高等數(shù)學-同濟第六版-（上冊）課后復(fù)習題全解

..習題111.設(shè)3>,寫出AB,及解5>.2.BC.證明CxACB所以 BC.3.f:XBX.證明使xA或y所以 使xA且xB>y所以 ..4.:若存在一個映射g:使gofIX,fogIY,..xx;yy.f是雙射,且g是f的逆映射:gf證明yyy,元素的像,因此f即fYX,因為對每個yY,yyy,定義,5.f:.有f...證明f所以 f使f.射,這就證明了f因此f.6...y3x2;..解由得x2.函數(shù)的定義域為[2,>...y3 31 ;x2..解由函數(shù)的定義域為..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..解因為對應(yīng)法則不同時因為定義域、對應(yīng)法則均相相同.因為定義域不同...x|8.設(shè)<x>?|3求<>,<>,<>,并作出函數(shù)..?|?36 4 4....解<>1<>2<>2<2>0...6 6 2 4 4 2 4 4 29...yx1x

,1>;..x,證明1>,有1x..y21x21x2x2 0,x2>..所以函數(shù)yx1x..有..y2<x1ln><x2lnx2><x1x2x>ln10,x2..所以函數(shù)x在區(qū)間<0,10.設(shè)若調(diào)增加.證明因為所以證明..兩個奇函數(shù)的乘積是偶函數(shù),奇函數(shù).證明如果則所以如果則所以則所以如果則所以如果而則所以12.哪些既非奇函數(shù)又非偶函數(shù)？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的反函數(shù)為yx...xyxx..yaxb得xb所以yaxb的反函數(shù)為yb...cxdcyacxdcxa..得x1arcsiny所以3x的反函數(shù)為y1arcsinx.3 2 3 2..y2x2x得xlogy2y所以y2x2x的反函數(shù)為ylogx.2x..15.X試證XX上既有上界又有下界...證明X上有界,M,這這就證明了X上有下界M和上界M.再證充分性.設(shè)函數(shù)f<x>在X上有下界K1和上界K2,即K1f<x>.取Mmax{|K1|,則 ,即 這就證明了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.設(shè)解1].得所以函數(shù)f<sin>.且當0a1時當a1時無解.因此當0a1時2 2 2函數(shù)的定義域為[a當a1時函數(shù)無意義.2..?1 |x1..?18.設(shè)f<x>?0?1|并作出這兩個函數(shù)的圖形.|..??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.已知水渠的橫斷面為等腰梯形斜角40<圖當過水斷面ABCD的面積為定圖..解 AbDC1hsino又從..<BC2cot40h>]S 得2 0BCcoth所以hoLS02cos40h...h自變量hsin..確定0h0Scot.0S0coth0h..20.元,成本為購量超過臺以上的,每多訂購1臺,1分,元.pxPx臺,解時,令0.當時,..?pp??900.01x750x100x.x..?30x?0x100..P<p60>x?31x0.01x2100x1600...??15x?x....習題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.問limxnn n..,當時,..解limxnn0.....|0||cos|21>0,要使|x,只要1,n1

