80120x y ⎡⎤=⨯+⎢⎥⎣⎦.综上所述有,02000;2520200.80,02000.1202020x xx x y x x x x ⎧⎡⎤<≤=⎪⎢⎥⎪⎣⎦=⎨⎡⎤⎡⎤⎪⨯<≤≠+⎢⎥⎢⎥⎪⎣⎦⎣⎦⎩且且其中20x ⎡⎤⎢⎥⎣⎦,120x ⎡⎤+⎢⎥⎣⎦分别表示不超过20x ,120x +的最大整数.15. 证: (1)由e e sinh 2x xy x --==得2e 2e 10x xy --= 解方程2e 2e 10x xy --=得e x y =因为e 0x>,所以e x y =ln(x y = 所以sinh y x =的反函数是arcsin h ln(().y x x x ==+-∞<<+∞(2)由e e tanh e e x x x xy x ---==+得21e 1x y y +=-,得1112ln ,ln 121y yx x y y ++==--; 又由101yy +>-得11y -<<,所以函数tanh y x =的反函数为11arctan h ln (11).21xy x x x +==-<<-16. 解: 011()(2cot )(cot )22S h AD BC h h BC BC h BC h ϕϕ=+=++=+ 从而 0cot SBC h h ϕ=-.000()22cot sin sin 2cos 2cos 40sin sin 40L AB BC CD AB CD S h hBC h hS S h h h h ϕϕϕϕϕ=++==+=+---=+=+由00,cot 0S h BC h h ϕ>=->得定义域为. 17. 解:1(1),1n n x n -=+当n →∞时,1n x →. 1(2)cos π2n n x n -=,当n 无限增大时,有三种变化趋势:趋向于+∞,趋向于0,趋向于-∞.21(3)(1)21nn n x n +=--,当n 无限增大时,变化趁势有两种,分别趋于1,-1.18. 解: (1)lim 0n n a x →∞==,0ε∀>,要使11π0sin 2n n x n n ε-=<<,只须1n ε>.取1N ε⎡⎤=⎢⎥⎣⎦,则当n N >时,必有0n x ε-<. 当0.001ε=时,110000.001N ⎡⎤==⎢⎥⎣⎦或大于1000的整数.(2)lim 0n n a x →∞==,0ε∀>,要使0n x ε-==<=<1ε>即21n ε>即可.取21N ε⎡⎤=⎢⎥⎣⎦,则当n N >时,有0n x ε-<. 当0.0001ε=时, 821100.0001N ⎡⎤==⎢⎥⎣⎦或大于108的整数.19. 证: (1)0ε∀>,要使22110n n ε=<-,只要n >取N =,则当n>N 时,恒有210n ε<-.故21lim 0n n →∞=.(2) 0ε∀>,要使555313,2(21)4212n n n n n ε-=<<<-++只要5n ε>,取5N ε⎡⎤=⎢⎥⎣⎦,则当n>N 时,恒有313212n n ε-<-+.故313lim 212n n n →∞-=+. (3) 0ε∀>,要使2221a n ε=<<,只要n >,取n =,则当n>N 时,1ε<,从而lim 1n n →∞=.(4)因为对于所有的正整数n ,有10.99991n <-个,故0ε∀>,不防设1ε<,要使1,0.999110n n ε=<-个只要ln ,ln10n ε->取ln ,ln10N ε-⎡⎤=⎢⎥⎣⎦则当n N >时,恒有,0.9991n ε<-个故lim 0.9991n n →∞=个.20.证:lim 0n n x →∞=,由极限的定义知,0,0N ε∀>∃>,当n N >时,恒有nx a ε-<.而n n x x a a ε-<-<0,0N ε∴∀>∃>,当n N >时,恒有n x a ε-<,由极限的定义知lim .n n x a →∞=但这个结论的逆不成立.如(1),lim 1,n n n n x x →∞=-=但lim nn x →∞不存在.21. 解:1111(1)0(1)(1)1(1)1k k k kk k n n n n n n n -⎡⎤⎡⎤<+-=<=+-+-⎢⎥⎢⎥⎣⎦⎣⎦而lim 00n →∞=,当1k <时,11lim0k n n -→∞=lim[(1)]0k k n n n →∞∴+-=.(2)记12max{,,,}m a a a a =则有n <<即1na m a <⋅而 1lim , lim ,nn n a a m a a →∞→∞=⋅=故n a=即12lim max{,,,}m n a a a =.(3)111(3)(123)(33)n nn n nn n<++<⋅即 113(123)3n nn nn+<++< 而 1lim33,lim33n nn n +→∞→∞==故1lim(123)3nn nn →∞++=.