微积分(曹定华)(修订版)课后题答案第二章习题详解

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372limlim(2)73x0x0x2x022732x2xsin7sin3212limxx0limx217xx03222xxX6.limx

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47=lim(lxxO(x)(x))=1lim(x)(x)xxO1xxOHim(x)(x)110.ºᡠ»Ax->xOB(x)¸P(x)Ḅᐙ⌕ᩩµlim(x)(x)(x)=0.xxO2.4B(x)W0,limB(x)=07limxxO(x)(x)xxOᙠ9:lim(x)=0.xxO

489!lim(x)limxxO(x)(x)xxO(x)lim(x)(x)xxOlim(x)limxxO(x)(x)xxO00ᓽlim(x)0.xx0123.9:!4Axf஺B,f(x)=o(xa),g(x)=o(xb),ᑣf(x)•g(x)=o(xab),ᐸ]a,b¾ᜧz0,pᵫÀᑨÂAx—0Btanx—sinxxḄÃ▤`.9!:Ax-0Bf(x)=o(xa),g(x)=o(xb)limf(x)A(A0),limg(x)xOxaxbB(B0)x0z!limf(x)g(x)limf(x)g(x)f(xxOxabxOxaxblim)limg(x)ABOxOxaxOxb••AxfOBf(x)g(x)O(xab),Vtanxsinxtanx(1cosx).AxfOBtanx0(x),1cosx0(x2),

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51132(y/2+\J\+cos,v)(5)limarctanxlimx1.x0arcsinxx0xbxax(6)limeaxe1)(ebxl)x0sinaxsinbxlim(ex02cosabxsinab22xeax

52bxlim1lx02cosabxsinablimex0ab2x2cos2xsinab22xlimaxlimbxx02cosabx02xab2x

532cosab2xab2xlimabx0(ab)cosablimx02x(ab)cosab2xaabbababab1.(7)limlncos2xllnlimIn1(cos2x1)x0cos3xIn1(cos3x1)

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62V3+2x0-O2G(2)limx0(3)limx21n(x1)In(21)InlO;(4)limarcsinarcsinarcsin>x1223;(5)lim(lnx)xeexe(Ine)11.?⚪2-81.BCDEx5-x4-x2-3x=lFGᨵIJKL1M2N'Ḅ᪷.BPf(x)x5x4x23x1,ᑣf(x)ᙠR1,2S#17Tfপ50,f(2)50ᵫW)XᙠYᳮ[FGXᙠI)x0(1,2),\]f(xO)0.ᓽx0x0x03x01,ᓽDExxx3x1FGᨵIJKL1M2N'Ḅ᪷.2.BCDEIn(1+ex)-2x=0FGᨵIJ_L1Ḅ`᪷.BPf(x)ln(lex)2x,ᑣ£6)ᙠ()#abᙠR0,1S#Tf(0)ln(leO)20ln20f(l)ln(le)20542542

63ᵫW)XᙠYᳮ[FGXᙠI)xO(0,1)\]f(xO)0.ᓽDEIn(1e)2x0FGᨵIJ_L1Ḅ`᪷.3.f(x)wC(-8,+oo),Tlimf(x)=A,limf(x)=B,A•B<0,jᵫ᩽▲kW)XᙠYxxXxᳮḄmnopqCFGXᙠI)xOe(—8,+8),\]f(xO)-0.BᵫA•B<0[AsBtu,v■Aு0,B<0ᵫlimf(x)AO,limf(x)B0,kyᦪ᩽▲Ḅ{uឋ[XI0,\|xxXXI,ᨵf(x)0,X20,\|xX2}ᨵf(x)0.~xaXI,ᑣf(a)0,xbX2,>ijf(b)0,Tab,ᵫ⚪[f(x)ᙠRa,bS#ᵫW)XᙠYᳮ,FGXᙠI)xO(a,b)\f(xO)0,ᓽFGXᙠI)xO(,)\0)0.4.⚗Pn(x)=xn+alxFGᨵI᪷.BPn(x)x1nn1+…+an.,ᑭᵨ3⚪BC|n$᜻ᦪ}DEPn(x)=0alxa2x2annx18limPn(x)xnx10,ᵫ᩽▲Ḅ{uឋ[.Pn(x)xnX0,\|xX}ᨵ0,}Pn(x)sxua$n$᜻ᦪᡠ(2X)snn(-2X)ntuLPn(2X)sPn(2X)tuPn(x)ᙠR2X,2XS#ᵫW)XᙠYᳮ,FGXᙠI)XO(2X,2X),\Pn(xO)0,ᓽPn(x)0FGᨵI᪷.19

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