高等數(shù)學(xué)極限與連續(xù)(習(xí)題)

第二章極限與連續(xù)習(xí)題2-11、觀察下列數(shù)列的變化趨勢限.,判別哪些數(shù)列有極限,如有極限,寫出它們的極1(a1)；有.limx0.(1)xnannn(1)n11；有.limx0.(2)xnnnn(3)x(1)1；無.nnnnsin；無.(4)x2n(5)xn1；有.limx1.n1nnnx2(1)n；無.n(6)1cos；有.limx1.(7)xnnnn1ln.無.n(8)xn2、設(shè)u0.91,u0.99,,u0.999,問2nn個(1)limu?n(2)應(yīng)為何值時,才能使nu與其極限之差的絕對值小于n0.0001？n11,可見limu1；10解：(1)顯然,unnnn11,只需104u(2)欲使|1|n0.0001n5即可.10n{x}n1n0.01;,(n1,2,),給定(1)0.1;(2)、對于數(shù)列n|x1|成(3)0.001時,分別取怎樣的nN時,不等式N,才能使當(dāng)n立,并利用極限定義證明此數(shù)列的極限為1.n1n110k1,只需n10k1.|x1|1解：欲使n1n0.1,此時k1,取N1019即可;(1)若給定(2)若給定(3)若給定10.01,此時k2,取N10199即可;0.001,此時k3,取N101999即可;231n1n11n.limx.欲使|x1|1,只需下面證明nnn110,取N[]1N,當(dāng)nN時,恒有|x1|,nn所以limlimx1.nn1nn4、用極限定義考查下列結(jié)論是否正確,為什么？(1)設(shè)數(shù)列{x},當(dāng)n越來越大時,|xa|越來越小,則limxa.nnnn1|xa|11x1,a0,顯然解：結(jié)論錯誤.例如取越來越小,nnnn但limx10a.nn(2)設(shè)數(shù)列{x},當(dāng)n越來越大時,|xa|越來越接近于0,則limxa.nnnn1|xa|11越來越接近于nnx1,a0,顯然解：結(jié)論錯誤.例如取nn0,但limx10a.nn0,N,當(dāng)nN時,有無窮多個滿足|xa|,(3)設(shè)數(shù)列{x},n則limxa.xnnnn解：結(jié)論錯誤a1x(1)n,,顯然|xa|0,(k1,2,),.例如取n2k0,N1,當(dāng)nN時,有無窮多個|xa|,x,滿足nnlimx不存在.但顯然nn|xa|,則{x},若對0,{x}(4)設(shè)數(shù)列中僅有有限個x不滿足nnnnlimxa.nn解：結(jié)論正確0,假設(shè)僅有.,|xa|x,x,,x不滿足nnnn12kNmax{n,n,,n}N,那么當(dāng)nN時,|xa|,于是取所以12knlimxa.nn5、用極限性質(zhì)判別下列結(jié)論是否正確(1)若{x}收斂,則limxlimx,為什么？(為正整數(shù))；knnnknn解：結(jié)論正確.顯然limxlimx.nkn{x}{x}是的子數(shù)列,故nknnn(2)有界數(shù)列解：結(jié)論錯誤{x}必收斂；nx(1)n,雖然.例如取{x}有界,但顯然{x}發(fā)散.nnn(3)無界數(shù)列解：結(jié)論正確{x}必發(fā)散；n.因收斂數(shù)列必有界,那么無界數(shù)列必發(fā)散.(4)發(fā)散數(shù)列解：結(jié)論錯誤{x}必?zé)o界.nx(1)n,雖然.例如取{x}發(fā)散,但顯然{x}有界.nnnN”分析定義證明下列極限：6、利用數(shù)列的“(1)lim10；n2n11110,只需n或n[]1即可.,欲使|x0|分析：n2nn110N[]1N,當(dāng)nN時,恒有證明：,取|x0|11n,n2nlim1limx0.所以nn2nn2n13n123；(2)limn22n12110,欲使x,33n133(3n1)nn或n[1]1即可.分析：n1只需110N[]1N,當(dāng)nN時,恒有證明：,取x2133(3n1)n,1n2n1n3n1lim所以limx2.3nn(3)lim(11)1；3nn11110,只需分析：,欲使xn|1|n或n[]1即可.3nn110N[]1N,當(dāng)nN時,恒有證明：,取11|x1|,3nnnlim(11)limx1.