(4)11111n n <+<+而 1lim10,lim(1)1n n n →∞→∞=+=故1n=.22. 证: (1)122x =<,不妨设2k x <,则12k x +<=.故对所有正整数n 有2n x <,即数列{}n x 有上界.又1n n nx xx +-=0>,又由2n x <从而10n n x x +->即1n nx x +>,即数列{}n x 是单调递增的.由极限的单调有界准则知,数列{}n x 有极限.设lim n n x a→∞=,则a =于是22a a =,2,0a a ==(不合题意,舍去),lim 2n n x →∞∴=.(2) 因为110x =>,且111nn n xx x +=++, 所以02n x <<, 即数列有界又111111111(1)(1)n n n n n n n n n n x x x x x x x x x x --+---⎛⎫⎛⎫++-=-=⎪ ⎪++++⎝⎭⎝⎭ 由110,10n n x x -+>+>知1n n x x +-与1n n x x --同号,从而可推得1n n x x +-与21x x -同号, 而1221131,1,022x x x x ==+=->故10n n x x +->, 即1n n x x +>所以数列{}n x 单调递增,由单调有界准则知,{}n x 的极限存在. 设lim n n x a→∞=, 则11a a a =++,解得a a ==(不合题意,舍去).所以1lim 2n n x →∞+=23. 证:(1)0ε∀>,要使1sin sin 0x xx x x ε=≤<-,只须1x ε>,取1X ε>,则当x X >时,必有sin 0xx ε<-,故sin limx xx →+∞=. (2)0ε∀>,要使 22221313313||44x x x x ε-=<<-++,只须x >取X =X x >时,必有223134x x ε-<-+,故2231lim 34x x x →∞-=+. (3) 0ε∀>,要使24(4)22x x x ε-=<--++,只要取δε=,则当02x δ<<+时,必有24(4)2x x ε-<--+,故224lim 42x x x →--=-+. (4) 0ε∀>,要使21142221221x x x x ε-==<+-++,只须122x ε<+,取2εδ=,则 当102x δ<<+时,必有214221x x ε-<-+ 故21214lim 221x x x →--=+.(5) 0ε∀>,要使11sin0sin x x x x x ε=≤<-,只要取δε=,则 当00x δ<<-时,必有1sin0x x ε<-,故 01lim sinx x x →=.24. 解:()()2232233lim 33933(1)lim 1lim 9151x x x x x x x →→→---===+++.2221424242112222333422424lim()11(2)lim 2.31lim(31)13111111(3)limlim .1121221111lim (4)lim lim 0.3131311lim 1(5x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x →→→→∞→∞→∞→∞→∞→∞+++===--+-+-⨯+--==----⎛⎫-- ⎪-⎝⎭===-+⎛⎫-+-+ ⎪⎝⎭222222121lim 21)lim lim 01111lim 1x x x x x x x x x x x x →∞→∞→∞→∞⎛⎫++ ⎪+⎝⎭===+⎛⎫++ ⎪⎝⎭由无穷大与无穷小的关系知, 21lim 21x x x →∞+=∞+.3(1)(2)(3)1123(6)limlim 1115511123lim lim lim .11155n n n n n n n n n n n n n n n →∞→∞→∞→∞→∞+++⎛⎫⎛⎫⎛⎫=+++ ⎪⎪⎪⎝⎭⎝⎭⎝⎭⎛⎫⎛⎫⎛⎫=⋅⋅=+++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭(7)因为221(1)()(1)11x a x a b x b ax b x x +--++---=++ 由已知211lim 21x x ax b x →∞⎛⎫+=-- ⎪+⎝⎭知,分式的分子与分母的次数相同,且x 项的系数之比为12,于是10a -= 且 ()112a b -+=解得31,2a b ==-. 25.