.取N[1],..n.當nN[1]n ..3...lim10;..nn2lim3n13;n2n12....limn2a2..nnn4>lim999nn個..要使|10|1,只須n21,即n1...n2 n2 ..證明N[1],當時,有|10,所以lim10...n2nn2..要使|311,只須1,即n1...2n12n>n 4n ..證明N[1],當時,有|3|,所以lim13...2 2 2 22n122n2n122 a2..要使|nannannn<a an2a2n> n,只須n .....2證明N[a2],時,有|n2a2,limn2a2...n nn..110n,只須110n1<,即n1lg1...證明N1],時,,所以lim0.9999.n＿＿＿n個4.limuna,lim|un||a|...極限.nn..證明limuna,當n>N有|una|,從而n..這就證明了lim|un.n...數(shù)列|有極限,數(shù){x}未必有極限.例如lim|>n1,但lim>n不存在...nn....5.又limy0,證明:limxy0...nnnnn..證明..又limynn0,當時,有|ynM.從而當時,有..所以limxnynn0.|xnynxnynM|ynM,M..6.證明:證明 當有||<;時,|<..|<...習題131.lim<3x8;x3lim<5x;x2..2limx244;..x2x2..3lim14x32...x12x2證明,只須|x1.3證明0,1,當時,,所以lim<3x8.3 x3,只須|x2|1.5證明0,1,當時,,所以lim<5x...52x4<>2x24x4xx<2>|,要使x22x4<4>2,只須..|x<2>|.x2x2x22 2..證明0,,當時,有x4<4>,limx44...x23 3x2x2..14x22x2|2|x<1>|,要使14x2,只須|x<11...2x12 2x1 2 2..證明0,1,當0|x<1>|時,有14x32,lim4x32.3..2 22.32x1x12x12..lim1x1;..x2x3 2..limsinx0.xx證明1證明1x3 1 1x3x31 ,要使1x3 1,只須 11.2x322x32|x|32x322|x|3..|x|3..證明0,X1,時,有1x3 1,所以lim1x1...x證明x32x3 2x2x3 23sinx3sinx0|sinx|1,要使sinx0,只須1,即x1xx xX1,當xX時,x有sinxx0,所以lim2sinx0...2 xx3.時, |y4|<0.解即|x2|0.0010.0002,則當5..4.當yx1,問X等于多少,時,..2x232..解要使x4 0.01,只|x|43397,X397...2x232x230.01..5.當..6.求f<x>x,x<x>|x|當并說明它們在x..證明..limf<x>limxlim,..x0x0xx0..limf<x>limxlim11,..x0x0xx0..limx0f<x>limx0f<x>,..所以極限limf<x>存在.x0因為lim<x>lim|x|limx1,..x0x0xx0x..lim<x>lim|x|limx,..x0x0xx0x..lim<x>lim<x>,..x0x0..所以極限lim<x>不存在.x07.若x時,A,則limf<x>A.x..證明limxf<x>A,limxf<x>A,..;.,即limf<x>A.x8.自存在并且相等.證明使當時,有...時都有....有|;.時,,從而有|,9.解x時函數(shù)極限的局部有界性的定理如果x時的極限存在則存在X0及M0時證明設(shè)X0時所以這就是說存在X0及時其中..習題141.解例如,當時,但lim<x>2,<x>不是無窮小...2.2x0<x>3<x>..yx9當..x3yxsin1當x..證明x3時|y|x29x3|.0,,當時,有..x3|y|x29x3|,..2所以當時yx29為無窮小.x3..x3時|y||x||sin1x0|.因為0,,當時,有x..所以當時yxsin1為無窮小.x|y||x||sin1x0|,x..3.y12x為當x0時的無窮大.問x應(yīng)滿足什么條件,能使x..證明|y|12x2112,只須12M,即|x|1 ...x x |x||x|M2..證明1M2