所以3nnnn(4)limsinn0.nnsinn1110,只需分析：,欲使xn|0|n或n[]1即可.nn110N[]1N,當(dāng)nN,取時,恒有|x0|sinn1,nnnlimsinnlimx0.所以nnnn7、若limu0,證明lim|u|0,并舉例說明,如果數(shù)列{|u|}有極限,但nnnn{u}n數(shù)列未必有極限.nNNs.t.nN時,|u0|,limu0,有0,證明：因nnn|u|0|u0|,所以lim|u|0.于是nnnnu(1)n,顯然lim|u|1,但顯然.而若取{u}沒有極限nnnn{x}xa,(k),xa,(k),8、對于數(shù)列,若n2k12k證明xa,(n).ns.t.kN時,|x|a,2k1limx0,有k0證明：因,NN2k111limx0,對,NNs.t.kN時,|xa,222kNmax{2N,2N}N,當(dāng)nN時,|0又因取2kk12n1N2Na|若n2k1,有kN,12|xa||x2k1,221nnN2N若n2k,有kN,|xa||xa|,n222222k總之,當(dāng)nN時,xa||,所以xa,(n).nn習(xí)題2-21、用極限定義證明：(1)lim(5x2)12；x2|f(x)12|5|x2|,只需|x2|0,欲使分析：即可.5000|x2|,取,當(dāng)時,恒有5|f(x)12|5|x2|,所以lim(5x2)limf(x)12.x2x2x244；x2(2)limx2|f(x)(4)||x2|,只需0,欲使0,取0,當(dāng)0|x(2)|時,0|x2|即可.分析：證明：x24|f(x)(4)|x2(4)|(x2)4||x2|恒有所以,limx24limf(x)4.x2x2x2(3)lim(3x1)8.x3|x3||f(x)8|3|x3|,只需0,欲使分析：即可.300|x3|0證明：,取,當(dāng)時,恒有3|f(x)12|3|x3|,所以lim(3x1)limf(x)8.x3x32、用極限定義證明：(1)lim6x56；xx0,欲使|f(x)6|55,只需|x|分析：即可.xK505時,恒有|f(x)6|5x0,取,,當(dāng)|x|證明：lim6x5limf(x)6.所以xxx2)limsinx0.xx0,欲使|f(x)0|sinx1x,只需x1x分析：即可.20,取K10,當(dāng)xK時,恒有證明：|f(x)0|1,x2limsinxlimf(x)0.所以xxx等于多少,則當(dāng)0|x2|時,x2時,yx4,問3、當(dāng)2|y4|0.001？(提示：因為x2,所以不妨設(shè)1x3).|y4||x24||x2||x2||(x2)4||x2|解：欲使1(|x2|4)|x2|5|x2|0.00110,31|x2|51030.0002即可.只需0.0002,當(dāng)0|x2|時,有|y4|0.001.因此,取x,x3,f(x)x3時,f(x)的左右3x1,x3.作f(x)的圖形1題(3)的結(jié)果).4、設(shè),并討論極限(利用第解：(1)f(x)的圖形(2)令g(x)x,h(x)3x1,limg(x)limx3,limh(x)lim(3x1)8,.已知x3x3x3x3limg(x)3,limh(x)8.于是x33xx3時,f(x)g(x),于是x3時,f(x)h(x),于是limf(x)limg(x)3；顯然,當(dāng)當(dāng)x3x3limf(x)limh(x)8.x3x3f(x)|x|,當(dāng)x0時的極限為零0,取0,當(dāng)0|x|5、證明證明：.時,恒有|f(x)0||x|0|x|,lim|x|limf(x)0.x0x0|x|f(x)6、函數(shù),回答下列問題：xf(x)在x0處的左右極限是否存在？f(x)在x0處的左右極限是均存在(1)函數(shù)答：.limf(x)limxlim(1)1；這是因為：xx0x0x0limf(x)limxlim11.xx0x0x0f(x)在x0處是否有極限？f(x)在x0處是沒有極限(2)函數(shù)答：.