高等數(shù)學(xué)-同濟(jì)第六版-（上冊）課后復(fù)習(xí)題全解

..習(xí)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(gè)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?yàn)閷γ總€(gè)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?yàn)閇2,>...y3 31 ;x2..解由函數(shù)的定義域?yàn)?.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?yàn)閷?yīng)法則不同時(shí)因?yàn)槎x域、對應(yīng)法則均相相同.因?yàn)槎x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?yàn)樗宰C明..兩個(gè)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(dāng)0a1時(shí)當(dāng)a1時(shí)無解.因此當(dāng)0a1時(shí)2 2 2函數(shù)的定義域?yàn)閇a當(dāng)a1時(shí)函數(shù)無意義.2..?1 |x1..?18.設(shè)f<x>?0?1|并作出這兩個(gè)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(dāng)過水?dāng)嗝鍭BCD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(tái)以上的,每多訂購1臺(tái),1分,元.pxPx臺(tái),解時(shí),令0.當(dāng)時(shí),..?pp??900.01x750x100x.x..?30x?0x100..P<p60>x?31x0.01x2100x1600...??15x?x....習(xí)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(dāng)時(shí),..解limxnn0.....|0||cos|21>0,要使|x,只要1,n1

.取N[1],..n.當(dāng)nN[1]n ..3...lim10;..nn2lim3n13;n2n12....limn2a2..nnn4>lim999nn個(gè)..要使|10|1,只須n21,即n1...n2 n2 ..證明N[1],當(dāng)時(shí),有|10,所以lim10...n2nn2..要使|311,只須1,即n1...2n12n>n 4n ..證明N[1],當(dāng)時(shí),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(shí),有|n2a2,limn2a2...n nn..110n,只須110n1<,即n1lg1...證明N1],時(shí),,所以lim0.9999.n＿＿＿n個(gè)4.limuna,lim|un||a|...極限.nn..證明limuna,當(dāng)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(dāng)時(shí),有|ynM.從而當(dāng)時(shí),有..所以limxnynn0.|xnynxnynM|ynM,M..6.證明:證明 當(dāng)有||<;時(shí),|<..|<...習(xí)題131.lim<3x8;x3lim<5x;x2..2limx244;..x2x2..3lim14x32...x12x2證明,只須|x1.3證明0,1,當(dāng)時(shí),,所以lim<3x8.3 x3,只須|x2|1.5證明0,1,當(dāng)時(shí),,所以lim<5x...52x4<>2x24x4xx<2>|,要使x22x4<4>2,只須..|x<2>|.x2x2x22 2..證明0,,當(dāng)時(shí),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(dāng)0|x<1>|時(shí),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(shí),有1x3 1,所以lim1x1...x證明x32x3 2x2x3 23sinx3sinx0|sinx|1,要使sinx0,只須1,即x1xx xX1,當(dāng)xX時(shí),x有sinxx0,所以lim2sinx0...2 xx3.時(shí), |y4|<0.解即|x2|0.0010.0002,則當(dāng)5..4.當(dāng)yx1,問X等于多少,時(shí),..2x232..解要使x4 0.01,只|x|43397,X397...2x232x230.01..5.當(dāng)..6.求f<x>x,x<x>|x|當(dāng)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?yàn)閘im<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(shí),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(dāng)時(shí),有...時(shí)都有....有|;.時(shí),,從而有|,9.解x時(shí)函數(shù)極限的局部有界性的定理如果x時(shí)的極限存在則存在X0及M0時(shí)證明設(shè)X0時(shí)所以這就是說存在X0及時(shí)其中..習(xí)題141.解例如,當(dāng)時(shí),但lim<x>2,<x>不是無窮小...2.2x0<x>3<x>..yx9當(dāng)..x3yxsin1當(dāng)x..證明x3時(shí)|y|x29x3|.0,,當(dāng)時(shí),有..x3|y|x29x3|,..2所以當(dāng)時(shí)yx29為無窮小.x3..x3時(shí)|y||x||sin1x0|.因?yàn)?,,當(dāng)時(shí),有x..所以當(dāng)時(shí)yxsin1為無窮小.x|y||x||sin1x0|,x..3.y12x為當(dāng)x0時(shí)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(dāng)時(shí),有12xM,x..所以當(dāng)時(shí),函數(shù)y12x是無窮大.x..則11042