解:22123(1)(1)111(1)limlim lim .1222n n n n n n n n n →∞→∞→∞++++--⎛⎫===- ⎪⎝⎭1221112244411112(2)lim lim 2.11221221(1)(3)limlim lim(1)0.1168(2)(4)22(4)lim lim lim .54(1)(4)13n n n n x x xx x x x xx x x x x xx x x x x xx x +→∞→∞→→→→→→⎛⎫- ⎪⎛⎫⎝⎭==+++ ⎪⎝⎭--+-==-=---+---===-+---322000(5)lim lim lim2.lim(1 2.x x x x x x x →+∞→→→=====-+=-5555x x x x →→→→=====3333ππ4422π422π41cot 1cot (8)lim lim 2cot cot (1cot )(1cot )(1cot )(1cot cot )lim (1cot )(11cot cot )1cot cot 3lim .2cot cot 4x x x x x xx x x x x x x x x x x x x x →→→→--=---+--++=-+++++==++122222(9)lim(1)(1)(1)(1)(1)(1)(1)(1)lim 111lim .11nnn x x x x x x x x x x x xx +→∞→∞→∞+++<-+++=--==-111211211(1)(1)(10)lim(1))(1))(1)11.234!n n x n n n n x n n n n x n x x x x x x x x n n -→--→-→--=++++=++++==⨯⨯⨯⨯ 22223111221113213(11)lim lim lim (1)(1)(1)(1)11(1)(2)(2)lim lim 1.(1)(1)1x x x x x x x x x x x x x x x x x x x x x x x x x →→→→→++-+-⎛⎫==- ⎪-++-++--⎝⎭-+-+===--++++2212211221lim(1)(1)(12)lim 01lim(1)1lim .(1)x x x x x x x x x x x x x →→→→--==-+-+-+∴=∞-1log (1)(13)log (1)a xa x x x +=+而10lim(1).xx x e →+= 而1limlog log ln a a u eu e a →==0log (1)1lim.ln a x x x a →+∴=(14)令1,xu a =-则log (1),a x u =+当0x →时,0u →. 所以00011lim lim ln log (1)log (1)limx x u a a u a u a u x u u →→→-===++(利用(13)题的结果).1122000336ln(12)ln(12)sin sin 2sin 0lim 6ln(12)6limlimln(12)sin sin 61ln e 6(15)lim(12)lim elim ee ee e .x xx x x xx x xxx xx x x xxx x xx x →→→++→→→⋅⋅+⋅⋅+⨯⨯+======(16)令sin x u x =, 则00sin lim lim 1x x xu x →→==而1limln 0u u →= 所以0sin limln 0.x x x →=26. 解:232200lim lim 022x x x x x x x x x →→--==--∴当0x →时,23x x -是比22x x -高阶的无穷小量.27.解:211111(1)lim lim 112x x x x x →→-==-+ ∴当1x →时,1x -是与21x -同阶的无穷小.2111(1)12(2)lim lim 112x x x xx →→-+==-∴当1x →时,1x -是与21(1)2x -等价的无穷小. 28. 解:(1)因为当0x →时,sin ~,sin ~,mx mx nx nx所以00sin limlim .sin x x mx mx mnx nx n →→== 000002000limcos cos (2)lim cot lim cos lim 1.sin sin sin lim 1cos 22sin sin (3)lim lim 2lim 2.sin sin x x x x x x x x x x x x x x x xx x xx x x x x x x x →→→→→→→→=⋅===-===(4)因为当0x →时,2221ln(1e sin)~e sin 1~2x x x x x+,所以22200002e sin sin lim lim 2e lim 2.12x x x x x x x x x x x →→→→⎛⎫==⋅= ⎪⎝⎭(5)因为当0x →时,arctan3~3,x x 所以00arctan 33limlim 3x x x xx x →→==. sin sin 22(6)lim 2sin lim lim .222n nn n n n n n n x x x x x x x x →∞→∞→∞=⋅==(7)因为当12x →时,arcsin(12)~12x x --,所以22111122224141(21)(21)lim lim lim lim(21) 2.arcsin(12)1212x x x x x x x x x x x x →→→→---+===-+=----(8)因为当0x →时,22arctan ~,sin~,arcsin ~,22x xx x x x 所以2200arctan lim lim 2sin arcsin 22x x x x xx x x →→==⋅.