這是因為：limf(x)11limf(x).x0x0f(x)在x1處是否有極限？f(x)在x1處有極限.(3)函數(shù)答：limf(x)limxlim11；這是因為：xx1x1x1limf(x)limxlim11.xlimf(x)limf(x)1,故limf(x)1.x1x1x1由于x1x1x1limf(x)A的充要條件是limf(x)limf(x)A.7、證明xx0證明:“必要性”xx0xx0limf(x)A00s.t.0|xx|,時,0xx0|f(x)A|,從而,時,|f(x)A|；0xx時,|f(x)A|,0xx當(dāng)0也有,當(dāng)0limf(x)lim()fxA.所以0xxxx00,0,12limf(x)limf(x)As.t.“充分性”xxxx00時,|f(x)A|；0xx01時,|f(x)A|,0xx當(dāng)021時,有|f(x)A|,min{,}0,當(dāng)0|xx|取20fxA.lim()所以xx0|A|2fx.|()|8、設(shè)limf(x)A(A0),證明當(dāng)xx充分大時|A|0K0,,當(dāng)xK時,證明:因limf(x)A(A0),對于x20|A||f(x)A|.20|f(x)||A(f(x)A)||A||f(x)A||A||A2||A2|所以.習(xí)題2-31、根據(jù)定義證明：(1)yx1為當(dāng)x1時的無窮小；0,取0,當(dāng)0|x1|證明：時,恒有|y||x1|,yx1為當(dāng)x1時的無窮小所以.(2)yxcos1為當(dāng)x0時的無窮小.x0,取0,當(dāng)0|x0|時,恒有|y||x|,證明：所以yxcos1為當(dāng)x0時的無窮小.x12x為當(dāng)xyx0時的無窮大,問x應(yīng)滿足什么2、根據(jù)定義證明：函數(shù)|y|104條件,能使(1)分析：？|y|12x11K2K0,欲使2K,只需0|x||x|即可.x1K2K0,取0,當(dāng)0|x|時,恒有|y|12x1212K,|x|xxlim12xlimy.所以xx0x011042100021(2)欲使|y|104K,取,1則x滿足0|x|即可.100023、利用有界量乘無窮小依然是無窮小求下列極限：1(1)limx2sin.xx01解：因limx0,sinx01(x0),x1有xo(1)(無窮小),sinO(1)(有界),(x0),x11則x2sino(1)o(1)O(1)o(1),(x0),所以limx2sin0.x0xx(2)limarctanx.xxlim10,xx1o(1)(無窮小),arctanxO(1)(有界),(x),arctan解：因,2有則xarctanxx(x),所以limarctanxxo(1)O(1)o(1),0.xyxsinx在區(qū)間(0,)內(nèi)是否有界?又當(dāng)x時,這個函數(shù)是4、函數(shù)否為無窮大?為什么?x2k,則y(2k)sin(2k)2k,k1,2,,(1)取2222可見,函數(shù)yxsinx在區(qū)間(0,)內(nèi)無界.(2)取xk,則yksin(k)0,k1,2,,x時,函數(shù)yxsinx不是無窮大.可見,當(dāng)1yxsin在區(qū)間(0,)內(nèi)是否有界?又當(dāng)x時,這個函數(shù)是4’、函數(shù)x否為無窮大?為什么?111||1x0時,xsin|x|sinx,解：(1)當(dāng)xxx1可見,函數(shù)yxsin在區(qū)間(0,)內(nèi)有界.x1yxsin在區(qū)間(0,)內(nèi)有界(2)因函數(shù)可見,當(dāng),xx時,函數(shù)yxsinx不是無窮大.習(xí)題2-41、填空題：an2bn2(1)已知a,b為常數(shù),limnb3,則a0,6;2n1b2ann21n1an2bn2003limnn2n1limn2a,有a0.2解：由于b2nb,有b6.12an2bn22n1bn2而3limnlimn2n1limn2nx21(2)已知a,b為常數(shù),lim(xbaxb)1,則a1,-1;x11x210limlim(xxxaxb)lim(11ab)1a,xxxxx2有a1.