数分5_微分中值定理及其应用

16.求证：arcsinx?arccosx??2证明：令f(x)?arcsinx?arccosx,由于 f'(x)?11?x2.?11?x2?0,

?2于是可知f(x)?c,c为常数。 又因为f(0)?0??2??2,于是可知f(x)?arcsinx?arccosx?,得证。第二节 洛比达法则

1.求下列待定性的极限：

tanax洛比达法则ab(1).lim?lim?.x?0sinbxx?0bcos2axcosbxa1?cosx2(2)lim3x?0xsinx2xsinx2sinx22x?lim2?limx?03xsinx?x3cosxx?0x23sinx?xcosx洛比达法则2x21 ?lim?lim?.x?03sinx?xcosxx?03cosx?cosx?xsinx2洛比达法则1?1洛比达法则ln(1?x)?x洛比达法则x11?x(3)lim?lim?lim?lim?1.x?0x?0?sinxx?0(1?x)sinxx?0sinx?(1?x)cosxcosx?11?12tanx?x洛比达法则1?cos2x1?cosxcosx(4)lim?lim?lim?lim?2.x?0x?sinxx?01?cosxx?0cos2x?cos3xx?0cos2x11ex?1?x洛比达法则ex?1(5)lim(?x)?lim?limxx?0xx?0e?1?xexe?1x?0x(ex?1)洛比达法则?ex1limx?.x?02e?xex2lncosax洛比达法则?asinax?cosbxa2cosbxbxsinaxa2(6)lim?lim?lim2?2.x?0lncosbxx?0cosaxx?0bsinbxbcosaxaxsinbxb12tanx?6洛比达法则1(7)lim?limcosx?lim?1.?secx?5?sinx?sinxx?x?x?2222cosx1洛比达法则11x?1?lnx洛比达法则x?111x(8)lim(?)?lim?lim?lim?lim?.x?1lnxx?1x?1lnx?2x?1x?1xlnx?x?1x?1x?1(x?1)lnx2lnx?x1? 壹拾壹

x??x洛比达法则?1x(9).lim(??x)tan?lim?lim?lim2sin?2.x??x??x???12x??cotx2x22sin2(10).limxx?111?x?limex?11ln(x1?x)?limex?1lnx1?x?elnx洛比达法则limx?11?x?e1limxx?1?1?e?1.xb(11).limaxx???e洛比达法则?bxb?1洛比达法则洛比达法则b(b?1)(b?2)?(b?[b])limax???lim?0.x???aex???a[b]?1x?b?[b?1]eax1??arctanx洛比达法则2x211?x2(12).lim?lim?lim?lim?1.x???x???x???111x???111sin?2cos(1?x2)cos(2?1)cosxxxxxxlncx洛比达法则clnc?1x洛比达法则洛比达法则c(c?1)(c?2)?(c?[c])(13).limb?lim???lim?0.x???xx???x???bxbb[c]?1xbln[c?1]?cx(14)lim?xblncx,(b,c?0);x?0?11?lim()blnct???ttcclnt?lim(?1)b?0.t???tx?1tlnx;x?0cotx6洛比达法则洛比达法则?2coxssin2x ?li?m ??lim?

x?0x3xx??3sin6??limsinx?0.x?0?32*2?3.?33;1?2sixn(15)lim?cosx3x?(16)lim?(17)lim(1?x)?e;x?0x1x1xx?ln(1?x)ln(1?x)y'1?x解：令y?(1?x),lny?,于是?.xyx2x?ln(1?x)11?x 因此有y'?(1?x)x.于是2xx?ln(1?x)1x11?xx(1?x)?ln(1?x)12(1?x)x?e洛比达法则x lim?lim?lim1?x2(1?x)xx?0x?0x?0x1x1x?(1?x)ln(1?x)x?(1?x)ln(1?x)x ?lim(1?x)?lime22x?0x?0x(1?x)x(1?x)1?洛比达法则1?ln(1?x)?1洛比达法则1?x??e. ?elim?elimx?02x(1?x)?x2x?02?6x2

