第一章函數(shù)與極限第一節(jié)函數(shù)教材習(xí)題1-1答案（上冊(cè)P17）.（2）x2,69,21.(3).11.解:（1）.x(,100)(100,)x1010(4).x(0.99,1)(1,1.01).2.解:由.又因,即該鄰域以1為中心,為半徑,所以2x2x12xU(1,).當(dāng)0.1時(shí),0.05;當(dāng)0.01時(shí),0.0052.3.解:(1)不同.f(x)的定義域?yàn)閤0,而g(x)的定義域?yàn)閤0.(2)不同.對(duì)應(yīng)法則不同:f(x)x,而g(x)x.(3)相同.f(x)x3x1g(x).(4)不同.對(duì)應(yīng)法則不同:f(x)sinx而g(x)sinx..(3)x2x4.(2)x1且x24.解(1)x1x1且x0.(4)xx3且x0.(5)xx1.(6).xx05.解:f(0),4022f(1)415,2111f(1)4(1)25,f()4()24,aaa2f(x)4x2,f(xh)4(xh)2.0000()()sin6.解:=1,22,2(2)0.()6sin6444112517.證:f()2()25()f(t).#(1)2(1)ttttt8.證：(1)左邊=F(x)F(y)exeyexyF(xy)=右邊(2)左邊=F(x)eexyf(xy)=右邊.#xF(y)ey9.證：(1)左邊=G(x)G(y)lnxlnyln(xy)G(xy)=右邊xG()x=右邊G(x)G(y)lnxlnyln()(2)左邊=.#yy10.解(1)偶函數(shù).(2)既非奇函數(shù)又非偶函數(shù).(3)奇函數(shù).(4)偶函數(shù).(6)既非奇函數(shù)又非偶函數(shù)(5)既非奇函數(shù)又非偶函數(shù)..11.證:(1)設(shè)f(x),f(x)都是偶函數(shù)1,都是奇函數(shù).g(x),g(x)122令F(x)f(x)f(x),G(x)g(x)g(x),1212則()f(x)f(x)f(x)1f(x)F(x),所以為偶函數(shù).F(x)Fx122G(x)g(x)g(x)=g(x)(g(x))=(g(x)g(x))=G(x),所以G(x)121212為奇函數(shù).#12.證:x,x,0,不妨設(shè),,則x,x0,,且xx,因?yàn)閘xxl12121212f(x)在0,l內(nèi)單調(diào)更加,所以1()f(x).又因?yàn)闉槠婧瘮?shù),所以fx()fx2f(x)f(x),即f(x)f(x).所以()在,0內(nèi)單調(diào)更加.#fxl12122.(2)T22.(3)不是13.解:(1)周期T周期函數(shù).22(4)ysin2x1cos2x,T2.14.解(1)由y1x得x1y,則y1x的反函數(shù)為y1x.1y1x1x1x(2)由y2sin3x得x1arcsiny,則y2sin3x的反函數(shù)為y1arcsin3x.232得xe2,則所求的反函數(shù)為2.yeyy1ln(x2)(3)由x1e2xyx.(4)由y得xlog,則所求的反函數(shù)為ylog2x121y21x15.解(1)復(fù)合函數(shù)為yf(x)sin2x,則yf()sin2=1,yf()sin2346643312(2)復(fù)合函數(shù)為yf(x)1x2,則yf(1)1122,yf(2)122512(3)復(fù)合函數(shù)為yf(x)ex2,則yf(0)e01,yf(1)e1e.12(4)復(fù)合函數(shù)為e2x,則yf(1)e2,yf(1)e2.yf(x)(ex)2120.15x(x50)(x16.解:此函數(shù)為分段函數(shù)1:.圖形略.為正整數(shù))y0.1550(x50)(x50)17.解:總數(shù)為一年期存款為A0.042AA(10.042)時(shí)：一年后連本帶息共有;將AA(10.042)再存一年即兩年后連本帶息共有;A(10.042)(10.042)A(10.042)2半年期存款時(shí)：半年后連本帶息共有A(10.02),一年后連本帶息共有A(10.02)(10.02)A(10.02)2，一年0兩半后可取出年0后2可)取出A(1A(100.002A)(1，.(10.03.2)(1A04.，所以存一年期的定期收益較多，02)(10.02)多了A(10.042)2A(10.02)40.0033A.第二節(jié)數(shù)列的極限教材習(xí)題1-2答案（上冊(cè)P27）11.解(1)收斂,limxlim0.2nnnn(2)收斂,limxlim(1)n10.nnnn(3)收斂,limxlim(21)2.nn2nnlimn1lim(12(4)收斂,)1.limxnnn1nn1n(5)發(fā)散,因?yàn)楫?dāng)n為偶數(shù)時(shí),x=n,n時(shí),x;nn當(dāng)n為奇數(shù)時(shí),x=-n,n時(shí),x.nn2.解:limxlim1cosn0.2nnnn1cosnn01n,只需使n1,即取N1,當(dāng)nN時(shí),對(duì)0,要使2N1.所以當(dāng)0.001時(shí),有x0n1000.0.00111,只需使n21,即n13.證:(1)對(duì)0,要使a0n0.于是對(duì)n2n21取N10,,當(dāng)nN時(shí),都有a00.由數(shù)列極限的定義n2n1lim0.#nn233n1311,要使,只需1,即n1.n3n13(2)a22n122n1n2n12n131n3于是對(duì)取N,當(dāng)nN時(shí),都有,由0,2n12lim3n1.#32n12nn2a2n2a2nna21a2n(3)a1nnn(n2a2n)n2a2a2a21,,.故對(duì)要使只需即于是對(duì)0,取0,nnnna221.#naa2,當(dāng)nN時(shí),都有n22n1,.limnn11a1,只需,即(4)a10.99991,故對(duì)0,要使n10n10nnn個(gè)11.于是對(duì)0(1),取n,當(dāng)nN時(shí),都nlglg有,a10.99991,lim0.99991.#nnn個(gè),N當(dāng)nN時(shí),Z,ua,4.證:a,對(duì)0limuuannnnlimua.#nn:若1n,則u1,limu1,而數(shù)列沒(méi)有極限u例如.unnnnn第三節(jié)函數(shù)的極限教材習(xí)題1-3答案（上冊(cè)P36）1.證:(1)fxA()(3x1)83x3,要想使33,即x3x,3,取3,當(dāng)0x300時(shí),總有3(x31)x8由函3數(shù)極3限的定義,lim(3x1)8.#x3x24(4)x24x(2)(2)f(x)A,要想使x2x24x2(4),即x(2),00,當(dāng)0x(2),取x42時(shí),總有(4)x24x(2),由函數(shù)極限的定義x2limx244.#x2x21x311,要想使1,即,1x132232x2.證:(1)f(x)A2x322x3x3110,當(dāng)xM時(shí),總有亦即x31x31,0,取M3221x1113,由函數(shù)極限的定義,lim.#2x322M32x322x3x1,要想使sinx0,即,1(2)f(x)Asinx0xxxx10,取0,當(dāng)xM時(shí),總有sinx0,x1亦即x,M22limsinx0.#xx3.解:limx24x24x2x2,要想使,即x42x2limx211x21144x2x2,(此時(shí)1x3),xx3x3x23x222x250,取亦即5x2,0x2.#時(shí),總有x2x25x4min(1,),當(dāng)25若取0001,則min(1,0001)00002.5limx211x21144x211,即4.解:,要想使xx3x3x