而1lim(x211axb)lim(x1xb)lim(b)bxxxxxx有b1.axbx12,則a2,-2.b(3)已知a,b為常數(shù),limx1axb,有ba.ablim(axb)lim(x1)x1020解：由于x1x1而alimaxalimaxb,有b22x1x1x1x12、求下列極限：313n2n303404limn1(1)limn4n21.n4n221()n55n(2)n105(2)051.(2)limn5(2)limn12n1n5(2)()n511112222nlim111111310342n1(3)lim.103n1111111111n3323n23n122n11122n1(4)lim(13)lim()nnnn2nnnnn221112n1limn(2n1)lim(2)1.nnn2n111n(n1))lnim(1223lim[()(11)(1nn11)]lim(1)1.111n11223nnn111.1012(6)lim(n1n)nlimn1nlim111nnnn3、求下列極限：x23x4x24(1)limx2.x24224x23x4.limx2x23x4223240,所以x24解：由于limx2(xh)3h33x2h3xh2h3(2)limh0limh0lim(3x23xhh2)h0hh3x23x0023x2.351xx2300limx134xx2(23)20(32)303x25x1xx23x4(3)lim3.100(2x3)20(3x2)30220330(4)limxlimxx(5x1)501550(5)50xx11(5)lim(1)(2x)122.x2x231limxxxx4272x33x150000.24(6)limx4x52x7400xx543x1xlim(3x)(1x)limx1x1x1(x1)(x1)(3xx1)2222(x1)(3x1x)(11)(3111)4.limx111x1x33131x2x2(8)lim(x1)limx1(11x1xx2)limx11x1xx2(x2)(12)1.limx11xx11122(9)limx1lim(x1)/(x1)limxm1xm21mmx1x1(x1)/(x1)x1xn1xn21nx1n11121m.(m,n是自然數(shù)mm).111nn1n23x1lim(3x1)(3x23x1)(x1)(10)limx1x1(3x31)(x1)x1(x1)(x1)(3x2x1)3x11123limlim.x1(x21)(3x23x1)x13x23x1312113(11)lim(1x)(12x)(13x)1lim(1x)(15x6x2)1x0xx0xlim(611x6x2)61106026.x0(12)lim((x2)(x1)x)lim(x2)(x1)x2(2)(1)xxxxx12x2lim(x2)(x1)xlimx(12)(11)1xxxx101(10)(10)12.4、求下列極限：x23x(1)limx3(x3)2;lim(x3)2(33)2x23x.x3x32330,所以lxim3(x3)解：由于22x3x32lim(2)x3x4;343x4limxlim3x4000,10limxx322x32解：由于xx32x1x3x32x3x4.lim所以(3)2lim(5x2x3);x11x5x22x3023500x20,limlimx解：由于5xx2所以2lim(5x2x3).xlimf(x)A,limg(x)不存在lim[f(x)g(x)]不存在.5、設(shè),證明xx0xxxx00fxgxB,則lim[()()]證明：反證.假設(shè)xx0limg(x)lim[f(x)g(x)f(x)]lim[f(x)g(x)]limf(x)xxxxlimg(x)存在,這與條件xxlimg(x)不存在沖突xx0,BA,可見000xx0xx0lim[f(x)g(x)]不存在.xx0習(xí)題2-51、求下列極限：2sin2x2x212(1)limsin2xlim.x0sin5x515sin5xx055xcos2xsin2x2x11.