高等数学求极限方法小结及举例ppt课件

1、求极限方法小结及举例:. 常用办法一:. 利用初等函数的连续性1)()(lim00 xfxfxx能代则代:.型型和洛必达法则003.等价无穷小代换4乘除法运算中的,时当0 xxxxxxarctanarcsintansin,)ln(11xexx.lnaxax1:. 幂指函数取极限2.)(lim)(lim)(lim)(xvxvxuxu)(lim)(limxufxuff 连续. 约去公因子5.,.如取对数等代数变换6:. 常见未定式及对策二.,.常见型001.,.常见型2,.型03.)()()()()()()()(xgxfxgxfxfxgxgxf11或,cos212xx.,)(21111xxxx ,

2、时当0 x)(ln)()(xfxfxf,.,.,)(.型型型00706015,.型4.)()()()()()(xgxfxgxfxgxf11)(ln)()()(xuxvxvexu)(ln)(lim)()(limxuxvxvexu,)(.型015,)()(ln)()(xfxgxgexf11)(幂指型)(指数型,)(lim)(ln)(lim)(xfxgxgexf11)(经验公式cxgexf)()(lim 1)()(limxgxfc )()(limlim.121123111221xxxxxxxxx例.lim2211xxx.limlim.111210 xxxexexxxx例ttetextxxx)ln(l

3、imlim11100ttt110)ln(limttt110)(limln.ln1e.lim110 xexx?,xexx10证明时当xxxxx223lim.例xxxxxx222lim?,用洛必达法则.lim111112xxx112324xxxxlim.例)(0112122121221xxxxx)(lim幂指函数取极限11221xxxlim12122121221xxxxxx)(limlim12122121221xxxxx)(lim.ee 1可用经验公式验证10000522xxxxlim.例.lim1100001112xxxxxxtanlim.2162 例ttxttcotlim120 .tanlim

4、10tttxxxcotlim.07例0.tanlim10 xxx.coslim010 xxx)( 有界量乘无穷小.coslimlimcoslim011000 xxxxxxx!错.sin)(.正整数例nxxxxxxxfnn000018.)(的取值范围的连续性及讨论nxf 0000 xfxffx)()(lim)(.解xxxnx10sinlim.sinlim0110 xxnx1n如果0000 xfxffx)()(lim)(xxnx0 lim.01n如果,1n如果.)(00 f则,时当1n000011121xxnxxxxxxnxfnnncossin)(xxxxnxfnnxx112100cossinli

5、m)(lim.02n如果100nxxxnxflim)(lim1n如果.0,)(,存在时当xfn1.)(,连续时当xfn 2.,.yfxfy 求二次可导例29 .2222xfxxxfy解xxfxxfy22222 .22242xfxxf .)(,.的二阶导数求二次可导设例xfyf110. )()(.yfxxfy的直接函数是解1,)(yfxdyd1)(yfxddxdyd122xdydyfyf 2)()(.)()(3yfyf ,)()(, )()()(.次可导内在例111naUxxaxxfn . )()(afn求)()( !)(.)(xaxnxfn 1解)()(!)(xaxnn 221, )()()(xaxnn1 .)()(01afnaxafxfafnnaxn)()(lim)()()()(11)()()(xaxnn11 )()(!)()(!limxaxnnxnax 21. )(!an 012 )()()()(.tftftftytfx例.22xdyd求)()(.txtyxdyd解)()()()(tftftfttf t txddxdyd221?xdtdt )(.)(tf 1., )tan(.yyxy 求例1