高等数学-同济第六版-（上册）课后复习题全解

..习题111.设3>,写出AB,及解5>.2.BC.证明CxACB所以 BC.3.f:XBX.证明使xA或y所以 使xA且xB>y所以 ..4.:若存在一个映射g:使gofIX,fogIY,..xx;yy.f是双射,且g是f的逆映射:gf证明yyy,元素的像,因此f即fYX,因为对每个yY,yyy,定义,5.f:.有f...证明f所以 f使f.射,这就证明了f因此f.6...y3x2;..解由得x2.函数的定义域为[2,>...y3 31 ;x2..解由函数的定义域为..y1xx2;..解且1]...y1 ;4x2..解由得2>.ysinx;解由..解由x12x1,2..解4]...y3xarctan1;x..解由且x03>.解由1yex.解由7...x;..x2..f<x>3x4x3,g<x>x..解因为对应法则不同时因为定义域、对应法则均相相同.因为定义域不同...x|8.设<x>⎪|3求<>,<>,<>,并作出函数..⎨|⎩36 4 4....解<>1<>2<>2<2>0...6 6 2 4 4 2 4 4 29...yx1x

,1>;..x,证明1>,有1x..y21x21x2x2 0,x2>..所以函数yx1x..有..y2<x1ln><x2lnx2><x1x2x>ln10,x2..所以函数x在区间<0,10.设若调增加.证明因为所以证明..两个奇函数的乘积是偶函数,奇函数.证明如果则所以如果则所以则所以如果则所以如果而则所以12.哪些既非奇函数又非偶函数？yx2;x2xcosyaxax.2解f<x>1<x>21x2f<x>所以..11x2..所以xcosx1..fa<x>a<x>2xaxa2af<x>所以..13.4x;xx;x.解l2.l.2l2.l.14.y3x1yxxyaxbcxdx..y2 .2x1..解y3x1y3x1yx得x1y所以yx的反函数为yx...xyxx..yaxb得xb所以yaxb的反函数为yb...cxdcyacxdcxa..得x1arcsiny所以3x的反函数为y1arcsinx.3 2 3 2..y2x2x得xlogy2y所以y2x2x的反函数为ylogx.2x..15.X试证XX上既有上界又有下界...证明X上有界,M,这这就证明了X上有下界M和上界M.再证充分性.设函数f<x>在X上有下界K1和上界K2,即K1f<x>.取Mmax{|K1|,则 ,即 这就证明了X上有界.16...usinxxx..1 6 2 3..uxx..12 4..yu..0..解ysin2<1>21ysin2<3>23...1 6 2 4 23 2 4..ysin<2>sin2ysin<2>sin...1 8 4 2 2 4 2..y1x221225.....yex2ye02ye...121217.设解1].得所以函数f<sin>.且当0a1时当a1时无解.因此当0a1时2 2 2函数的定义域为[a当a1时函数无意义.2..⎧1 |x1..⎨18.设f<x>⎪0⎨1|并作出这两个函数的图形.|..⎨⎧1 |ex⎨⎧1 x0..⎨解f[g<x>]⎪0⎨|ex即f[g<>]⎪0x0...1|ex⎧⎨|⎨1x0⎧e|..⎨fx]ef<x>⎪e0⎨|x|1即fx]⎪1|...e1|e1|..19.已知水渠的横断面为等腰梯形斜角40<图当过水断面ABCD的面积为定图..解 AbDC1hsino又从..<BC2cot40h>]S 得2 0BCcoth所以hoLS02cos40h...h自变量hsin..确定0h0Scot.0S0coth0h..20.元,成本为购量超过台以上的,每多订购1台,1分,元.pxPx台,解时,令0.当时,..⎧pp⎪⎩900.01x750x100x.x..⎧30x⎪0x100..P<p60>x⎨31x0.01x2100x1600...⎩⎪15x⎩x....习题121.1xn ;2n..xn<1>n1;n..xnxn21;n2n1;n1..解x10,lim10...n 2nn2n..x<>n10,lim<>n10...n n nn..1xn2n2lim<2n1>2.n2..xn1n n12n1limn1.nn1..cos..2.xn2.问limxnn n..,当时,..解limxnn0.....|0||cos|21>0,要使|x,只要1,n1