(2)limxcot2xlim212x0x02(3)lim1cos2xlim2sin2xlim(2xsinxxsinxsinx)212.xx0x0x01t2(4)lim2nsinxlim(xsintx)x1x,(x為不等于零的常數(shù)n).2txnnn1sinxxsinxx0xsinx110.limx(5)lim11x01sinxxsinx2sin2x2tanxsinxlimsinx(1cosx)lim(6)limx0x3x3cosxx3cosxx0x02sinxsinx11cos21112.2limx011212xxx2t2atlimsin(at)sinalim2sincos(7)limsinxsina2txa2xattxat0t0sintlimt02limcos(at)cosa.t02t2sin(x)3limtxsint312cosx(8)lim)t012cos(tx33sintsint1cost3sintlimlimt012(costcossintsin)t033sintsint13ltim01cost3sintlimtt23sint01313t0sin2t2tt()22.tlimttan()x(1t)(9)lim(1x)tanx1t1xlimttant02222t0tu2ucotulimu02cosu2.tlimtcott02limsinu2u0u2、求下列極限：22111e(10)e.t11x1(1)lim(1)lim(1t)t0x2tx1xlim(1t)t(1t)t02x2tx11.1)2lim(1t)t0(2)lim(xt2e1x0lim(1t)tt0(11)1xe1.x1)limx(3)lim(xxx11ee2x(1)xx(4)lim(11)xtxlim(11)tlim[(11)t(11)t]1e1.xt2ttexttx2111(5)lim(x21)xlimx(11)xlim[(11)x(11)x]211.xexxxex31t3tan2xlim(1t)[lim(1t)t]3e3.(6)lim(13tan2x)cot2xtx0t0t0112limln(12x)2x(7)limln(12x)lim2xln(12x)sin3xx03x3x2x2lne2x0.3limsin3xsin3x313x03xx0(8)limn[ln(n2)lnn]limnln(12)lim2ln(1)22lne2.2nnnnnn3、利用極限存在準則證明：1nn211n2n1n221)1.(1)lim(n21n2n2，n2nn(n2n22n2n)證明：由于n21n21lim1,limnnlim21,而limnn2nn1n1nn211n221n2n)1.lim(nn2所以Amax{a,a,,a},(a0,i1,2,,m),則有(2)設(shè)12milimnnana1anA.n2mAnAnnana1annmAnAnm，證明：由于n2m而limAnmAlimnmA1A,nnlimnnana1anA.所以n2mx2,x2x,n2,3,,證明數(shù)列{x}存在極限并求之(3)設(shè).1nn1nx22,假設(shè)x2,有x2x222,證明：①顯然1n1n1n0x2,n1,2,3,；因此,nx2x2x,假設(shè)xx,有②由于2111nnx2x2xxnn1nn1{x}因此,為單調(diào)遞增數(shù)列；n③由①②知,數(shù)列{x}必存在極限.nlimxa，顯然有0a2，且④假設(shè)nnalimxlim2x2a，nn1nn即a2a20，得a2(a1舍去),limx2.nnx1(x1)的極限存在x2,1(4)數(shù)列.2xn1nnx1(x)2x11,112x2xn1nnx21,而1證明：①顯然nnx1x2xx1(x1)xn2x22x1120,n211②由于2xn1nnnnnn即xx,因此,{x}為單調(diào)遞減數(shù)列；n1③由①②知,nn1x2,n1,2,3,,因此數(shù)列{x}的極限必存在.nn4、某企業(yè)計劃發(fā)行公司債券,規(guī)定以年利率6.5%的連續(xù)復(fù)利計算利息,10年后每份債券一次償還本息1000元,問發(fā)行時每份債券的價格應(yīng)定為多少元？