,使當時,有12xM,x..所以當時,函數(shù)y12x是無窮大.x..則11042

.當0|x0|11042時,..4.lim2x1;nx2lim1x.x01x..解2x121,而當時1是無窮小,所以lim2x12...x x x2nx2..1x1xx時x所以lim1xx01x1...5.填寫下表:6.yxcosxx么？解函數(shù)yxcosxx,例如1,2,當k當時,函數(shù)xN,N的x,y<2k><2k>cos<2k>01,2,2 2 2對任何大的N,當k總有x2kN,27.函數(shù)y1sin1在區(qū)間<0,x x證明y1sin1x x在<0,例如當..時,有xk12k21,2,..當k充分大時,y<xk>2k,2..時,函數(shù)y1sin1不是無窮大.x x使例如可取1xk1,2,2k當k充分大時,..習題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<分子次數(shù)低于分母次數(shù),....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解因為lim<x00,所以limx32x2...x2x32x216x2<x..limx2;..x2x1解limx2x2x..3>lim2x3x>.x解lim<2x3x>x3.limx21;x0x解limx210時,而1x0 x xlimarctanx.xx..解limarctanxlim1arctanx0x時,1是無窮小,而x..xxxx x..4.3中的<2>...習題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為正整數(shù)xx..解lim1>kxlim1><x><k>ek...xxxx..3.解4...lim1;n..證明因為111,..n n而 lim11且lim1>1,..由極限存在準則limn1.n..lim1n21n21n2n1;..證明n2n2n1n21n21n2n2 n2..而 lim n21,limn2,..n2nn2..所以 lim1n21n21n2n....2222, 222,....證明2,22,3,....時22,時,....2,3,2222,..xxn n2xn nx222,..n1 n22..limnx;證明時,則有,,..從而有x|n1xx|...因為 x|>x|>,..x0根據(jù)夾逼準則,有x0..limlimnx.limx1.x0x證明1111,所以1x1.x x x x又因為limx>lim,根據(jù)夾逼準則,有l(wèi)imx1...x0x0x0x..習題171.時..解 limx2x3limxx20,..x02xx2x02x..所以當x0時2.時無窮小<2>1x2>是否同階？是否等價？222..3解 lim1x3limxx>xx2>3,..x11x1x..所以當x1時,1x2>..lim2x11x1x>1,2..所以當時,和1x2>是同階的無窮小,23.當時有:x2secx.2..證明limarctanxlimy令x,則當時,y..x0xy0tany..所以當時limsecx2lim1cosxlim2sin2x2lim2sinx2,..x0所以當時,1x22secxx0x2cosxx2.2x0 x22x0 x2..4.limtan3x;..x0lim2xsin<xn>m為正整數(shù)>;..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~,證明<1>lim,所以~;若則lim,從而lim.;..若~,limlimlim.....習題181...?x2f<x>??2x0x1;x2....?xf<x>?x1.|x|1..解..在x1處,因為limf<x>limx21,limf<x>lim<2x>....所以limf<x>1,處是連續(xù)的.和..在處,因為函數(shù)在處間斷,limx1f<x>lim11f,x1limx1f<x>limx1x1f,所以..在x1處,因為limf<x>limxf<1>,limf<x>lim11f<1>,x1處..連續(xù)...在處間斷,2.說明這些間斷點屬于哪一類,則補充或改變函數(shù)的定義使它連續(xù):2..yx,x2;..x23x2..yxtanx,xk,x2>;..ycos21,x....?xy?xx1x,x....解<1>yx2<x.因為函數(shù)在和x2和是函數(shù)..x23x2<x2><x....的間斷點.因為limylimx2,x2..x2x2x23x2..因為limylim<x2,x1在..<x2>..令x1處成為連續(xù)的.x2..因limx故..xktanx..因為limx,limx0所以和x是第一類間斷點且是可..x0tanx去間斷點.xktanx 22..令x時,則函數(shù)在x處成為連續(xù)的.2 2ycos21在x0處無定義,x0ycos21的間斷點.又因為x xlimcos21不存在,x0..x0xlimf<x>lim<x0limf<x>lim<3x>2,所以x1是函數(shù)的第一類不可去間斷..點.2n..3.f<x>lim1xn1x2nx的連續(xù)性,..2n ?x|x|1..解f<x>limxn1x2nx?0???x??