.取N[1],..n.当nN[1]n ..3...lim10;..nn2lim3n13;n2n12....limn2a2..nnn4>lim999nn个..要使|10|1,只须n21,即n1...n2 n2 ..证明N[1],当时,有|10,所以lim10...n2nn2..要使|311,只须1,即n1...2n12n>n 4n ..证明N[1],当时,有|3|,所以lim13...2 2 2 22n122n2n122 a2..要使|nannannn<a an2a2n> n,只须n .....2证明N[a2],时,有|n2a2,limn2a2...n nn..110n,只须110n1<,即n1lg1...证明N1],时,,所以lim0.9999.n＿＿＿n个4.limuna,lim|un||a|...极限.nn..证明limuna,当n>N有|una|,从而n..这就证明了lim|un.n...数列|有极限,数{x}未必有极限.例如lim|>n1,但lim>n不存在...nn....5.又limy0,证明:limxy0...nnnnn..证明..又limynn0,当时,有|ynM.从而当时,有..所以limxnynn0.|xnynxnynM|ynM,M..6.证明:证明 当有||<;时,|<..|<...习题131.lim<3x8;x3lim<5x;x2..2limx244;..x2x2..3lim14x32...x12x2证明,只须|x1.3证明0,1,当时,,所以lim<3x8.3 x3,只须|x2|1.5证明0,1,当时,,所以lim<5x...52x4<>2x24x4xx<2>|,要使x22x4<4>2,只须..|x<2>|.x2x2x22 2..证明0,,当时,有x4<4>,limx44...x23 3x2x2..14x22x2|2|x<1>|,要使14x2,只须|x<11...2x12 2x1 2 2..证明0,1,当0|x<1>|时,有14x32,lim4x32.3..2 22.32x1x12x12..lim1x1;..x2x3 2..limsinx0.xx证明1证明1x3 1 1x3x31 ,要使1x3 1,只须 11.2x322x32|x|32x322|x|3..|x|3..证明0,X1,时,有1x3 1,所以lim1x1...x证明x32x3 2x2x3 23sinx3sinx0|sinx|1,要使sinx0,只须1,即x1xx xX1,当xX时,x有sinxx0,所以lim2sinx0...2 xx3.时, |y4|<0.解即|x2|0.0010.0002,则当5..4.当yx1,问X等于多少,时,..2x232..解要使x4 0.01,只|x|43397,X397...2x232x230.01..5.当..6.求f<x>x,x<x>|x|当并说明它们在x..证明..limf<x>limxlim,..x0x0xx0..limf<x>limxlim11,..x0x0xx0..limx0f<x>limx0f<x>,..所以极限limf<x>存在.x0因为lim<x>lim|x|limx1,..x0x0xx0x..lim<x>lim|x|limx,..x0x0xx0x..lim<x>lim<x>,..x0x0..所以极限lim<x>不存在.x07.若x时,A,则limf<x>A.x..证明limxf<x>A,limxf<x>A,..;.,即limf<x>A.x8.自存在并且相等.证明使当时,有...时都有....有|;.时,,从而有|,9.解x时函数极限的局部有界性的定理如果x时的极限存在则存在X0及M0时证明设X0时所以这就是说存在X0及时其中..习题141.解例如,当时,但lim<x>2,<x>不是无穷小...2.2x0<x>3<x>..yx9当..x3yxsin1当x..证明x3时|y|x29x3|.0,,当时,有..x3|y|x29x3|,..2所以当时yx29为无穷小.x3..x3时|y||x||sin1x0|.因为0,,当时,有x..所以当时yxsin1为无穷小.x|y||x||sin1x0|,x..3.y12x为当x0时的无穷大.问x应满足什么条件,能使x..证明|y|12x2112,只须12M,即|x|1 ...x x |x||x|M2..证明1M2

,使当时,有12xM,x..所以当时,函数y12x是无穷大.x..则11042

.当0|x0|11042时,..4.lim2x1;nx2lim1x.x01x..解2x121,而当时1是无穷小,所以lim2x12...x x x2nx2..1x1xx时x所以lim1xx01x1...5.填写下表:6.yxcosxx么？解函数yxcosxx,例如1,2,当k当时,函数xN,N的x,y<2k><2k>cos<2k>01,2,2 2 2对任何大的N,当k总有x2kN,27.函数y1sin1在区间<0,x x证明y1sin1x x在<0,例如当..时,有xk12k21,2,..当k充分大时,y<xk>2k,2..时,函数y1sin1不是无穷大.x x使例如可取1xk1,2,2k当k充分大时,..习题151.limx25;x2x3解limx252259...x2x3limx23;x3x223....解limx23<3>230...x3x2<limx22x13>2..x2;..解limx22xlim<xlimx00..x2<xx2 ...lim4x32x2x..x02x ;..解lim4x32x2xlim4x22x11...x03x22xx02 2..lim<xh>2x2;h0 h解lim<xx2limx2h2x2li2xh>2x...h0 hh0 hh0..lim<211>;..xx x2..解lim<211>2lim1lim12...xlimx x2x2;xxxx2..x2x2x1

11..解limx2limx2 1...x2x2xx2112..x x2..limx2x ;..xx43x2解lim x2xxx43x20<分子次数低于分母次数,....x2x11x2x3..或lim limx43x22 10...xx1x2 x4..2limx26x8;..x4x25x4解limx26x8lim<x4>limx2422..x4x25x4x4<xx4x413...0>lim1>21>;xx x2解lim11>21>lim1>lim21>22...xxx2 xxxx2..1>lim111>;..n24 2n1<1>n1..111..解lim1n 24