(9)因为当0x →时,2331sin ~,1cos ~,sin ~2x x x x x x -,所以233300001tan sin sin (1cos )2lim lim lim sin sin cos cos 11lim .2cos 2x x x x x x x x x x x x xx x x →→→→⋅--==⋅== (10)因为当0x →时,sin ~,sin ~2222x x x xαβαβαβαβ++--,所以220020222sin sin cos cos 22lim lim 222lim 1().2x x x x xx x x x x xxαβαβαβαβαββα→→→+---=+--⋅⋅==- (11)因为当0x →时,arcsin ~)~,x x --所以000 1.x x x →→→==-=-(12)因为当0x →时,sin ~,sin 2~2,x x x x 所以2222200222200201cos 42sin 2lim lim 2sin tan sin (2sec )2(2)8lim lim (2sec )2sec 84.lim(2sec )x x x x x x x x x x x x x x x x x x xx x →→→→→-=++⋅==++==+(13)因为ln cos ln[1(cos 1)],ln cos ln[1(cos 1)],ax ax bx bx =+-=+- 而当0x →时,cos 10,cos 10ax bx -→-→故 ln[1(cos 1)]~cos 1,ln[1(cos 1)]~cos 1,ax ax bx bx +--+--又当x →0进,2222111cos ~,1cos ~,22ax a x bx b x --所以22220000221ln cos cos 11cos 2lim lim lim lim .1ln cos cos 11cos 2x x x x a xax ax ax a bx bx bx b b x→→→→--====--(14)因为当0x →时,222sin 0,0e e x xx x→→故 222222sin sin ln ~,ln ~,11e e e e xx x x x x x x ⎛⎫⎛⎫++ ⎪ ⎪⎝⎭⎝⎭ 所以22222222200022222000020sin ln 1ln(sin e )ln(sin e )ln e e lim lim lim ln(e )2ln(e )ln e ln 1e sin sin sin e lim lim e lim e lim e e 1 1.x x x x x x x x x x x x x xx x x x xx x x x x x x x xx x x x x →→→→→→→⎛⎫+ ⎪+-+-⎝⎭==+-+-⎛⎫+ ⎪⎝⎭⎛⎫⎛⎫==⋅=⋅ ⎪ ⎪⎝⎭⎝⎭=⋅=29.解:1112222111(1)lim lim e 1lim 11xxxx x x x x x →∞→∞→∞⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫====+++ ⎪⎢⎥⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎣⎦⎣⎦1022121553555(2)lim lim lim 1112222x x x x x x x x x x x -++→∞→∞→∞⎡⎤+⎛⎫⎛⎫⎛⎫⎛⎫==⋅++⎢⎥ ⎪ ⎪ ⎪+ ⎪---⎝⎭⎝⎭⎝⎭⎢⎥-⎝⎭⎣⎦102551051055lim e 1e .1lim 122x x x x x -→∞→∞⎡⎤⎡⎤⎛⎫⎛⎫=⋅=⋅=+⎢⎥ ⎪+⎢⎥ ⎪-⎝⎭⎣⎦⎢⎥-⎝⎭⎣⎦22233112cot323tan 23tan 000(3)lim(13tan )lim e .lim(13tan )(13tan )xx x x x x x x x →→→⎡⎤⎡⎤+===+⎢⎥+⎢⎥⎣⎦⎣⎦ [][][]cos 211cos 212221cos 2121cos 2120220333ln ln cos21(cos21)03(cos21)ln 