壹拾贰

(18)limx?x?0sinx?lime?x?0ln(xsinx)?ex?0?limln(xsinx)?ex?0?limsinxlnx?ex?0?limxlnxx??elim1tt???lim?lntt?e0?1.1x(19)lim(ln)?limex?0?x?0?x11ln[(ln)x]x?lime?x?011xln(ln)x?e11x?limxln(ln)txx?0??et????lnlntt?e0?1.(20)lim(x?0tanxx2)?limex?0xlimtanxx2ln[()]xln(?limex?0tanx)xx2?ex?0limlntanx?lnxx2?elim111?2tanxcosxxlimx?02xx?sinxcosx2 ?ex?02xsinxcosx?e1x?sin2xlim2x?0x2sin2x?e1x?sin2xlim23x?02x1?cos2x6x2?ex?0?ex?0lim2sin2x6x2?e.1311sin2x?x2sin2x?x22sinxcosx?2x(21)lim(2?2)?lim22?lim?limx?0xx?0xsinxx?0x?0sinxx44x32cos2x?2?4sin2x1 ?lim?lim??.22x?0x?012x12x3x?x?0x?0(22)limsinxlnx?limxlnx?lim??1t?lnt?0.t???t2.对函数f(x)在[0,x]上应用拉格朗日中值定理有f(x)?f(0)?f'(?x)x,??(0,1).试证对下1列函数有lim??:x?0?2(1)f(x)?ln(1?x);(2)f(x)?ex.证明：只要证得lim?x?0 2?xln(1?x)?f(x)?f(0)ln(1?x)?ln(1?0)(2?x)ln(1?x)1?x?1;(1)lim?lim?lim??lim???x?01x?01x?0x?02x2f'(x)xxx21?2f(x)?f(0)ex?1ex(2)lim?lim??lim??1.xxx?0?1x?0x?0xf'(x)xxe2xe222e?2f(x)?f(0)?1即可。1f'(x)x2 壹拾叁

3.设f(x)二阶可导,求证：f(x?2h)?2f(x?h)?f(x)?f''(x).h?0h2证明：当h?0时,分子f(x?2h)?2f(x?h)?f(x)?0,分母h2?0,因此我们可以对h使用 lim洛比达法则：f(x?2h)?2f(x?h)?f(x)h?0h22f'(x?2h)?2f'(x?h) ?limh?02h4f''(x?2h)?2f''(x?h) ?limh?022f''(x?h) ?lim?f''(x).h?02f(x?2h)?2f(x?h)?f(x)于是有lim?f''(x).h?0h2 lim4.试说明下列函数不能用洛比达法则求极限：

x2sin(1)limx?0sinx1x;x2sin1112xsin?cosx的分子分母分别求导之后得xx,这个函数在x?0的时 答：对函数sinxcosx1x2sinx?limxxsin1?0.候极限是不存在的;但是原函数的极限是存在的：limx?0sinxx?0sinxx(2)limx?sinx;x??x?cosxx?sinx1?cosx的分子分母分别求导之后得,这个函数在x??时是没有x?cosx1?sinxx?sinx极限的;而原函数是有极限的:lim?1.x??x?cosx 答：对函数(3)lim2x?sin2x;x??(2x?sinx)esinx2x?sin2x?1本身极限就不存在。但是函数有聚点e和e.x??(2x?sinx)esinx

答：这是因为函数lim 壹拾肆

(4)limx?1(x2?1)sinxln(1?sin?2.x)(x2?1)sinx2xsinx?(x2?1)cosx 答：对函数的分子分母分别求导之后可得,这个函??1?ln(1?sinx)cosx?221?sinx22(x2?1)sinx数在x?1时候是没有极限的,但是原函数是有极限的：lim?0.x?1?ln(1?sinx)2第三节 函数的升降、凸性和函数作图

1.应用函数的单调性证明下列不等式：x?sinx?x, x?(0,);?2sinxxcosx?sinx证明：取函数y?,可求得y'?;取函数z?xcosx?sinx,可以知道z(0)?0,xx2z'??xsinx. 由于在区间(0,)上z'?0恒成立,于是函数z?xcosx?sinx在区间(0,)上单调递减,于22?sinx是有z?xcosx?sinx?z(0)?0成立;那么y'在区间(0,)上小于零成立,因此函数y?在2x此区间上单调递减。 因此有limy(0)?y(x)?y(),即1?x?02(1)2????sinx22??,整理即得x?sinx?x, x?(0,).x??2x3(2)x?sinx?x?, x?0;6证明：取函数y?sinx?x,可求得y'?cos?1;在区间(??,0)上可以知道恒有y'?cos?1?0,于是函数y?sinx?x单调递减,因此有y?sinx?x?y(0)?0,即得sinx?x,(x?0).x3x2 取函数z?sinx?x?,可求得z'?cosx?1?;那么有z'(0)?0,z''??sinx?x.由于在62x3区间(??,0)上恒有z''??sinx?x?0,因此z'满足z'(x)?z'(0)?0,于是函数z?sinx?x?单6x3调递减,即z(x)?z(0)?0,于是有sinx?x?, x?0.6x3 综上可得x?sinx?x?, x?0.6 壹拾伍

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