>lim2n n2112...2lim123<n;..nn2..解lim123<nlim2 1limn1...nn2nn22nn 2..lim3>;..n..解limn>n>n>1..5..或limn>n>n>1lim11>12>13>1...n13>;5nnn n 5..xx3..解13>limxx23limlimx21...xx3x1>xx2>x1>xx2>x11xx2..2.x32x2lim ;x2<x2>2解因为lim<x00,所以limx32x2...x2x32x216x2<x..limx2;..x2x1解limx2x2x..3>lim2x3x>.x解lim<2x3x>x3.limx21;x0x解limx210时,而1x0 x xlimarctanx.xx..解limarctanxlim1arctanx0x时,1是无穷小,而x..xxxx x..4.3中的<2>...习题161.limsinx;x0 x..limsinxlimsinx...解x0x x0..limtan3x;x0 x..解limtan3x3limsin13...x0 xx0cos3x..limsin2x;x0sin解limsin2xlimsin2x5x22...x0sin5xx02xsin55..limxcotx;x0..解limxcotxlimxcosxlimxlimcosx...x0x0sinxx0sinxx0..lim1cos2x;..x0解法一xsinxlim1cos2xlim1cos2xlim2sin2x2sinx2...x0xsinxx0 x2x0 x2x0 x..解法二lim1cos2xlim2sin2x2limsinx2...x0nxsinxxx0xsinxx0x..lim2nsin2nsinx..lim2nsinxlim 2n ...解2nx2nxx..2.11>li>x;x0..11..解lim1>xli1>]<x> li1>]<x>1e1...x0x0x0..12>li2>x;x0..111..解li2>xli2>2x2li2>2x2e2...x0x0x0..x>2x;xx解limx>2xlim1>x2e2...xxx x..4>lim1>kxk为正整数xx..解lim1>kxlim1><x><k>ek...xxxx..3.解4...lim1;n..证明因为111,..n n而 lim11且lim1>1,..由极限存在准则limn1.n..lim1n21n21n2n1;..证明n2n2n1n21n21n2n2 n2..而 lim n21,limn2,..n2nn2..所以 lim1n21n21n2n....2222, 222,....证明2,22,3,....时22,时,....2,3,2222,..xxn n2xn nx222,..n1 n22..limnx;证明时,则有,,..从而有x|n1xx|...因为 x|>x|>,..x0根据夹逼准则,有x0..limlimnx.limx1.x0x证明1111,所以1x1.x x x x又因为limx>lim,根据夹逼准则,有limx1...x0x0x0x..习题171.时..解 limx2x3limxx20,..x02xx2x02x..所以当x0时2.时无穷小<2>1x2>是否同阶？是否等价？222..3解 lim1x3limxx>xx2>3,..x11x1x..所以当x1时,1x2>..lim2x11x1x>1,2..所以当时,和1x2>是同阶的无穷小,23.当时有:x2secx.2..证明limarctanxlimy令x,则当时,y..x0xy0tany..所以当时limsecx2lim1cosxlim2sin2x2lim2sinx2,..x0所以当时,1x22secxx0x2cosxx2.2x0 x22x0 x2..4.limtan3x;..x0lim2xsin<xn>m为正整数>;..x0<sinx>mlimtanxsinx;x0 sin3xlimsinxtanx ...x0<31x21sinx..解<1>limtan3xlim3...x02xx02x2⎧nm..limsin<xn>limxn⎪nm...x0<sinx0xmnm....limtanxsinxlimsin1cosxlim1cosxlim1x221...x0sin3xx0sin3xx0cosxsin2xx0x2cosx2..sinxtanxtanx<cosx2tanxsin2x~2x<x>21x32 2 2..231x2x2~1x2..3x2>231x23..1sinxsinx1sinx~sinx~x1x3..所以 limsinxtanxlim 23...x0<31x21sinxx01x2x3..5.~若~,证明<1>lim,所以~;若则lim,从而lim.;..若~,limlimlim.....习题181...⎧x2f<x>⎨⎩2x0x1;x2....⎧xf<x>⎨x1.|x|1..解..在x1处,因为limf<x>limx21,limf<x>lim<2x>....所以limf<x>1,处是连续的.和..在处,因为函数在处间断,limx1f<x>lim11f,x1limx1f<x>limx1x1f,所以..在x1处,因为limf<x>limxf<1>,limf<x>lim11f<1>,x1处..连续...在处间断,2.说明这些间断点属于哪一类,则补充或改变函数的定义使它连续:2..yx,x2;..x23x2..yxtanx,xk,x2>;..ycos21,x....⎧xy⎨xx1x,x....解<1>yx2<x.因为函数在和x2和是函数..x23x2<x2><x....的间断点.因为limylimx2,x2..x2x2x23x2..因为limylim<x2,x1在..<x2>..令x1处成为连续的.x2..因limx故..xktanx..因为limx,limx0所以和x是第一类间断点且是可..x0tanx去间断点.xktanx 22..令x时,则函数在x处成为连续的.2 2ycos21在x0处无定义,x0ycos21的间断点.又因为x xlimcos21不存在,x0..x0xlimf<x>lim<x0limf<x>lim<3x>2,所以x1是函数的第一类不可去间断..点.2n..3.f<x>lim1xn1x2nx的连续性,..2n ⎧x|x|1..解f<x>limxn1x2nx⎪0⎨⎪⎩x⎨⎪|x|1.|x|1..处,因为第一类不可去间断点.limx1f<x>lim<x>1,x1limx1f<x>limx1x1,所以为函数的..x1limf<x>limx,limf<x>lim<x>1,所以..类不可去间断点...4.证明所以limf<x>f<x0>0,由极限的局部保号性定理,xx0。 。存在x0的某一去心邻域U<x0>,使当xU<x0>这就是说,则存..5...1,,21,是n..R上处处不连续,R上处处连续;R上处处有定义,..解函数f<x>csc<x>csc在点1,2,x1,,21,且这些点是n..函数的无穷间断点...⎨f<x>⎧1⎨在R上处处不连续,在R上处处连续...⎩1 ..⎨f<x>⎧x⎨在R上处处有定义,x0..⎩x..习题1931.f<x>x33x2x3的连续区间,并求极限limf<x>,limf<x>及limf<x>...x2x6x0x3x2..3解f<x>x33x2x3<x3><x,x2和..x2x6<x3><x..的,..处,limf<x>f<0>1...x0和处,limf<x>lim<x3><x,2limf<x>lim<x8...x2x2<x3><x2>x3x3x2 5..2.证明函数..证明limxx0f<x>f<x0>,limg<x>g<x0>.xx0..<x>1[f<x>g<x>|f<x>g<x>|],2<x>1[f<x>g<x>|f<x>g<x>|].2..因此 <x0>1[f<x2 0>g<x0>|f<x0>g<x0>|],..<x0因为>1[f<x2 0>g<x0>|f<x0>g<x0>|]...lim<x>lim1[f<x>g<x>|f<x>g<x>|]..xx01[limxx02f<x>limg<x>|limf<x>limg<x>|]..2xx0xx0xx0xx0..1[f<x2 0>g<x0>|f<x0>g<x0>|]..3...limx0x22x5;..lim<sin2x>3;x4limln<2cos2x>x6..limx0limx1x;x5x4x;x..limxsina;..xalim<xxax2xx2x>.....解f<x>x22x5是初等函数,x0所以....limx0x22x5f<0>022055...有定义,4lim<sin2x>3f<><sin2>3.x4 44x有定义,所以6limln<2cos2x>f<>ln<2cos2>0.x6 66limxlim<xxlimx lim1 1 1...x0 xx0x<xx0x<xx0x012..lim5x4xlim<5x4x><5x4x>lim4x4 ..limx4x1<x45x4x>2.<x5x4x>..x15x4x41..2cosxaxalimxsinalim 2 2..xaxaxaxa..limcosxalimxa2cosaa1cosa...xa2 xaxa22..lim<x2xx2x>lim<x2xx2x><x2xx2x>..xlimx2x lim<x2x2x2x>...x<x2xx2x>x<1x1>x..4.1limex;xlimlnx;x0xxlim11>2;xx..3tan2x0x>cot2x;..lim<3xx>2;..x6xlimtanx1sinx...x0x1sin2xx..1 lim1..解limexexxe0.x....limlnxln<limx>ln10...x0x x0xx 11..