.當(dāng)0|x0|11042時(shí),..4.lim2x1;nx2lim1x.x01x..解2x121,而當(dāng)時(shí)1是無窮小,所以lim2x12...x x x2nx2..1x1xx時(shí)x所以lim1xx01x1...5.填寫下表:6.yxcosxx么？解函數(shù)yxcosxx,例如1,2,當(dāng)k當(dāng)時(shí),函數(shù)xN,N的x,y<2k><2k>cos<2k>01,2,2 2 2對任何大的N,當(dāng)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(dāng)..時(shí),有xk12k21,2,..當(dāng)k充分大時(shí),y<xk>2k,2..時(shí),函數(shù)y1sin1不是無窮大.x x使例如可取1xk1,2,2k當(dāng)k充分大時(shí),..習(xí)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?yàn)閘im<x00,所以limx32x2...x2x32x216x2<x..limx2;..x2x1解limx2x2x..3>lim2x3x>.x解lim<2x3x>x3.limx21;x0x解limx210時(shí),而1x0 x xlimarctanx.xx..解limarctanxlim1arctanx0x時(shí),1是無窮小,而x..xxxx x..4.3中的<2>...習(xí)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?yàn)?11,..n n而 lim11且lim1>1,..由極限存在準(zhǔn)則limn1.n..lim1n21n21n2n1;..證明n2n2n1n21n21n2n2 n2..而 lim n21,limn2,..n2nn2..所以 lim1n21n21n2n....2222, 222,....證明2,22,3,....時(shí)22,時(shí),....2,3,2222,..xxn n2xn nx222,..n1 n22..limnx;證明時(shí),則有,,..從而有x|n1xx|...因?yàn)?x|>x|>,..x0根據(jù)夾逼準(zhǔn)則,有x0..limlimnx.limx1.x0x證明1111,所以1x1.x x x x又因?yàn)閘imx>lim,根據(jù)夾逼準(zhǔn)則,有l(wèi)imx1...x0x0x0x..習(xí)題171.時(shí)..解 limx2x3limxx20,..x02xx2x02x..所以當(dāng)x0時(shí)2.時(shí)無窮小<2>1x2>是否同階？是否等價(jià)？222..3解 lim1x3limxx>xx2>3,..x11x1x..所以當(dāng)x1時(shí),1x2>..lim2x11x1x>1,2..所以當(dāng)時(shí),和1x2>是同階的無窮小,23.當(dāng)時(shí)有:x2secx.2..證明limarctanxlimy令x,則當(dāng)時(shí),y..x0xy0tany..所以當(dāng)時(shí)limsecx2lim1cosxlim2sin2x2lim2sinx2,..x0所以當(dāng)時(shí),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(xí)題181...?x2f<x>??2x0x1;x2....?xf<x>?x1.|x|1..解..在x1處,因?yàn)閘imf<x>limx21,limf<x>lim<2x>....所以limf<x>1,處是連續(xù)的.和..在處,因?yàn)楹瘮?shù)在處間斷,limx1f<x>lim11f,x1limx1f<x>limx1x1f,所以..在x1處,因?yàn)閘imf<x>limxf<1>,limf<x>lim11f<1>,x1處..連續(xù)...在處間斷,2.說明這些間斷點(diǎn)屬于哪一類,則補(bǔ)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?yàn)楹瘮?shù)在和x2和是函數(shù)..x23x2<x2><x....的間斷點(diǎn).因?yàn)閘imylimx2,x2..x2x2x23x2..因?yàn)閘imylim<x2,x1在..<x2>..令x1處成為連續(xù)的.x2..因limx故..xktanx..因?yàn)閘imx,limx0所以和x是第一類間斷點(diǎn)且是可..x0tanx去間斷點(diǎn).xktanx 22..令x時(shí),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(diǎn).又因?yàn)閤 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(diǎn).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?yàn)榈谝活惒豢扇ラg斷點(diǎn).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(diǎn)...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(dāng)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(diǎn)1,2,x1,,21,且這些點(diǎn)是n..函數(shù)的無窮間斷點(diǎn)...?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(xí)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?yàn)?gt;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?yàn)閘imx0f<x>limex,x0limx0f<x>lim<ax>a,所以只須取x0..習(xí)題1.至少有一個(gè)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(gè)不超過ab若則使x也是方程的根.總之,至少有一個(gè)正根,ab.3.設(shè)數(shù)fx對于閉區(qū)間[a,b的任意兩點(diǎn)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?yàn)榍沂沟?.,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(xí)題一1."充分"必要"充分必要三者中選擇一個(gè)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(gè)結(jié)論中一個(gè)正確的結(jié)論:時(shí),有< >.x是等價(jià)無窮小; x同階但非等價(jià)無窮小;x高階的無窮小; x低階的無窮小.解因?yàn)閘imf<x>lim2x3x2lim2x1lim3x..x0 xx0xx0 xx0 x..ln2lim tln3lim uln2ln3<令..t0ln1t>u0ln1u>..所以f<x>與x同階但非等價(jià)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?yàn)樗??xx0;..因?yàn)間<x>0,所以因?yàn)間<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(dāng)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(jià)無窮小換>.2....lim<axbxcxa1 xa>xlibxcx33>axbxcx33axbxcx33x

,因?yàn)?.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?yàn)?.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?yàn)閘imf<x>limax2>a,limf<x>limxsin10..x0x0x0x0x..所以當(dāng)a0時(shí),f<x>在處連續(xù).因此選取a0時(shí),1 ..11.設(shè)f<x>x1x0

