分析方法第二章演示文稿

第二章函数极限存在的证明与求法方法一利用函数极限的定义定义AxfaxxAxfax)(0:,0,0)(limAxfaxaxAxfax)(:,0,0)(limAxfaxaxAxfax)(:,0,0)(limAxfBxxBAxfx)(:,0,0)(limAxfBxxBAxfx)(:,0,0)(limAxfBxxBAxfx)(:,0,0)(lim以上为正常极限定义，非正常极限定义还有18个，请大家作为练习，自己写出.，使找出小正数，须对用定义证明,0)(limAxfax.)(0成立时，有Axfax为使解不等须解不等式的过程找后证明过程主要是任给,,,0再放大的取值范围或先限制经常采用放大绝对值式的过程简单,,,x.,,找出小正数然后再解不等式绝对值.0312lim11xxxx语言证明用例211112,11xxxx则有首先限制证明1122)1(3xxx于是有12312312xxxxxxx.21,12,0xx解得由对有时，则对于是取,10:2,1minxx12312xxxx0312lim1xxxx.1)(1lim,0)(lim2bxfbbxfaxax证明设例，首先取证明0)(lim,020bbxfbax时有则110:,0axx,2)(bbxf2)()()(bbxfbbbxfxf从而2)()()(1)(12bbxfbxfbxfbxf于是则再由,)(limbxfax2)(0:,0,0222bbxfaxx有时则取,0:,,min21axx2)()()(1)(12bbxfbxfbxfbxfbxfax1)(1lim语言证明（常数）用设例-A)(lim3xfax33)(limAxfax,0,0,0)(lim01则时，）当证明（xfAax3)(0:xfaxx3)(xf0)(lim3xfax3233233)()()()(0)2(AAxfxfAxfAxfA时，当323223343)(432)()(AAxfAAxfAxf时有则axxAxfax0:,0,0,)(lim3243)(AAxf时有从而axx0:323343)()(AAxfAxf33)(limAxfax.)(lim14xfx叙述定义：例.)(02为极限不以时Axfax.)(:,0,01MxfBxxBM）解（00)(:,0,0)2(Axfaxax,)(lim)()2(),0()(5Axfxfxfxfx，上满足在设函数例.,0,)(xAxf证明有任意取定证明,,0x)2()2()2()(2xfxfxfxfn,)(limxfx由海涅定理，Axfxfnn)2(lim)(.,0,)(,xAxfx有的任意性再由，上有定义且在设例0)(lim)1,0()(60xfxfx,02)(lim0xxfxfx.0)(lim0xxfx求证，则证明02)(lim0xxfxfx有,0:,0,0xxxxfxf2)(于是有以下不等式.2)(2)(xxfxfxfxf2)(xfxxf2222xfxxf322222xfxxf---------1222nnnxfxxf以上各式相加,得122222)(nnxfxxxxxf,2)(,xxfn得，对上式取极限令,2)(xxf.0)(lim0xxfx即且的任意闭区间上有界在上有定义在设例,,0,,0)(7xf.)(1lim有限数lxfxfx.)(limlxxfx求证0)(1lim,0(1)xfxflx时当证明有则对,,0,0AxA21xfxf则设对,10,,2,1,0,snsnAxAxsAfsAfsAfsnAfsnAfsnAfxf111snAfsnAfsnAfsnAf211sAfsAfsAf1sAfn2,)(,1,,0,1,MxfAAxMAAf有即上有界在MsAf)(xMxMxnxxf22)(,2,0,0,0limxMBxBxMx则对以上又有则取,,,maxCxBAC22)(xxf0)(limxxfx,)()(,0)2(lxxfxgl令时当以及的任意闭区间上有界且在上有定义在则,,0,,0)(xg0)(1lim)(1limlxfxfxgxgxx.)