lim1>2lim1>x2e2e...xxxx..12lim<13tan2x>cotxlim<13tan2x>3tan2x2e3...x0x0..3xx36x3x1..<>26x6x>36x2

...lim136x>3e,limx3,..x6xx6x 2 2..3xx3..所以lim< >2x6xe2...1tanx1sinx<1tanx1sinx><1sin2xlim lim..x0x1sin2xxx0x<1sin2x1tanx1sinx>..lim<tanxsinx><1sin2xlimtanx2sin2x2lim2x<x>221...x0xsin2x<1tanx1sinx>x0xsin2xx0 x3 2..⎧ex5.f<x>⎨⎩axx0x0a,..数？解只须处连续,即只须..limx0f<x>limx0f<x>f<0>a.....因为limx0f<x>limex,x0limx0f<x>lim<ax>a,所以只须取x0..习题1.至少有一个根介于1和2之间.证明即x1和2之间的根.1和2之间.2.其中证明设则若则说明的一个不超过ab若则使x也是方程的根.总之,至少有一个正根,ab.3.设数fx对于闭区间[a,b的任意两点xy,恒|fxfy||x|,其中L为正常数,且证明:使得证明0lim|f<x>f<x0>|limL|xx00,..xx0所以 lim|f<x>f<x0>|0,xx0xx0....即 limxx0f<x>f<x0>...因此a处左连续,b所以因为且使得4.,使..f>f<x1>f<x2>f<xn>.n..证明nmf<x1>f<x2>f<xn>nM,..mf<x1>f<x2>f<xnn>M...由介值定理推论,使..f>f<x1>f<x2>f<xn>.n..5.若且limf存在,则x证明令limf<x>A,0,存在X0,就有x,即.又由于存在取即6...总习题一1."充分"必要"充分必要三者中选择一个正确的填入下列空格内:条件.有界的的条件.limf<x>存在的条件.xx0limf<x>条件.xx0<3>limf<x>的条件.xx0limf<x>条件.xx0limf<x>存在xx0的条件.解 <1>必要,充分.<2>充分.<3>充分.<4>3.选择以下题中给出的四个结论中一个正确的结论:时,有< >.x是等价无穷小; x同阶但非等价无穷小;x高阶的无穷小; x低阶的无穷小.解因为limf<x>lim2x3x2lim2x1lim3x..x0 xx0xx0 xx0 x..ln2lim tln3lim uln2ln3<令..t0ln1t>u0ln1u>..所以f<x>与x同阶但非等价无穷小.故应选4.设求下列函数的定义域:x>;x>.解 <1>由得x0,<2>由0lnx1得1xe,即函数f<ln<3>由0arctanx1得即函数f<arctantan1].<4>由0cosx1得x<n0,1,2,>,2 2..即函数,2n2<n0,1,2,..5.设⎨f<x>⎧0⎨⎩xx0,x0⎨g<x>⎧0⎨x2x0,x0..求x0..解因为所以⎨⎩xx0;..因为g<x>0,所以因为g<x>0,所以x0..2<x>⎨.x2 x06.利用ysinx的图形作出下列函数的图形:x|;<3>y2sinx.27.把半径为R的一圆形铁片,自中心处剪去中心角为的一扇形后围成一无底圆锥.试将这圆锥的体积表为的函数.解设围成的圆锥的底半径为r,高为h,依题意有>,r..h圆锥的体积为R2r2R2R2>2R22.....V1R2>2R2..3 2 ..R3R>2a2..28.根据函数极限的定义证明limx2x65...x3x3..证明 对于任意给定的0,要使|x2x65|,只需|x3|,,x3..当就有|x3|,即|x2x65|,所以limx2x65...9.求下列极限:x2xx3x3x3..limx1;<x>2....limxx21;..<3>lim<2x3>x1;..x2x..limtanxsinx;..x0lim<axx0x3bx3cx1>x<a0,b0,..lim<sinx>tanx.x2..解lim<x>20,所以limx2x...x2xx1<x....<2>limx21limx21x21..xlimxx lim<x211 1...xx21x2x3x12x22 22x11..<3>>x1li>x1li>2 2..x2x2x2x2x2 1x2x22x2 1..li>2>2li>2li>2e...x2x2xx2xx2x..limtanxsinxlimsin1cosxlimsincos>..x0x3sinx2sinx02x2x32x<x>22x01x3cosx..limx0x3cosxlimx0 x3 <提示:用等价无穷小换>.2....lim<axbxcxa1 xa>xlibxcx33>axbxcx33axbxcx33x

