高数课后习题答案及其解析.pdf

该【高数课后习题答案及其解析 】是由【小屁孩】上传分享，beplayapp体育下载一共【7】页，该beplayapp体育下载可以免费在线阅读，需要了解更多关于【高数课后习题答案及其解析 】的内容，可以使用beplayapp体育下载的站内搜索功能，选择自己适合的beplayapp体育下载，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此beplayapp体育下载到您的设备，方便您编辑和打印。:..:①定义域不同;②定义域对应法则相同同;()?2,f(?)?1??①x2?1?y2,x??1?y2,0?y?1yyyycea②lnea?ln(bx?c),ea?bx?c,bx?c?ea,x??,???y???bby?lnu,u?v2?3,v?sinxy?arctanu,u?2x3?①②:①无穷大②无穷小③负无穷大④负无穷大⑤无穷小⑥:⑴lim(3x3?2x?1)?3limx3?2limx?1?2x?1x?1x?1lim(x2?3)x2?31⑵lim?x?2?x?22x?1lim(2x?1)5x?2atanx⑶lim?0x??x2x?32x?2?12(x?1)1lim?lim?lim?limx?1x2?5x?4x?1(x?4)(x?1)x?1(x?4)(x?1)x?1(x?4)(x?1)⑷2???????3x2?5x?6(x?3)(x?2)x?31⑸lim?lim?lim??x?2x2?4x?2(x?2)(x?2)x?2x?24x2x2(1?1?x2)⑹lim?limx?01?1?x2x?0(1?1?x2)(1?1?x2)x2(1?1?x2)?lim??lim(1?1?x2)??2x?0?x2x?011?x2?1x21⑺lim?lim?x???3x?1x???133?x2x??2?3???2?3x?3?2x?3?1⑻lim?lim?x???4?3x?5?2xx????2?x24?5????3??13?x2?x?2(x?1)(x?2)?3⑼lim????lim?lim???1x??1?x?1x3?1?x??1x3?1x??1(x?1)(x2?x?1)3(n?1?n)(n?1?n)1⑽lim(n?1?n)?lim?lim?0n??n??(n?1?n)n??n?1?n:..?132n?1?n2⑾lim???????lim?1x???n2n2n2?x??n2111???111?2n2⑿lim?1??????lim?2?2222n?1n??n??1?⑴lim?x?0tan4x41sin1x⑵limxsin?lim?1x??xx??1x1?cos2x2sin2x2sinx⑶lim?lim?lim?2x?0xsinxx?0xsinxx?0xx⑷lim2n?sin?xn??2narcsinx1y1⑸lim?lim?x?02x2y?0siny21sin11x2⑹limxsin??limx2sin??lim??1x???x2x???x2x???1x2k11?2?(?k)?2?(?k)⑺lim(1?x)x?lim(1?x)x?lim[(1?x)x(1?x)?2]?e?kx?0x?0x?01?x2x1x?2????⑻lim???lim?1???e2x???x?x???x?13⑼lim(1?3tanx)cotx?1?lim[(1?3tanx)3tanx(1?3tanx)1]?e3x?0x?022x?323x424???(?)?⑽lim?1???lim[(1?)23?(1?)3]?e3x???3x?x??3x3x1??1lim(1?3?x)x?lim(1?3?x)3x?3?x?x?1?lim(1?3?x)3xx?e0?1⑾x?3x???x???x???x?3??x?3???x?111⑿lim???lim?????e?1x???1?x?x???1?1x?3e1?????lim?1???x?xx????(x?1)(x?1)1、⑴因为函数在x=1点处无定义,f(x)?,但是limf(x)??2,x=1(x?1)(x?2)x?1点是函数的第一类间断点(可去)。limf(x)??,故x=2是二类间断点。x?2:..⑵虽然在x=0处,函数无定义,但是极限存在,故x=0是第一类间断点(可去)⑶?limf(x)?lim(1?1)?0,?limf(x)?lim(3?1)?2,极限存在但是不等,x?1?x?1?x?1?x?1?x=1是第一类间断点。⑷函数在x=0处无定义,上下摆动,极限不存在,是第二类间断点。x2(x?3)?(x?3)(x?3)(x2?1)2、解:f(x)??(x?3)(x?2)(x?3)(x?2)函数的定义域是:(??,?3)?(?3,2)?(2,??)19?184?1limf(x)?,limf(x)???,limf(x)???x?02x??3?3?25x?203、lim(a?x2)?1,故a?1,limln(b?x?x2)?lnb?1,故b?ex?0?x?0?4、证:设f(x)?x5?3x?1,则f(x)在[1,2]上连续,f(1)??3,f(2)?16?6?1?9,?f(1)?f(2)?0,由推论1,知函数在[1,2]内有一个正根。????5、证:设f(x)?sinx?x?1,f(?)