1(cos21)0cos213limlim ln 1(cos21)2sin 3limln lim (4)lim(cos 2)lim elim elim ee e x x x x x x x x xx x x x x x x x x x x x x x x x x ----→→→→⎧⎫⎪⎪⎨⎬+-⎪⎪⎩⎭→→→-+-→-⋅+--⋅=====[]1cos 212201(cos21)sin 6ln e lim 6116ee e .x x x x x -→⎧⎫⎪⎪⎨⎬+-⎪⎪⎩⎭⎛⎫-⋅⋅ ⎪-⨯⨯-⎝⎭===22222(5)lim [ln(2)ln ]lim 2ln lim 2ln 12222lim ln 2ln 1lim 12ln e 2.x x x x xxx x x x x x x x x x x →∞→∞→∞→∞→∞+⎛⎫+-=⋅⋅=+ ⎪⎝⎭⎛⎫⎛⎫⎛⎫==⋅+ ⎪ ⎪+ ⎪ ⎪⎝⎭⎝⎭⎝⎭==(6)令1x t =+,则当1x →时,0t →.1110001111lim lim 1.ln ln(1)ln e ln lim ln(1)lim(1)x t tt t t x t x t t t →→→→-=-=-=-=-=-+⎡⎤++⎢⎥⎣⎦30. 解:(1)令1(e )x xy x =+,则1ln ln(e )x y x x =+于是:()0000ln e ln 111e lim ln lim ln lim ln e lim1e e x x x x x x x x x x x y x x x x →→→→⎛⎫++ ⎪⎛⎫⎝⎭===++ ⎪⎝⎭e 0001e 1lim 1lim lim ln 1ln 11e e e e 11ln e 2xx xx x x x x x x x x x →→→⎡⎤⎛⎫⎛⎫==+⋅+⋅++ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦=+⋅=即()lim ln 2x y →= 即2lim ex y →= 即()120lim e e xxx x →=+.(2)令13xxxxa b c y ⎛⎫++= ⎪⎝⎭,则1ln ln 3x x xa b c y x ++=于是00333303300001lim(ln )lim ln 313lim ln 1333lim lim ln 1331111lim ln lim 13x x x x x x xxx x x xx x a b c x x x a b c x xxxxxxa b c x x x x x x x x x x a b c y x a b c x a b c a b c x a b c a b c x x x →→++-++-→++-→→→→++=⎡⎤⎛⎫++-=⎢⎥+ ⎪⎢⎥⎝⎭⎣⎦++-⎛⎫++-=⋅+ ⎪⎝⎭⎛⎫---++=⋅++ ⎪+⎝⎭33331(ln ln ln )ln e ln 3x x x a b c a b c ++-⎡⎤⎛⎫-⎢⎥ ⎪⎢⎥⎝⎭⎣⎦=++⋅=即0lim(ln )x y →= 即()lim ln ln x y →=故0lim x y →=即1lim 3x x xxx a b c →⎛⎫++= ⎪⎝⎭.(3)令11sin cos xy x x ⎛⎫=+ ⎪⎝⎭,则11ln ln sin cos y x x x ⎛⎫=+ ⎪⎝⎭ 于是11sin cos 1111sin cos 1111sin cos 111lim ln lim ln 1sin cos 11111lim ln 1sin cos 1sin cos 111sin 1cos lim ln lim 11xx x x x x x xx x y x x x x x x x x x x x x ⎛⎫+- ⎪⎝⎭+-→∞→∞+-→∞→∞⎧⎫⎪⎪⎡⎤⎛⎫=⎨⎬++- ⎪⎢⎥⎝⎭⎪⎪⎣⎦⎩⎭⎡⎤⎛⎫⎛⎫=⋅++-+- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦⎛⎫- ⎪=-⋅ ⎪ ⎪⎝⎭111sin cos 1111sin cos 1x x x x x +-→∞⎧⎫⎪⎪⎡⎤⎛⎫⎨⎬++- ⎪⎢⎥⎝⎭⎪⎪⎣⎦⎩⎭2111sin 2ln e (10)ln e 1limlim 11x x x x x x →∞→∞⎛⎫⎛⎫ ⎪⎪⎝⎭=⋅=-⋅= ⎪- ⎪ ⎪⎝⎭即limln 1x y →∞= 从而()lim ln 1x y →∞= 故lim e x y →∞=即 11lim e sin cos xx x x →∞⎛⎫=+ ⎪⎝⎭.