,因为..x0 3x03..x x x3..liax0bc33>axbxcx3e,..limaxbxcx31axbxcx..x0 3x0 x x x..1[lnalim 1lnblim 1lnclim 1 ]..3 t0ln1t>u0ln1u>v0ln1>..1<lnalnblnln3abc,3..所以 lim<axbxcx13>xeln3abc3abc...x03提示:..1lim<sin>tanxli<sinx>sinx1x

,因为..x2x2..1li<sinx>sinx1e,x2lim<sinxtanxlimsinx..x2x2cosx..,limsin<sin2x>limsinxcosx0,..xcos<sinx>2所以 lim<sinx>tanxe01.x2x2sinx1..⎪xsin110.设f<x>⎨xx0,要使应怎样选择数a?..ax2 x0解要使函数连续,必须使函数在x0..因为limf<x>limax2>a,limf<x>limxsin10..x0x0x0x0x..所以当a0时,f<x>在处连续.因此选取a0时,1 ..11.设f<x>x1x0

,求并说明间断点所属类形...ln>x0..解因为函数x1所以x11..因为limf<x>limex10<提示lim1>,..1x..limf<x>limex1<提示lim1>,..xx1x..所以x1..又因为limf<x>limln<x0,lim11f<x>limex1 ,1..x0x0x0x0e..所以x0且为第一类间断点...证明n1 n21n221n2n1...证明因为n n2n1 n21n221 n2nn ,且n2..limnnn2nlimn11n1,limnnn2limn11n21,..所以n1 n21n221n2n1...13.证明方程sin,>内至少有一个根.22证明设则函数f<x>在[,]上连续.22..因为f>11,f<>12,f>f<>0,..2 2 2 2 22 2 2..所以由零点定理,在区间,>内至少存在一点,使这说明方程sin22xx10,>内至少有一个根.2214.如果存在直线使得当曲线yf<x>上的动点L的距离d<M,则称L为曲线当直线L的斜率k0时,称L为斜渐近线.直线ykxb..k limxf<>,bxlimx[f<x>kx]...<x,x><x,x>..1<2>线y2xex的斜渐近线.证明<1>仅就按渐近线的定义f<x><kxb>]0x必要性设是曲线则f<x><kxb>]0x..于是有limf<x>kb]0limf<x>k0klimf<x>..xx xxxxx..同时有f<x>kxb]0bf<x>kx]..x充分性如果klimf<x>xbf<>k],则..xxx..f<x><kxb>]f<x>kxb]f<x>kx]bbb0..xxx..因此是曲线xklimylim2x1e12x..xxxx..11..bli[y2]li[2xex2]2limex>12lim t11..xxxt0lnt>..1所以曲线y2xex的斜渐近线为y2x1..习题21y3x2y1xy1x2yx35xx x232x xyx5解..y<3x2><x3>223222x3231x3....y<1><xx112>1211x2123x2..y<1><x2>2x3x2..y<x35x><x5>161651616x516511x5..y<x23x2x5><x6>116111x6165x6..89如果且证明证明当所以flimf<x>flimfflimfff..x0x0x0x0x0x0..从而有2f即f10x解因为x23..kcos21kcos1..1 3 2 2x<,1>处的切线方程和法线方程式32..解sin3..x3 3 2..故在点<,1>处y13<x>..32 2 2 3法线方程为y12<x>2 3 3..12即13的两点一点的切线平行于这条割线？解割线斜率为k914..令得31 2..14x|..⎨1xy⎧x2sin⎨1x0解x0x0..