r6.5%,解：設(shè)A為發(fā)行時每份債券的價格0k10年后每份債券一次償還本息,年利率為A1000元,k,則AAekr,即1000Ae100.065,得若以連續(xù)復(fù)利計算利息k00A1000e100.065552.05(元).0習(xí)題2-6x0時,下列各函數(shù)都是無窮小1、當(dāng),試確定哪些是x觀的高階無窮?。客A無窮??？等價無窮小？(1)x2x；limx2x解：因為lim(x1)1,x0xx0所以x2x~x,(x0).(等價無窮小)(2)xsinx；limxsinxlim(1sinx解：因為)112,xxx0x0x2xO(x),(x0).(同階無窮小)(3)xsinx；limxsinxlim(1sinxx0解：因為)110,xxx0x2xo(x),(x0).(高階無窮小)所以1cos2x；lim1cos2xx0x2xo(x),(x0).(高階無窮小lim2sin2xx0lim(2sinxsinx)2010,x0解：因為所以xxx)(5)tanx；limtanxx0lim(sinx1x0tanx~x,(x0).(等價無窮小))111,xcosx解：因為x所以(6)tan2x.limtan2xlim(sin2x2x0)122,解：因為x2xcos2xx0tan2xO(x),(x0).(同階無窮小所以)x0時,有：2、證明當(dāng)arctanx~x；(1)limarctanxx0limt0cost11,sint1ttarctanxlim證明：因為xt0tantt所以arctan~(x0).xx,(2)secx1~1x2；222sin2x2limsin2xlimsecx1lim2(1cosx)lim證明：因為21,x0()212xx2cosxx2x0x0x0x22secx1~1x(x0).所以2,2(3)1xsinx1~1x2；22sinxlim1xsinx1lim211,x11xsinx1101x0x0x221xsinx1~1x2,(x0).所以21x21x2~x2.(4)1x21x221,limx02limx0證明：因為10101x21x2x21x21x2~x2,(x0).所以3、利用等價無窮小的性質(zhì),求下列極限：1x2(1)lim1xsinx1limlim11.x0211cosxx0x0x221xsinx1~1x2,1cosx~1x2,(x0).其中：22sin2x(ex1)tanx2lim2xxlim22.(2)limx0x2x0x0sin2x~2x,ex1~x,tanx2~x2(x0).其中：(3)limln(12x)limsin5x2x5xlim(2)x0ln(12x)~2x,sin5x~5x,(x0).25.5x0其中：x01x2(4)limtanxsinxlim1cosxlimsin2xcosxx0x2cosxlimx011.2sinx2cosx23x0x01cosx~1x,sin~(xxx0).其中：2212x111)lim1cosxlim2lim11.xsinx222(5)lim(x0xsinxtanxxx0x0x01cosx~1x,sinx~x(x0).其中：221(mx)2(6)lim1cosmxlim2limx02mm2.222x2xx0x0m0時,1cosmx~1(mx)2,(x0),其中：2m0時,1cosmx1(mx)20.而24、證明無窮小的等價關(guān)系具有下列性質(zhì)：(1)~(自反性)；~.limlim11,所以證明：因(2)若~,則~(對稱性)；lim11~,因lim~.證明：已知1,所以1(3)若~~(傳遞性).~,,則~~證明：已知,,~因limlim()limlim111,所以習(xí)題2-7、研究下列函數(shù)的連續(xù)性,并畫出函數(shù)圖形：1,x1,(1)f(x)x,1x1,21,x1.f(x)在(,1),(1,1)以及(1,)連續(xù).解：顯然斷；,函數(shù)limf(x)limx211limf(x),則f(x)在x1間由于由于x1x1x1limf(x)limx2f(1)1limf(x),則f(x)在x1連續(xù).x1總之,函數(shù)11xxf(x)在(,1),(1,)連續(xù),在x1間斷.x2,0x1,2x,1x2.(2)f(x)limf(x)limx21,有f(x)在[0,1),(1,2]連續(xù).