|x|1.|x|1..處,因為第一類不可去間斷點.limx1f<x>lim<x>1,x1limx1f<x>limx1x1,所以為函數(shù)的..x1limf<x>limx,limf<x>lim<x>1,所以..類不可去間斷點...4.證明所以limf<x>f<x0>0,由極限的局部保號性定理,xx0。 。存在x0的某一去心鄰域U<x0>,使當xU<x0>這就是說,則存..5...1,,21,是n..R上處處不連續(xù),R上處處連續(xù);R上處處有定義,..解函數(shù)f<x>csc<x>csc在點1,2,x1,,21,且這些點是n..函數(shù)的無窮間斷點...?f<x>?1?在R上處處不連續(xù),在R上處處連續(xù)...?1 ..?f<x>?x?在R上處處有定義,x0..?x..習題1931.f<x>x33x2x3的連續(xù)區(qū)間,并求極限limf<x>,limf<x>及l(fā)imf<x>...x2x6x0x3x2..3解f<x>x33x2x3<x3><x,x2和..x2x6<x3><x..的,..處,limf<x>f<0>1...x0和處,limf<x>lim<x3><x,2limf<x>lim<x8...x2x2<x3><x2>x3x3x2 5..2.證明函數(shù)..證明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因為>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是初等函數(shù),x0所以....limx0x22x5f<0>022055...有定義,4lim<sin2x>3f<><sin2>3.x4 44x有定義,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,..數(shù)？解只須處連續(xù),即只須..limx0f<x>limx0f<x>f<0>a.....因為limx0f<x>limex,x0limx0f<x>lim<ax>a,所以只須取x0..習題1.至少有一個根介于1和2之間.證明即x1和2之間的根.1和2之間.2.其中證明設(shè)則若則說明的一個不超過ab若則使x也是方程的根.總之,至少有一個正根,ab.3.設(shè)數(shù)fx對于閉區(qū)間[a,b的任意兩點xy,恒|fxfy||x|,其中L為正常數(shù),且證明:使得證明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處左連續(xù),b所以因為且使得4.,使..f>f<x1>f<x2>f<xn>.n..證明nmf<x1>f<x2>f<xn>nM,..mf<x1>f<x2>f<xnn>M...由介值定理推論,使..f>f<x1>f<x2>f<xn>.n..5.若且limf存在,則x證明令limf<x>A,0,存在X0,就有x,即.又由于存在取即6...總習題一1."充分"必要"充分必要三者中選擇一個正確的填入下列空格內(nèi):條件.有界的的條件.limf<x>存在的條件.xx0limf<x>條件.xx0<3>limf<x>的條件.xx0limf<x>條件.xx0limf<x>存在xx0的條件.解 <1>必要,充分.<2>充分.<3>充分.<4>3.選擇以下題中給出的四個結(jié)論中一個正確的結(jié)論:時,有< >.x是等價無窮小; x同階但非等價無窮小;x高階的無窮小; x低階的無窮小.解因為limf<x>lim2x3x2lim2x1lim3x..x0 xx0xx0 xx0 x..ln2lim tln3lim uln2ln3<令..t0ln1t>u0ln1u>..所以f<x>與x同階但非等價無窮小.故應(yīng)選4.設(shè)求下列函數(shù)的定義域:x>;x>.解 <1>由得x0,<2>由0lnx1得1xe,即函數(shù)f<ln<3>由0arctanx1得即函數(shù)f<arctantan1].<4>由0cosx1得x<n0,1,2,>,2 2..即函數(shù),2n2<n0,1,2,..5.設(shè)?f<x>?0??xx0,x0?g<x>?0?x2x0,x0..求x0..解因為所以??xx0;..因為g<x>0,所以因為g<x>0,所以x0..2<x>?.x2 x06.利用ysinx的圖形作出下列函數(shù)的圖形:x|;<3>y2sinx.27.把半徑為R的一圓形鐵片,自中心處剪去中心角為的一扇形后圍成一無底圓錐.試將這圓錐的體積表為的函數(shù).解設(shè)圍成的圓錐的底半徑為r,高為h,依題意有>,r..h圓錐的體積為R2r2R2R2>2R22.....V1R2>2R2..3 2 ..R3R>2a2..28.根據(jù)函數(shù)極限的定義證明limx2x65...x3x3..證明 對于任意給定的0,要使|x2x65|,只需|x3|,,x3..當就有|x3|,即|x2x65|,所以limx2x65...9.求下列極限: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:用等價無窮小換>.2....lim<axbxcxa1 xa>xlibxcx33>axbxcx33axbxcx33x

