高等數(shù)學(xué)求極限方法小結(jié)及舉例ppt課件

1、求極限方法小結(jié)及舉例:. 常用辦法一:. 利用初等函數(shù)的連續(xù)性1)()(lim00 xfxfxx能代則代:.型型和洛必達法則003.等價無窮小代換4乘除法運算中的,時當(dāng)0 xxxxxxarctanarcsintansin,)ln(11xexx.lnaxax1:. 冪指函數(shù)取極限2.)(lim)(lim)(lim)(xvxvxuxu)(lim)(limxufxuff 連續(xù). 約去公因子5.,.如取對數(shù)等代數(shù)變換6:. 常見未定式及對策二.,.常見型001.,.常見型2,.型03.)()()()()()()()(xgxfxgxfxfxgxgxf11或,cos212xx.,)(21111xxxx ,

2、時當(dāng)0 x)(ln)()(xfxfxf,.,.,)(.型型型00706015,.型4.)()()()()()(xgxfxgxfxgxf11)(ln)()()(xuxvxvexu)(ln)(lim)()(limxuxvxvexu,)(.型015,)()(ln)()(xfxgxgexf11)(冪指型)(指數(shù)型,)(lim)(ln)(lim)(xfxgxgexf11)(經(jīng)驗公式cxgexf)()(lim 1)()(limxgxfc )()(limlim.121123111221xxxxxxxxx例.lim2211xxx.limlim.111210 xxxexexxxx例ttetextxxx)ln(l

3、imlim11100ttt110)ln(limttt110)(limln.ln1e.lim110 xexx?,xexx10證明時當(dāng)xxxxx223lim.例xxxxxx222lim?,用洛必達法則.lim111112xxx112324xxxxlim.例)(0112122121221xxxxx)(lim冪指函數(shù)取極限11221xxxlim12122121221xxxxxx)(limlim12122121221xxxxx)(lim.ee 1可用經(jīng)驗公式驗證10000522xxxxlim.例.lim1100001112xxxxxxtanlim.2162 例ttxttcotlim120 .tanlim

4、10tttxxxcotlim.07例0.tanlim10 xxx.coslim010 xxx)( 有界量乘無窮小.coslimlimcoslim011000 xxxxxxx!錯.sin)(.正整數(shù)例nxxxxxxxfnn000018.)(的取值范圍的連續(xù)性及討論nxf 0000 xfxffx)()(lim)(.解xxxnx10sinlim.sinlim0110 xxnx1n如果0000 xfxffx)()(lim)(xxnx0 lim.01n如果,1n如果.)(00 f則,時當(dāng)1n000011121xxnxxxxxxnxfnnncossin)(xxxxnxfnnxx112100cossinli

5、m)(lim.02n如果100nxxxnxflim)(lim1n如果.0,)(,存在時當(dāng)xfn1.)(,連續(xù)時當(dāng)xfn 2.,.yfxfy 求二次可導(dǎo)例29 .2222xfxxxfy解xxfxxfy22222 .22242xfxxf .)(,.的二階導(dǎo)數(shù)求二次可導(dǎo)設(shè)例xfyf110. )()(.yfxxfy的直接函數(shù)是解1,)(yfxdyd1)(yfxddxdyd122xdydyfyf 2)()(.)()(3yfyf ,)()(, )()()(.次可導(dǎo)內(nèi)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