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极限存在的夹逼准则《高等数学》.0,)()(lim;)]()(lim[;)]()(lim[,)(lim,)(lim3BBAxgxfBAxgxfBAxgxfBxgAxf其中则设定理.0,lim;][lim;][lim,lim,lim4bbababababababbaannnnnnnnnnnnn其中则设定理一、回顾⑵⑴⑶⑵⑴⑶.sinlim0xxx二、问题,12111222nnnnan(1)设求极限.limnna(2)求极限定理1如果函数及满足下列条件:)(),(xgxf)(xh)(xf.)(lim0Axfxx那么函数的极限存在,且三、夹逼准则),()()(),,(0xhxfxgrxUxo当,)(lim,)(lim00AxhAxgxxxxAAA)(xf)(xg20x20xoxy0x10x10xrx0rx0)(xh⑴⑵证明因,)(lim0Axgxx所以由极限的定义,当,01时,有10||0xx,|)(|Axg).(xgA则,0①,02当时,有,20||0xx|)(|Axh又因为所以,)(lim0Axhxx则.)(Axh②取},,,min{21r当时,||00xx由条件(1)知,当时,有rxx||00).()()(xhxfxg③①,②,③式同时成立.,)()()(gAxhxfxA即.|)(|Axf.)(lim0Axfxx所以故注当时,定理1类似成立.x定理2如果数列及满足下列条件:}{},{nnyx}{nz;nnnzxy,lim,limazaynnnn}{nx.limaxnn那么数列的极限存在,且(1)当时,有,0NN0Nn(2)定理1和定理2称为夹逼准则(也称为两边夹法则).利用夹逼准则求极限关键是构造出合适的,ny,nz),(xg).(xh或nannn2,12nnnnnn2lim1lim2nnn,12111222nnnnan例1设解而所以,由夹逼准则得.1limnna求极限.limnna因为nn111lim2111limnn,1,1四、应用nannn212nn.sinlim0xxx例2求极限xABoyCDx解.tan,,sinACxABxBDx,tansinxxx02x1coslim0xx.1sinlim0xxx设由图知,即,tan2121sin21xxx因为,所以对不等式进行变形有此式对也成立.因与,11lim0x由夹逼准则知,AOCAOBAOBSSS扇形xABoyCDx,1sincosxxx,20x四、小结1.夹逼准则2.一个重要极限:五、作业.)2(),1(456P定理1如果函数及满足下列条件:)(),(xgxf)(xh)(xf.)(lim0Axfxx那么函数的极限存在,且)()()(),(0xhxfxgrxUxo,)(lim,)(lim00AxhAxgxxxx.1sinlim0xxx⑴当时,;⑵
本文标题:极限存在的夹逼准则
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