,因為..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

,因為..x2x2..1li<sinx>sinx1e,x2lim<sinxtanxlimsinx..x2x2cosx..,limsin<sin2x>limsinxcosx0,..xcos<sinx>2所以 lim<sinx>tanxe01.x2x2sinx1..?xsin110.設(shè)f<x>?xx0,要使應(yīng)怎樣選擇數(shù)a?..ax2 x0解要使函數(shù)連續(xù),必須使函數(shù)在x0..因為limf<x>limax2>a,limf<x>limxsin10..x0x0x0x0x..所以當a0時,f<x>在處連續(xù).因此選取a0時,1 ..11.設(shè)f<x>x1x0

,求并說明間斷點所屬類形...ln>x0..解因為函數(shù)x1所以x11..因為limf<x>limex10<提示lim1>,..1x..limf<x>limex1<提示lim1>,..xx1x..所以x1..又因為limf<x>limln<x0,lim11f<x>limex1 ,1..x0x0x0x0e..所以x0且為第一類間斷點...證明n1 n21n221n2n1...證明因為n n2n1 n21n221 n2nn ,且n2..limnnn2nlimn11n1,limnnn2limn11n21,..所以n1 n21n221n2n1...13.證明方程sin,>內(nèi)至少有一個根.22證明設(shè)則函數(shù)f<x>在[,]上連續(xù).22..因為f>11,f<>12,f>f<>0,..2 2 2 2 22 2 2..所以由零點定理,在區(qū)間,>內(nèi)至少存在一點,使這說明方程sin22xx10,>內(nèi)至少有一個根.2214.如果存在直線使得當曲線yf<x>上的動點L的距離d<M,則稱L為曲線當直線L的斜率k0時,稱L為斜漸近線.直線ykxb..k limxf<>,bxlimx[f<x>kx]...<x,x><x,x>..1<2>線y2xex的斜漸近線.證明<1>僅就按漸近線的定義f<x><kxb>]0x必要性設(shè)是曲線則f<x><kxb>]0x..于是有l(wèi)imf<x>kb]0limf<x>k0klimf<x>..xx xxxxx..同時有f<x>kxb]0bf<x>kx]..x充分性如果klimf<x>xbf<>k],則..xxx..f<x><kxb>]f<x>kxb]f<x>kx]bbb0..xxx..因此是曲線xklimylim2x1e12x..xxxx..11..bli[y2]li[2xex2]2limex>12lim t11..xxxt0lnt>..1所以曲線y2xex的斜漸近線為y2x1..習題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如果且證明證明當所以flimf<x>flimfflimfff..x0x0x0x0x0x0..從而有2f即f10x解因為x23..kcos21kcos1..1 3 2 2x<,1>處的切線方程和法線方程式32..解sin3..x3 3 2..故在點<,1>處y13<x>..32 2 2 3法線方程為y12<x>2 3 3..12即13的兩點一點的切線平行于這條割線？解割線斜率為k914..令得31 2..14x|..?1xy?x2sin?1x0解x0x0..

limylim|sinlim<sinx>0

limylim|sinlimsinx0..x0所以函數(shù)在處連續(xù)又因為x0x0x0x0x0..limy<x>lim|sinx||sin0|limsinx1..x0x0x0x0x0 x..<0>limy<x>lim|sinx||sin0|limsinx1..x0x0x0x0x0x..解因為limy<x>limx2sin10又處連續(xù)..又因為xx2sin10..limy<x>limx limxsin10..x0x0x0 xx0 x..所以函數(shù)在點且..?15f<x>?x2?x1為了使函數(shù)處連續(xù)且可導ab..解因為?axbx....limf<x>limx21limf<x>limb>ab..所以要使函數(shù)在必須時..f>limx22flimaxblima<xablima<xa..xx10xxx10x..所以要使函數(shù)在必須此時..?16已知f<x>?x2?xx0x0又f..解因為limf<x>flimx01limf<x>flimx200..x0xx0 xx0xx0 x..所以f..?17x?x0求f..?x x0解當時xfx當時f..因為limf<x>flimsinx01..x0xx0 x..limf<x>flimx01所以f從而..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..習題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這就是物體達到最高點的時刻0 g5解時所以所求的切線方程為所求的法線方程為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>的導數(shù)....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設(shè)求下列函數(shù)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>..習題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的二階導數(shù)解fd2y:dx2....y1f<x>f..yff<x>ffff<x>[f..[f4試從dx1導出[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求物體運動的加速度并驗證d2s..dt22s0..解dstdtd2sA2sintdt2d2s就是物體運動的加速度dt2..d2sdt22sA2sint2Asint0..6..解 7解 xx8n階導數(shù)的一般表達式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則有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..習題241y的導數(shù)dydx解2y0于是 ..yy yx..于是 yayx2y2axyx于是 yexyyxexy于是 ..yey xey222..2x3y3a3在點<解242處的切線方程和法線方程4..1 12x32y3y03 311于是 yx31y3..在點<242處4..所求切線方程為..y2a<x42即xy42a2..所求法線方程為..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..解兩邊求導得1ylnxx1ln1>x1..y于是 y<xx>x[lnx1]1x..xxx..lny1ln|x5|1ln<x25 25兩邊求導得..1y1112x..y 5x525x22..于是 y155x55x22[1x5152x]x22..lny1ln<x2>4ln<3x>5ln<x2..兩邊求導得1yy143x5x1..于是 yx2<3x>4[1 45]..<x2<xx3x1..lny1lnx1lnsinx1lnex>2 2 4兩邊求導得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.求當t3時dydx的值..解dytcosttsintcostsint..dx sintcost13sintcost..當t時dy223

