《高数双语》课件section 3-2.pptx

1、L Hospitals Rules1L Hospitals Rules2LHospital,Guillaumede(1661-1704)French mathematician 0(,)U x (L Hospitals rules I 洛必达法则洛必达法则)()0g x and.Suppose that f,g,are both differentiable in 00lim()lim()0;xxxxg xf x(1)(A is a finite number or infinity),If the following conditions hold0()lim()xxfxAg x (2)00

2、()()limlim.()()xxxxf xfxAg xg x then0 xx 0 xx()x This rule is also valid for,and.L Hospitals Rules300(),(,),()0,f xxU xF xxx 00(),(,),()0,g xxU xG xxx Proof0 xx discontinuous at.We define and 00lim()lim()0 xxxxf xg xSince,when f(x)and(or)g(x)are0(,)xU x Choose any point.It is easy to see that the co

3、nditions of theHence 00()()()()()()()()()()F xF xF xFfG xG xG xGg ,Cauchy theorem are all satisfied for the functions F(x)and G(x)on the0 xinterval between the points and x.where lies between 0 xx and.L Hospitals Rules40 xx0 x 0()lim()xxfxAg x Let,so that,and the assumption means0()lim()xxfAg .Proof

4、(continued)00()()()()()()()()()()F xF xF xFfG xG xG xGg So,000()()()limlimlim()()()xxxxxxf xF xfAg xG xg .Finish.L Hospitals Rules532000sin1 cossin1limlimlim663xxxxxxxxxxSolutionsinxx 3.x and By LHospitals rules,Finish.30sinlim.xxxx Find It is easy to check that the conditions of LHospitals rulesare

5、 satisfied for the function we haveL Hospitals Rules6111lnlimlim111xxxxx。Solutionlnx1.x and By LHospitals rules,Finish.1lnlim.1xxx Find It is easy to check that the conditions of LHospitals rulesare satisfied for the function we haveL Hospitals Rules730.tanlimxxxx Find 23200tansec1limlim3xxxxxxx 202

6、sectanlim6xxxx 011.33tanlimxxxSolutionFinish.L Hospitals Rules80(,)U x (L Hospitals rules II)()0.g x and Suppose that f,g are bothIf the following conditions hold(1)00lim()lim(),xxxxg xf xA(is a finite number or infinity),0()lim()xxfxAg x (2)00()()limlim.()()xxxxf xfxAg xg x then0 xx 0 xx()x This ru

7、le is also valid for,and.differentiable in L Hospitals Rules911loglnlimlimaaaxxxxaxax 11limlnaxaax loglim(1).aaxxax Find lim(1,0).xxxaa Find 0.limxxxa 1limlnxxxaa L(1)(1)lim(ln)nnnxnxnxaa L222(1)lim(ln)xxxaa SolutionSolution11(1)(1)()lim(ln)nxnnxnaax L0.L Hospitals Rules10sin1coslim()lim()1xxxxxx si

8、nsinlim()lim(1)1.xxxxxxx A(is a finite number or infinity)()lim()fxAg x sinlim().xxxx Find SolutionNoteDoes not exist.Indeterminate Forms11Besides the types 00 and ,there are five other types of indeterminate form,namely ,0,1,00,0.For the type and 0,we can always transform them to indeterminate form

9、s of type 00 and and then calculate them;for 1,00 and 0,we can transform them to the type 0.We will give some examples to show how to do with these types of indeterminate form.Indeterminate Forms12sectanxx 1sincoscosxxx1sincosxx 2coslimsinxxx 2lim(sectan).xxx Find )(2x 221sinlim(sectan)limcosxxxxxx

10、Solution00,this becomes the type as.By LHospitals rule we haveFinish.1sinsectancosxxxx Since 0 Indeterminate Forms1300lnlimlnlimxxxxxx 0limxx 0.101limxxx 0limln(0).xxx Find Solutionwe haveFinish.0 （）.lnlnxxxx Since ,and this type is.By LHospitals ruleIndeterminate Forms1401.e20limsinxxx 00lnsinlimln

11、sinlim1xxxxxx 02cossinlim1xxxx 0limlnsin0lim(sin)xxxxxxe Find 0.lim(sin)xxx lnsin(sin)xxxxe Solution Since we haveThus,Finish.00()lnsinxx0，,and is of type 0.Indeterminate Forms1501limln(1)xxx 01lim1xx 1.ee1()10lim(1).xxx Find 11ln(1)(1)xxxxe 1.011limln(1)0lim(1)xxxxxxe Solution Since,and we haveThus,Finish.001ln(1)xx is of type ,