limylim|sinlim<sinx>0

limylim|sinlimsinx0..x0所以函数在处连续又因为x0x0x0x0x0..limy<x>lim|sinx||sin0|limsinx1..x0x0x0x0x0 x..<0>limy<x>lim|sinx||sin0|limsinx1..x0x0x0x0x0x..解因为limy<x>limx2sin10又处连续..又因为xx2sin10..limy<x>limx limxsin10..x0x0x0 xx0 x..所以函数在点且..⎨15f<x>⎧x2⎨x1为了使函数处连续且可导ab..解因为⎩axbx....limf<x>limx21limf<x>limb>ab..所以要使函数在必须时..f>limx22flimaxblima<xablima<xa..xx10xxx10x..所以要使函数在必须此时..⎨16已知f<x>⎧x2⎨xx0x0又f..解因为limf<x>flimx01limf<x>flimx200..x0xx0 xx0xx0 x..所以f..⎨17x⎨x0求f..⎩x x0解当时xfx当时f..因为limf<x>flimsinx01..x0xx0 x..limf<x>flimx01所以f从而..x0⎨fx⎨xx0x0x..⎩1 x018解ya2kya2x x2x0 2 0yya2<xx>x0 2 00yx2..解得x00x2x..a2 0 0..2解得ya2x2y0..0S1|2x||2y2|xy2a22 0 0 00..习题221; <csccscxcotx解<cotx><cosx>sinxsinxcosxcosxsin2xcos2x1csc2x..<cscx><sinx1sinxsin2x>cosxcscxcotxsin2xsin2xsin2x..2y472x5 x4 xxxcosxxxylnxxyexln3x2xcosxssint1cost解<1>y<47212><4x57x42x1x5 x4 x20x628x52x220282x6 x5 x2xsecxsecx<2secxsin2x1x<2lnx1xlnx..y<lnx>xx x21lnxx2..y<exx2ln3>exx2ex2xx4ex<xx3..xx1lnx>x2xlnxxxlnxsinxs<1sint>costcost>sint><sint>1sintcost..1costcost>2cost>2..3..xcosx求y和yxx6 4..sin1,求d2 4..f<x>35xx2求ff<2>5..解x..ycossin

31

31..x6 6

6 2 2 2..ycossin222..x4 4d

4 2 211..sinsinsind2 2..d1 1sin cos 222>..4d2 4 44

4 22 42 4 2..f3<5x>22x5f325f17151..4s2gt2求解即v得t这就是物体达到最高点的时刻0 g5解时所以所求的切线方程为所求的法线方程为y1x即26ye3x2..ya2x2..y<arcsin..x解ye3x2<3x2>e3x2<6x>6xe3x2..y1x2x21x22x2xx2..x<sinxsin111y<a2x2>21<a2x2>2<a2x2>1<a2x2>2<2x>x ..2 2ya2x2..y1<ex>ex ..1<ex>2 1e2xy2arcsinx<arcsinx>2arcsinx1x2..y1cosx<cosx>1cosx<sinx>tanx..7..y1 x2..xye2cos3xyarccos1xy1lnx1lnxysin2xxyarcsinx..yln<xa2x2>....解12>2212>21 xx2..1132>x2>21x2>21x21x2>22>x ..2 2 x2>x2....y<ex2>cos3xexx2<cos3x>exxx2>cos3xex2xx2<sin..1e2cos3x3e2sin1e2<cos3x6sin..y211<1>2x<1>x11<1>2x21>x2 x2|x| x2..1ln1ln1..yxlnx>2x2 lnx>2..ycos2x2xsin2x2xcos2xsin2x..y11<x2<x>2x>11<x21x>22x 21 xx2..yx1a2x2<xa2x2>x1a2x221a2x2<a2x2>]..1xa2x221a2x2<2x>]1 a2x2..1secxtanx<secxtanx>secxtanxsec2xsecxsecxtanx..1cscxcotx<cscxcotx>cscxcotxcsc2xcscxcscxcotx..8yarcsinx2ylntanx2..y1ln2x..yearctanxnxyarctanxx..yarcsinxarccosxyxx..xx..yarcsinxx..解<1>yarcsinxarcsinxarcsinx1<x>..2arcsinx21 122arcsinx21<x>2 22..2 1<x>2224x2....1x12xx12x1 x..ytanx<tan2>2tanxsec2<>22tanxsec2csc22..y1ln2x211ln2x1ln2211ln2x2lnx<ln..121ln2x2lnx1x xlnx 1ln2x..yearctanx<arctanx>earctanx1<<x>2x>..earctanx11earctanx..1<x>22x2x1>..xny1 <x1>1 <x<x1..1<xxx1<xx<xx2....1 arccosx1 arcsinx..1x2 1x21arccosxarcsinx..y<arccosx>21x2<arccosx>2..2x2<arccos..y1ln<ln1ln<ln1lnx<lnx>1ln<ln1lnx1x1 xlnxln<ln..<y21 1x 21 ><1x1x1x><1x1x><21 1x 21 >1x..1 1x2x2<1x1x>2..y1 <1x>1 11 ..11x1x1x11x1xx>221>....9.且fyf2<x>g2<x>的导数....解21f2<x>g2<x>f2<x>g2<x>]21f2<x>g2<x>f<x>f2g<x>g<x>]..f<x>fg<x>g<x>f2<x>g2<x>10设求下列函数ydydx解fffx..ylnchx1 2ch2x..ych2xx解x..y1ch2<ln<lnx>1 xch2<ln..x..y1ch21x2>1x2>2x ch21x2>..y11<x2<x22x x42x22..y1<e2x>2<e2x>2e2x e4x..y1 <thx>