32..3 dx133..2 27..?xsint??ycost??x在t處4..???y?1t2在3at21t2..解>dy2sint..dx當t時4dydxcost2sin<2>4cos4222222y2 00..所求切線方程為y2所求法線方程為2<x2>即222xy20..y12<x22>即22x4y10..att2>3at2t2>2t2>2t2>attt2>23at21t2>2..dyatt..dx 3at21t2..當dy224x6ay12a..dx所求切線方程為1223 05 0 5..y12a4<x6即5 3 5所求法線方程為y12a3<x6即..5 4 5d2y..8dx2..xt2?2;..?y.??xacost;??ybsint??x?..?y..?xft??ytft<t>f?設(shè)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問在2解對應(yīng)圓面積為S兩邊同時對t求導得S時故S其速率為其表面上升的速度為多少？解 h時r1hS12 4水的體積為V1hS1h1h33 3 4 12..dVdhdh4dV..dt 12dt dth2dt..dV4因此dtdhdt4h2

dVdt4416..1218cm直徑12cm的正圓錐形漏斗中漏入一直徑為10cm的圓柱形筒中開始時其表面下降的速率為問此時圓柱形筒中溶液表面上升的速率為多少？解設(shè)在ty圓柱形筒中水深為h1621r2y52h3 3..由ry得ry代入上式得..61831621<y>2y52h..3 3 3即 1621y352h3 兩邊對t1y22時1..23252160.6425..2711解..2y解3y12xxyxsin..yx x21..e2x..yarcsinx2..yarctanx2x2sA是常數(shù)>..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的長為s跨度為2lO與桿頂連線AB的距離為f..則電纜長可按下面公式計算s2f2>當f時..解dS2f28..6如果R減少改變了多少？又如果不變R增加問扇形面積大約改變了多少？解S1R22SdS<1R21R22 2..303360代入上式得..1<2360

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

310.87467..6180

6 6 180

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

4180180..8..解f當x時有..arcsin<xx>arcsinx所以11x2x..arcsin0.5002arcsin<0.5arcsin0.5110.520.0002..620.00023..f當x時有..x>arccosx所以11x2x..arccos0.4995arccos<0.50.0005>arccos0.5110.52..320.00053..9當x較小時<2>..1xx..并計算和的近似值x則有xxx則有取f<x>1則有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要求精確度在2%以內(nèi)問這時測量直徑D的相對誤差不能超過多少？解球的體積為V1dV1D2D因為計算球體體積時所6 2以其相對誤差不超過2%即要求1D2DdVD..2V 1D363D2%..所以 D2%3..也就是測量直徑的相對誤差不能超過2%312要求中心角為55測量弦長l問此而引起的中心角測量誤差..解由lRsin得2arcsinl2arcsinl..2 2 2R400..l2Rsin2 1 1 ..2400l..時1<l>2400..211<184.7>240014000.10.00056..總習題二1在充分"要"和"充分必要"三者中選擇一個正確的填入下列空格內(nèi)..的 條件條件....解必要條件..2設(shè)xa則xalimf1>f存在limff存在..hhh0 h..limff