(4)令211xy x ⎛⎫=+ ⎪⎝⎭,则21ln ln 1y x x ⎛⎫=+ ⎪⎝⎭ 于是:22221222211lim(ln )lim ln lim ln 111111lim ln lim lim ln 110ln e 0x x x x x x x x x x y x x x x x x x x →∞→∞→∞→∞→∞→∞⎡⎤⎛⎫⎛⎫==+⎢⎥ ⎪+ ⎪⎝⎭⎝⎭⎣⎦⎛⎫⎛⎫==⋅++ ⎪ ⎪⎝⎭⎝⎭=⋅=即()lim lim(ln )0,ln 0x x y y →∞→∞==lim 1x y →∞∴= 即21lim 11xx x →∞⎛⎫=+ ⎪⎝⎭.31.解:000(1)lim ()lim lim 1,x x x x x f x x x +++→→→=== 000lim ()lim lim 1x x x x xf x x x ---→→→-===-因为 0lim ()lim ()x x f x f x +-→→≠所以0lim ()x f x →不存在.(2)22221lim ()lim ,lim ()lim(2)42x x x x f x f x x x ++--→→→→==+∞=+=-因为2lim ()x f x +→不存在,所以2lim ()x f x →不存在.32. 解:(1)由初等函数的连续性知,()f x 在(0,1),(1,2)内连续, 又21111lim ()lim(2)1,lim ()lim 1x x x x f x x f x x ++--→→→→=-===1lim ()1,x f x →∴= 而(1)1f =,()f x ∴在1x =处连续,又,由2lim ()lim 0(0)x x f x x f ++→→===,知()f x 在0x =处右连续,综上所述,函数()f x 在[0,2)内连续. 函数图形如下:图1-2(2) 由初等函数的连续性知()f x 在(,1),(1,1),(1,)-∞--+∞内连续,又由1111lim ()lim 11,lim ()lim 1,x x x x f x f x x --++→-→-→-→-====-知1lim ()x f x -→-不存在,于是()f x 在1x =-处不连续.又由1111lim ()lim 1,lim ()lim11,x x x x f x x f x --++→→→→====及(1)1f =知1lim ()(1)x f x f →=,从而()f x 在x =1处连续,综上所述,函数()f x 在(,1)-∞-及(1,)-+∞内连续,在1x =-处间断.函数图形如下:图1-3(3)∵当x <0时,221()lim lim 1,1x x x x x x n n n n n f x n n n --→∞→∞--===-++ 当x =0时,0000()lim 0,n n n f x n n →∞-==+当x >0时,2222111()lim lim lim 1111xxxx x x x n n n xn n n n f x n n n n --→∞→∞→∞---====+++1,0,()lim 0,0,1,0.x xx xn x n n f x x n n x --→∞-<⎧-⎪∴===⎨+⎪>⎩由初等函数的连续性知()f x 在(,0),(0,)-∞+∞内连续,又由 00lim ()lim11,lim ()lim(1)1x x x x f x f x ++--→→→→===-=-知0lim ()x f x →不存在,从而()f x 在0x =处间断.综上所述,函数()f x 在(,0),(0,)-∞+∞内连续,在0x =处间断.图形如下:图1-4(4)当|x |=1时,221()lim 0,1nnn x f x x x →∞-==+当|x |<1时,221()lim ,1nnn x f x x x x →∞-==+当|x |>1时,2222111()limlim 111nnn nn n x x f x x x x x x →∞→∞⎛⎫- ⎪-⎝⎭==⋅=-+⎛⎫+ ⎪⎝⎭即,1,()0,1,,1.x x f x x x x <⎧⎪==⎨⎪->⎩ 由初等函数的连续性知()f x 在(-∞,-1),(-1,1),(1,+∞)内均连续,又由1111lim ()lim ()1,lim ()lim 1x x x x f x x f x x --++→-→-→-→-=-===-知1lim ()x f x →-不存在,从而()f x 在1x =-处不连续.又由 1111lim ()lim()1,lim ()lim 1x x x x f x x f x x ++--→→→→=-=-==知1lim ()x f x →不存在,从而()f x 在1x =处不连续.综上所述,()f x 在(-∞,-1),(-1,1),(1,+∞)内连续,在1x =±处间断.图形如下:图1-533. 解:22111(1)(1)(1)lim lim 232(1)(2)x x x x x x x x x →→--+==--+--2221lim 32x x x x →-=∞-+1x ∴=是函数的可去间断点.