(lim,0)(lim,)1(lxxfxxgxx即得由,0)()(lim,,)(8xfxfaxfx且上有连续导数在设例.0)(limxfx求证有则对证明,,0,0,0)()(limAxAxfxfx2xfxf)()()(xfxfexfexx得积分到对该式从,xAxAtxAtdttftfedttfe)()()(xAtAxdttftfeAfexfe)()()()(xAtAxdttftfeAfexfe)()()()(dttftfeeeAfexfxAtxxA)()(1)()(xAtxxAdteeeAfe12)(xAxxAeeeeAfe2)(2)(xAeAfe有则对以上因为,,,0,0)(limBxABeAfexAx2)(xAeAfe有从而,Bx22)(xf.0)(limxfx练习题.lim.1axax证明.,)(lim.2则极限唯一存在若xfax0)(lim,)(lim3xfxfaxax且存在若.0)(),(00xfaUx，有，使得证明使得则若,0),(lim)(lim.4xgxfaxax。有)()(),,(0xgxfaUx,0)(lim,,)(.5xfxfx且上的周期函数为若.,,0)(xxf证明,)(lim)(lim),()()()(0AxhxfxhxgxfaUaxax并且上有设在定理方法二用迫敛性定理.1lim10xxx求极限例,1111,0(1)xxxx有时当解,111xxx于是.11lim,,11lim00xxxxx由迫敛性定理而.)(limAxgax则,1111,0)2(xxxx有时,111xxx.11lim,11lim00xxxxx由迫敛性定理而.11lim)2(),1(0xxx得由.lim,0,020abxbaxbax证明设例,10)1(xbxbxbx时有证明,abxbaxaxab,lim0abaxabx而.lim0abxbaxx由迫敛性定理,1,0)2(xbxbxbx有时axabxbaxab，而abaxabx0lim.lim,0abxbaxx由迫敛性定理.lim)2(),1(0abxbaxx得由),(max,,)(3,xfMbaxfbax上为正值连续函数在设例.)(limMdxxfnbann证明bxMxf00,)(不妨设设证明,0,00xb则对,2)(,:00Mxfxxxxnxxnnbandxxfdxxf00)()(nnxxnMdxM12200又显然有nnbannbanabMdxMdxxf1)(nnbannabMdxxfM11)(2于是,22lim1MMnn再由于,lim1MabMnn及,0则对以上MabMNnNNMMNnNNnn122111,2,有取,,,max21NnNNNMdxxfMnban)(Mdxxfnbann)(lim练习题证明设,0,0.1ba0lim10bxxaxbxxax0lim2.)(lim,)(lim,2AxxxfAxfxx证明.sin1sinlim.4xxx求.01sinlim,0.30xxx证明设方法三利用洛必达法则在用此法则的过程中,若分子或分母某因子的非零极限易求,可先行求出,再用,~1,~sin0xexxxx时，例如,2~cos12xx,~tanxx,~arcsinxx,~arctanxx,~)1ln(xx等。,1~1nxxn满足以下条件：设函数例)(),(1xgxf,0lim)(lim1xgxfaxax,)()2(0可导在aU用此法则,同时分子或分母的无穷小因子,可用等价无穷小代替.，且lxgxfxgax)()(lim,0)(3。则lxgxfax)()(lim,0)()(axaxxfxF令证明.0)()(axaxxgxG且可导在连续在则,)(,)()(),(0aUaUxGxF0)(xG,,),(0xxaaUx之间存在一点与在根据柯西中值定理对，使得)()()()()()(xGFaGxGaFxFx，即)()()()(xxgfxgxflgfxgxfxxaxax)()(lim)()(lim.1lnlim220xxxexx求极限例xxexxx211)1(lim000原式解.232112lim20xexxx.2lim32tan1xxx求极限例xxxe2ln2tan1lim1原式解002cos2lnlim1xxxe2cos2ln2sinlim1xxxxe.22sin221lim1eexxxxxxx30sinsintanlim4例