1 1

1 1..1<thx>21th2xch2x1sh2xch2xch2x..1ch2xsh2x1 1x..y1chxx>12ch4x<ch2x>shxchx12ch4x2chxshx..shxshxch2xshxshx<ch2xsh3xth3x..chxch3xch3xch3xch3x..y2chxxxxx..xxxxx..sh2x<x<x2sh2x..x<x<xx..12yarctanx2ylnxxnyetetylncos1x..yesin21x..yxx..yxarcsinx24x2..yarcsin1t2..解..y2arctanx114 arctanx..2x2241xnlnxnxn1x24 2..yxx2n1nlnxxn1..y><etetet>et>2..ysec1<cos1>sec1<sin1><1>1tan1..xsin21x x x21sin21x2 x211x112sin21..yex<sin>exx<2sin>cosx<>x x2sinx2exx..y12x<xxx12x1x1>2x 42x1 xxx..yarcsinxx211x2412214x2arcsinx2..y1<>121t2>..1<>21t21<>21t2>2..1t21t2..1t21t2>21t2>1t2>21t2> |1t2|1t2>..习题231xyxxsint..ya2x2..x..y1 ..xyexxyxex2..yln<xx2>..解<1>4x1xy41x2..yxxxxxcosxx>xsint..y21a2x2<a2x2>x a2x2..ya2x2xa2x2xa2x2a2<a2x2>a2x2..1x22x..x2 x2yx2>2x<2x>x2>..x2>21x2>2..x..y<x3<x33x2 <x3>2..6x<x3>2x2<x3>x62x3>..<x39>2xarctanxx2>1<x32xarctanx..y2arctanx2xx2x2..yexxexex<xx2 x2..yex<x>exx2ex<x>2xex<x22x>x4 x3yex2xex2<2x>ex22x2>yex22x12x2>ex24x2xex2<32x2>..yx1x2<xx2>x11x2122x >1x21 x2..y1<x2x2>

1x222xx2> xx..2f解ffff..3若fy的二阶导数解fd2y:dx2....y1f<x>f..yff<x>ffff<x>[f..[f4试从dx1导出[f..dy ..d2xdy2<y>3..d3xdy3<y>5..解d2xddxd1d1dxy1..dy2dydydy dx dy<y>2<y>3..d3xdddx1..dy3 dyy3dx y3 dy<y>6y<y>5..5求物体运动的加速度并验证d2s..dt22s0..解dstdtd2sA2sintdt2d2s就是物体运动的加速度dt2..d2sdt22sA2sint2Asint0..6..解 7解 xx8n阶导数的一般表达式xx;解xy2cos2x2sin<2x>2y22>222>2 2y<4>2>233>2 2y>2n1sin2xn>]2..ylnxy1x>n2>!>n..9x,x,;xn1xn1..2x,.解x有xxx所以 xx..所以 u<100>vC1u<99>vC2u<98>vC98uv<98>C99uv<99>u..100x则有100100100..24848>248sin2x2..所以 y<50>u<50>vC1u<49>vC2u<48>vC48uv<48>C49uv<49>uv<50>..C50uv150uv50 50 50uv..48 49 <49>..50492228sin2x502x249cos2xx2sin2250<x2sin2x50xcos2x1225sin2..习题241y的导数dydx解2y0于是 ..yy yx..于是 yayx2y2axyx于是 yexyyxexy于是 ..yey xey222..2x3y3a3在点<解242处的切线方程和法线方程4..1 12x32y3y03 311于是 yx31y3..在点<242处4..所求切线方程为..y2a<x42即xy42a2..所求法线方程为..y2a<x42即xy04d2y..3y解xydx2..y<x>yyxxyy2x21..y y2 y2y3 y3..yb2x..a2yyb2x>..yb2yb2a2yb2a2y2b2x2b4..a2 y2 a2 y2a2 a2y3a2y3....ysec2<x1 sin2<xcos2<x11..1sec2<xy>cos2<xy>1sin2<xy2..y2y21>y2>..y3 y3 y2 y5yxe..yey1xeyey 1<yey2y..yeyeyeyy>ye2yy>..<2y>2<2y>2<2..4..y<x>x..x..y55x5x22..yx2<3x>4<x..yxsinx1ex..解两边求导得1ylnxx1ln1>x1..y于是 y<xx>x[lnx1]1x..xxx..lny1ln|x5|1ln<x25 25两边求导得..1y1112x..y 5x525x22..于是 y155x55x22[1x5152x]x22..lny1ln<x2>4ln<3x>5ln<x2..两边求导得1yy143x5x1..于是 yx2<3x>4[1 45]..<x2<xx3x1..lny1lnx1lnsinx1lnex>2 2 4两边求导得1y11cotxex ..y 2x 21ex>..于是 yxsinx1ex[11cotxex ]1xsinx1ex[22cotxex]..2x 21ex> 4x ex..5dydx..⎧xat2⎨⎩ybt2⎨..⎧xsin>⎨⎩y⎨..解dyt..dx ..dycosin..dx1sin..⎨6已知⎧xsint,⎨⎩ycost.求当t3时dydx的值..解dytcosttsintcostsint..dx sintcost13sintcost..当t时dy223