,求并說明間斷點(diǎn)所屬類形...ln>x0..解因?yàn)楹瘮?shù)x1所以x11..因?yàn)閘imf<x>limex10<提示lim1>,..1x..limf<x>limex1<提示lim1>,..xx1x..所以x1..又因?yàn)閘imf<x>limln<x0,lim11f<x>limex1 ,1..x0x0x0x0e..所以x0且為第一類間斷點(diǎn)...證明n1 n21n221n2n1...證明因?yàn)閚 n2n1 n21n221 n2nn ,且n2..limnnn2nlimn11n1,limnnn2limn11n21,..所以n1 n21n221n2n1...13.證明方程sin,>內(nèi)至少有一個(gè)根.22證明設(shè)則函數(shù)f<x>在[,]上連續(xù).22..因?yàn)閒>11,f<>12,f>f<>0,..2 2 2 2 22 2 2..所以由零點(diǎn)定理,在區(qū)間,>內(nèi)至少存在一點(diǎn),使這說明方程sin22xx10,>內(nèi)至少有一個(gè)根.2214.如果存在直線使得當(dāng)曲線yf<x>上的動(dòng)點(diǎn)L的距離d<M,則稱L為曲線當(dāng)直線L的斜率k0時(shí),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(shí)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(xí)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(dāng)所以flimf<x>flimfflimfff..x0x0x0x0x0x0..從而有2f即f10x解因?yàn)閤23..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(diǎn)<,1>處y13<x>..32 2 2 3法線方程為y12<x>2 3 3..12即13的兩點(diǎn)一點(diǎn)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?yàn)閤0x0x0x0x0..limy<x>lim|sinx||sin0|limsinx1..x0x0x0x0x0 x..<0>limy<x>lim|sinx||sin0|limsinx1..x0x0x0x0x0x..解因?yàn)閘imy<x>limx2sin10又處連續(xù)..又因?yàn)閤x2sin10..limy<x>limx limxsin10..x0x0x0 xx0 x..所以函數(shù)在點(diǎn)且..?15f<x>?x2?x1為了使函數(shù)處連續(xù)且可導(dǎo)ab..解因?yàn)?axbx....limf<x>limx21limf<x>limb>ab..所以要使函數(shù)在必須時(shí)..f>limx22flimaxblima<xablima<xa..xx10xxx10x..所以要使函數(shù)在必須此時(shí)..?16已知f<x>?x2?xx0x0又f..解因?yàn)閘imf<x>flimx01limf<x>flimx200..x0xx0 xx0xx0 x..所以f..?17x?x0求f..?x x0解當(dāng)時(shí)xfx當(dāng)時(shí)f..因?yàn)閘imf<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(xí)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(dá)到最高點(diǎn)的時(shí)刻0 g5解時(shí)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(dǎo)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(xí)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(dǎo)數(shù)解fd2y:dx2....y1f<x>f..yff<x>ffff<x>[f..[f4試從dx1導(dǎo)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(yùn)動(dòng)的加速度并驗(yàn)證d2s..dt22s0..解dstdtd2sA2sintdt2d2s就是物體運(yùn)動(dòng)的加速度dt2..d2sdt22sA2sint2Asint0..6..解 7解 xx8n階導(dǎo)數(shù)的一般表達(dá)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(xí)題241y的導(dǎo)數(shù)dydx解2y0于是 ..yy yx..于是 yayx2y2axyx于是 yexyyxexy于是 ..yey xey222..2x3y3a3在點(diǎn)<解242處的切線方程和法線方程4..1 12x32y3y03 311于是 yx31y3..在點(diǎn)<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(dǎo)得1ylnxx1ln1>x1..y于是 y<xx>x[lnx1]1x..xxx..lny1ln|x5|1ln<x25 25兩邊求導(dǎo)得..1y1112x..y 5x525x22..于是 y155x55x22[1x5152x]x22..lny1ln<x2>4ln<3x>5ln<x2..兩邊求導(dǎo)得1yy143x5x1..于是 yx2<3x>4[1 45]..<x2<xx3x1..lny1lnx1lnsinx1lnex>2 2 4兩邊求導(dǎo)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(dāng)t3時(shí)dydx的值..解dytcosttsintcostsint..dx sintcost13sintcost..當(dāng)t時(shí)dy223

32..3 dx133..2 27..?xsint??ycost??x在t處4..???y?1t2在3at21t2..解>dy2sint..dx當(dāng)t時(shí)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(dāng)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(shí)對t求導(dǎo)得S時(shí)故S其速率為其表面上升的速度為多少？解 h時(shí)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(shí)其表面下降的速率為問此時(shí)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(shí)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(jì)算s2f2>當(dāng)f時(shí)..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(dāng)x時(shí)有xx所以..cos<>cossin<>

310.87467..6180

6 6 180

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

4180180..8..解f當(dāng)x時(shí)有..arcsin<xx>arcsinx所以11x2x..arcsin0.5002arcsin<0.5arcsin0.5110.520.0002..620.00023..f當(dāng)x時(shí)有..x>arccosx所以11x2x..arccos0.4995arccos<0.50.0005>arccos0.5110.52..320.00053..9當(dāng)x較小時(shí)<2>..1xx..并計(jì)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(shí)測量直徑D的相對誤差不能超過多少？解球的體積為V1dV1D2D因?yàn)橛?jì)算球體體積時(shí)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(shí)1<l>2400..211<184.7>240014000.10.00056..總習(xí)題二1在充分"要"和"充分必要"三者中選擇一個(gè)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