??1,f()?1,故?f(?)?f()?02222????由推论1,知函数在?,内有一个正根。???22?6、证:?f(1)?1?3?7?10??5,?f(2)?16?12?14?10?8,?f(1)?f(2)?0,由推论1,知函数在[1,2]内有一个与x轴相交的交点。?解①lim(x2?1)tanx?(1?1)1??14②limlnsinx?lnlimsinx?ln1?0??x?x?22??sin?cossinx?cosx221③lim???????x?x??222211④limxln(1?)?lnlim(1?)x?lne?1??1x???xx???x****题1一、填空x?11?x1、lim?lim??1x?1?x?1x?1?x?12、?x2(x?1)?0,?连续区间是[1,+∞]x2?2x?k3、lim?4,必有32?2?3?k?0,k??3x?3x?34、:..?1?1???11?2x2x1???111?2x2x?1???x?0?,???,?2x???,lim?limlim??1,xx?0?1x?0??1?1x?0?1??1?2x?1?2x?2x2x?1????1111?2x?x?0?,???,?2x?0,lim?1xx?0?11?2x故x=0是第一类间断点。35、?limxsin?1?0?1?1,函数在x=0连续必有f(0)?1故k=1x?0x二、选择题1、如图,B\C\D均错误,选A。2、因为分子最高方幂项为:a5x100,故a5?8,a?58,选C。3、A的极限不为1,B的极限为零,C的极限为无穷大,选D。sin2(x3)4、?lim?1,故必须k=。x?0xk5、(0)??1,f(1)?1,在(0,1)内有实数根。(?1)?5,f(0)??1,在(-1,2)内有实数根。。、计算1.、?xn?1?(x?1)(xn?1?xn?2???1),xn?1?lim?lim(xn?1?xn?2???1)?nx?1x?1x?122、23:..x2?x?13、当x??时,?0是无穷小量,3?sinx是有界函数,故极限为0。x3?2(n?3n?n?n)(n?3n?n?n)4、lim(n?3n?n?n)?limn??n??(n?3n?n?n)n?3n?n?n4n?lim?limn??n?3n?n?nn??n?3n?n?n4?lim?2n??111?3?1?nnx(x2?1?x)(x2?1?x)x5、lim?limx???(x2?1?x)x???(x2?1?x)11?lim?x???121??1x21sin2x6、lim?atanx??0x??21x?1xsinaxsinaxaxa7、lim?lim?sinbxsinbxbx?0x?0bx1x2tanx?sinxtanx(1?cosx)x218、lim?lim?lim?x?0x3x?0x3x?0xx221?x?1?x?119、lim?lim?lim??x?04xx?04x?(1?x?1)x?04(1?x?1)8x2?x210、lim?lim?2?x?xx?0x?02?sin2sin2222x?3x?13x?1????x?1???1???2x?3?2x2x11、lim???lim???lim??,?2x?1??2x?1??1?x??x??x??1??????2x??2x?:..233?x?31??32??lim?1???lim?1??3x???2x?x???2x?e2???e211?1??x??1?112e2lim?1???lim?1??x???2x?x???2x?ln(1?3x)ln(1?3x)12、lim?lim3??3x?0xx?03xln(1?3x)3ln(1?3x)313、lim?lim?x?02x2x?03x2n11n14、?????n2?nn2?1n2?nn2?1nnlim?1,lim?1,n??n2?nn??n2?1?111?所以lim??????1??n???n2?1n2?2n2?n?四、分析计算题1、p??5,q?0时,y为无穷小量;q?0,p为任意实数时,y为无穷大量。2、由x2?ax?b?(x?1)(x?a?1)?(b?a?1)?0,?b?a?1?0而x?1时,x?a?1??5,所以a??7,b?6?x,x?11?x2n?3、f(x)?limxf(x)??0,x?1。2n,n??1?x??x,x?1??limf(x)?lim(?x)??1?limf(x)?lim(x)?1,x?1?x?1?x?1?x?1??limf(x)?limf(x)即x=1是第一类间断点,x?1?x?1??limf(x)?lim(x)??1?limf(x)?lim(?x)?1,x??1?x??1?x??1?x??1??limf(x)?limf(x)即x=-1是第一类间断点,x?1?x?1?故此函数只有两个是第一类间断点,它们分别是x=1与x=-1。1114、4、?x?0?,???,ex???,?0,f(0)?0,x1?ex:..111?x?0?,???,ex?0,?1,f(1)?0,在x=0处右连续。x1?ex