32..3 dx133..2 27..⎧xsint⎨⎩ycost⎨⎧x在t处4..⎪⎨⎪y⎩1t2在3at21t2..解>dy2sint..dx当t时4dydxcost2sin<2>4cos4222222y2 00..所求切线方程为y2所求法线方程为2<x2>即222xy20..y12<x22>即22x4y10..att2>3at2t2>2t2>2t2>attt2>23at21t2>2..dyatt..dx 3at21t2..当dy224x6ay12a..dx所求切线方程为1223 05 0 5..y12a4<x6即5 3 5所求法线方程为y12a3<x6即..5 4 5d2y..8dx2..xt2⎨2;..⎩y.⎨⎧xacost;⎨⎩ybsint⎨⎧x⎨..⎩y..⎧xft⎨⎩ytft<t>f⎨设f1..解dy1d2y<yxtt21..dx tdx2t t3..<bcsc2t ..dybcostbcottd2yyxtab ..dx asint adx2asinta2sin3t..<2..dydx3et23d2ydx2yxt33et49..dyft>tft>ft>td2y<xt1..dx fdx2fd3y..9⎧x1t2..⎨⎨⎩ytt3..⎧xln1t2> ⎨⎩ytarctant⎨..dyt3>13t2..dxt2<13t2..d2y1<13>..dx24t3 t1<1..d3ydx34t3t3t511t2>..dydxarctan[lnt21t2t21t2..<1..d2y2t> 1t2..dx2 1t2..<1t2..d3y> t4..dx31t2..10问在2解对应圆面积为S两边同时对t求导得S时故S其速率为其表面上升的速度为多少？解 h时r1hS12 4水的体积为V1hS1h1h33 3 4 12..dVdhdh4dV..dt 12dt dth2dt..dV4因此dtdhdt4h2

dVdt4416..1218cm直径12cm的正圆锥形漏斗中漏入一直径为10cm的圆柱形筒中开始时其表面下降的速率为问此时圆柱形筒中溶液表面上升的速率为多少？解设在ty圆柱形筒中水深为h1621r2y52h3 3..由ry得ry代入上式得..61831621<y>2y52h..3 3 3即 1621y352h3 两边对t1y22时1..23252160.6425..2711解..2y解3y12xxyxsin..yx x21..e2x..yarcsinx2..yarctanx2x2sA是常数>..解y1x21所以dy<1x x21x..所以..yx2x2xx21<x2x2所以dy1<x2x2dx..4>dyydx[ln21dx2ln1>12xln1dx....dy<arcsinx21x2><2x2|x|xx2dx....dydarctan1x21x211<1x2>21x2d<1x2>1x2..1 2x2>21x2>dx4xdx..1<1x2>21x21x2>21x4..t>]Acos<A4d< d< d< d< ..d< >1xdx..d< ..d< >1dxx..d< 解d<2xCd<3x2C2d<tCd<1cosxC..d<>1xdx..d<1e2xC2..d<2xC>1dxx..d<1tanC35的长为s跨度为2lO与杆顶连线AB的距离为f..则电缆长可按下面公式计算s2f2>当f时..解dS2f28..6如果R减少改变了多少？又如果不变R增加问扇形面积大约改变了多少？解S1R22SdS<1R21R22 2..303360代入上式得..1<2360

>43.63..12..dS<dR2 R..R1337解f当x时有xx所以..cos<>cossin<>

310.87467..6180

6 6 180

2 2180..x时..a36tan<>tansec2120.96509..4 180 4

4180180..8..解f当x时有..arcsin<xx>arcsinx所以11x2x..arcsin0.5002arcsin<0.5arcsin0.5110.520.0002..620.00023..f当x时有..x>arccosx所以11x2x..arccos0.4995arccos<0.50.0005>arccos0.5110.52..320.00053..9当x较小时<2>..1xx..并计算和的近似值x则有xxx则有取f<x>1则有x..11x1<1|xx..103996665解f<x>nx有fff1xn..3996310004103141000111341000

>9.987..f<x>nx有fff1x于是n..6656641261111>..64 664要求精确度在2%以内问这时测量直径D的相对误差不能超过多少？解球的体积为V1dV1D2D因为计算球体体积时所6 2以其相对误差不超过2%即要求1D2DdVD..2V 1D363D2%..所以 D2%3..也就是测量直径的相对误差不能超过2%312要求中心角为55测量弦长l问此而引起的中心角测量误差..解由lRsin得2arcsinl2arcsinl..2 2 2R400..l2Rsin2 1 1 ..2400l..时1<l>2400..211<184.7>240014000.10.00056..总习题二1在充分"要"和"充分必要"三者中选择一个正确的填入下列空格内..的 条件条件....解必要条件..2设xa则xalimf1>f存在limff存在..hhh0 h..limff