因为函数在x =1处无定义,若补充定义(1)2f =-,则函数在x =1处连续;x =2是无穷间断点.π0π2(2)lim1,lim 0tan tan x x k x x x x→→+==当0k ≠时,πlimtan x k xx →=∞. π0,π,0,1,2,2x x k k ∴==+=±±为可去间断点,分别补充定义f (0)=1,π(π)02f k +=,可使函数在x =0,及ππ2x k =+处连续.(0,1,2,k =±±);π,0,1,2,x k k k =≠=±±为无穷间断点(3)∵当0x →时,21cosx 呈振荡无极限,∴x =0是函数的振荡间断点.(第二类间断点). (4)11lim lim(3) 2.x x y x ++→→=-=11lim lim(1)0x x y x --→→=-=∴x =1是函数的跳跃间断点.(第一类间断点.)34.解:0003(1)lim ()2x x x f x →→→===∴补充定义3(0),2f =可使函数在x =0处连续. 000tan 22(2)lim ()lim lim 2.x x x x xf x x x →→→===∴补充定义(0)2,f =可使函数在x =0处连续.01(3)limsin sinx x x →=∴补充定义(0)0,f =可使函数在x =0处连续. 10(4)lim ()lim(1)exx x f x x →→=+=∴补充定义(0)e,f =可使函数在x =0处连续.35. 解:(1)()f x 在(,0),(0,)-∞+∞上显然连续,而00lim ()lim(),x x f x a x a ++→→=+=lim ()lim e 1,xx x f x --→→== 且(0)f a =,∴当(0)(0)(0)f f f -+==,即1a =时,()f x 在0x =处连续,所以,当1a =时,()f x 在(,)-∞+∞上连续.(2)()f x 在ππ(,),(,)22-∞+∞内显然连续.而ππ22ππ22lim ()lim (sin )1,πlim ()lim (1)1,2π()1,2x x x x f x x b b f x ax a f b ++--→→→→=+=+=+=+=+∴当π112b a +=+,即π2b a =时,()f x 在π2x =处连续,因而()f x 在(,)-∞+∞上连续. 36. 证:令()21xf x x =⋅-,则()f x 在[0,1]上连续,且(0)10,(1)10f f =-<=>,由零点定理,(0,1)ξ∃∈使()0f ξ=即210ξξ⋅-= 即方程21xx ⋅=有一个小于1的正根.37.证:令()sin f x x a x b =--,则()f x 在[0,]a b +上连续,且 (0)0,()(1sin )0f b f a b a x =-<+=-≥, 若()0f a b +=,则a b +就是方程sin x a x b =+的根. 若()0f a b +>,则由零点定理得.(0,)a b ξ∃∈+,使()0f ξ=即sin 0a b ξξ--=即sin a b ξξ=+,即ξ是方程sin x a x b =+的根,综上所述,方程sin x a x b =+至少有一个不超过a b +的正根.38. 证:令()()()F x f x f x a =-+,由()f x 在[0,2]a 上连续知,()F x 在[0,]a 上连续,且(0)(0)(),()()(2)()(0)F f f a F a f a f a f a f =-=-=-若(0)()(2),f f a f a ==则0,x x a ==都是方程()()f x f x a =+的根,若(0)()f f a ≠,则(0)()0F F a <,由零点定理知,至少(0,)a ξ∃∈,使()0F ξ=,即()()f f a ξξ=+,即ξ是方程()()f x f x a =+的根,综上所述,方程()()f x f x a =+在[0,]a 内至少有一根.39.证:令()()F x f x x =-,则()F x 在[0,1]上连续,且(0)(0)0,(1)(1)10,F f F f =≥=-≤ 若(0)0f =,则0,ξ=若(1)1f =,则1ξ=,若(0)0,(1)1f f ><,则(0)(1)0F F ⋅<,由零点定理,至少存在一点(0,1)ξ∈,使()0F ξ=即()f ξξ=.综上所述,至少存在一点[0,1]ξ∈,使()f ξξ=.. 12()()()()n f x f x f x f n ξ+++=.40证:已知()f x 在1[,]n x x 上连续,则()f x 在1[,]n x x 上有最大值M 和最小值m ,于是 12()()()n f x f x f x m Mn+++≤≤,由介值定理知,必有1[,]n x x ξ∈,使12()()()()n f x f x f x f nξ+++=.。