由于解：顯然,函數(shù)x1x1(),則fx在limf(x)lim(2x)211f(1)limf(x)x1連續(xù).x1總之,函數(shù)x1x1f(x)在[0,2]連續(xù).a,b使下列函數(shù)連續(xù)：2、確定常數(shù)ex,x0,xa,x0.(1)f(x)f(x)在(,0),(0,)連續(xù).解：顯然,函數(shù)limf(x)limexe01,由于x0x0limf(x)lim(xa)0aa,00xxf(x)在x0連續(xù),只需limf(x)limf(x)1f(0),即a1.欲使0xx0a1時,函數(shù)f(x)在(,)連續(xù).因此,僅當(dāng)ln(13x),x0,bx2,f(x)x0,x0.sinax,xf(x)在(,0),(0,)連續(xù).解：顯然,函數(shù)limf(x)limln(13x)lim3xlim3(b0),3由于,bbxbxsinaxbx0x0x0x0limf(x)limsinaxlim(a)a,a0,x0ax,xx0x00a,a0.f(x)在x0連續(xù),只需limf(x)limf(x)2f(0),欲使x0x03a2,即a2,b3有.2ba2,b3時,函數(shù)f(x)在(,)連續(xù).因此,僅當(dāng)23、下列函數(shù)在指出的點處間斷,說明這些間斷點屬于哪一類型.,如果是可去間斷點,則補充或改變函數(shù)的定義使它連續(xù)x24(1)yx25x6,x2,x3；yf(x)(x2)(x2)x2,x2.解：(x2)(x3)x3limf(x)limx2224,①由于x323x2x2limf(x)limx2224limf(x),x323x2x2x2,屬第一類間斷點可見,x2是函數(shù)yf(x)的可去間斷點..f(x)在x2連續(xù),只需定義f(2)4即可.欲使limf(x)limx2②由于,x3x3yf(x)的無窮間斷點x3可見,x3是函數(shù),屬第二類間斷點x,xk,(k0,1,2,)；(2)ysinxx,xk,(k0,1,2,).解：yf(x)sinxxlimf(x)lim1,①由于x0x0sinxxlimf(x)lim1limf(x),sinxx0x00xyf(x)的可去間斷點可見,x0是函數(shù),屬第一類間斷點.f(x)在x0連續(xù),只需定義f(0)1即可.欲使xlimf(x)lim,(k1,2,)②由于sinx(k1,2,)是函數(shù)xkxk可見,xk,yf(x)的無窮間斷點,屬第二類間斷點.(3)ycos1,x0；3x解：yf(x)cos31,x0.顯然函數(shù)xyf(x)有界,limf(x)limcos31由于不存在,xx0x0yf(x)的振蕩間斷點x1.可見,x0是函數(shù),屬第二類間斷點.(4)y2x1,x1,45x,x1.2x1,x1,45x,x1.yf(x)解：由于limf(x)lim(2x1)1,x1x0limf(x)lim(25x)31limf(x),11x1,屬第一類間斷點x可見,xx1是函數(shù)yf(x)的跳躍間斷點.fxx33x2x3()、求函數(shù)的連續(xù)區(qū)間,并求極限limf(x),x0x2x6limf(x)及l(fā)imf(x).x2x3f(x)x33x2x3(x21)(x3)x21x.解：(x2)(x3)x2,3x2x6顯然,函數(shù)f(x)在(,3),(3,2)以及(2,)連續(xù).limf(x)limx21x21,8,limf(x)limx2x25x2x2x3x3limf(x)limx211.x0x0x225、求下列極限：(1)limx0x22x3lim(x22x3)02203.x03(2)lim(cos2)(limcos2)[cos(2)]3(cos)300xx.33342x4x42t1e2(1)1e(3)limt11e2.t1sinsinxx2(4)limx2.226、求下列極限：1t1(1)limexxlimete01.xt01limcos[ln(12x1)]limcos[ln(12tt2)]txx2xt0cos{ln[lim(12tt2)]}cos[ln(12002)]cos01.t0(3)limexe2xl