Lhopitals Rule For Indeterminate Forms Homework In Spanish

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PROFESSOR: Sowe'rethroughwithtechniquesofintegration,whichisreallythemosttechnicalthingthatwe'regoingtobedoing.Andnowwe'rejustclearingupafewlooseendsaboutcalculus.Andtheonewe'regoingtotalkabouttodaywillallowustodealwithinfinity.Andit'swhat'sknownasL'Hôpital'sRule.Here'sL'Hôpital'sRule.And that'swhatwe'regoingtodotoday.L'Hôpital'sRuleit'salsoknownasL'Hospital'sRule.That'sthesamename,sincethecircumflexiswhatyouputinFrenchtoomitthes.Soit'sthesamething,andit'sstillpronouncedL'Hôpital,evenifit'sgotansinit.Alright,sothat'sthefirstthingyouneedtoknowaboutit.

Andwhatthismethoddoesis,it'saconvenientwaytocalculatelimitsincludingsomenewones.Soit'llbeconvenientfortheoldones.There aregoingtobesomenewonesand,asanexample,youcancalculatex lnxasxgoestoinfinity.Youcould,whoops,that'snotaveryinterestingone,let'stry xgoes to0fromthepositiveside.Andyoucancalculate,forexample,x e^(-x),as xgoestoinfinity.And,well,maybeIshouldincludeafewothers.Maybesomethinglikelnx/ xas xgoestoinfinity.So these aresomeexamplesofthingswhich,infact,ifyouplugintoyourcalculator,youcanseewhat'shappeningwiththese.Butifyouwanttounderstandthemsystematically,it'smuchbetter tohavethistoolofL'Hôpital'sRule.Andcertainlythereisn'taproofjustbasedonacalculationinacalculator.

Sonowhere'stheidea.I'llillustratetheideafirstwithanexample.Andthenwe'llmakeitsystematic.Andthenwe'regoingtogeneralizeit.We'llmakeitmuchmore--Sowhenitincludesthesenewlimits,therearesomelittlepiecesoftrickinessthatyouhavetounderstand.So,let'sjusttakeanexamplethatyoucouldhavedoneintheveryfirstunitofthisclass.Thelimitasxgoesto1of(x^10 - 1)/(x^2 - 1).Sothat'salimitthatwecould'vehandled.Andthethingthat'sinteresting,Imean,ifyoulikethisisinthiscategory,thatwementionedatthebeginningofthecourse,ofinterestinglimits.What'sinterestingaboutitisthatifyoudothissillything,whichisjustplugin x=1,atx=1you'regoing toget0/0.Andthat'swhatwecallanindeterminateform.It'sjustunclearwhatitis.Fromthatplugging,inyoujustcan'tgetit.

Now,ontheotherhand,there'satrickfordoingthis.Andthisisthetrickthatwedidatthebeginningoftheclass.AndtheideaisIcandivideinthenumeratoranddenominatorby x-1.Sothislimit isunchanged,ifItrytocancelthehiddenfactor x-1inthe numeratoranddenominator.Now,wecanactuallycarryouttheseratiosofpolynomialsandcalculatethembylongdivisioninalgebra.That'svery,verylong.Wewanttodothiswithcalculus.Andwealreadyhave.Wealreadyknowthatthisratioiswhat'scalledadifferencequotient.Andtheninthelimit,ittendstothederivativeofthisfunction.Sotheideaisthatthisisactuallyequalto,inthelimit,now let'sjuststudyonepieceofit.SoifIhaveafunctionf(x),which isx^10-1,andthevalueat1happenstobeequalto0,thenthisexpressionthatwehave,whichisindisguise,thisisindisguisethedifferencequotient,tendsto,asxgoesto1,thederivative,whichisf'(1).That'swhatitis.

Soweknowwhatthenumeratorgoesto,andsimilarlywe'llknowwhatthedenominatorgoesto.Butwhatisthat?Well,f'(x)=10x^9.Soweknowwhattheansweris.Inthenumeratorit's10x^9.Inthedenominator,it'sgoing tobe2x,that'sthederivative ofx^2-1.Andthenweregoing tohavetoevaluatethatatx =1.Andsoit'sgoing tobe10/2,whichis5.Sotheansweris5.Andit'sprettyeasytogetfromourtechniquesandknowledgeofderivatives,usingthisrathercleveralgebraictrick.Thisbusinessofdividingby x-1.

WhatIwanttodonowisjustcarrythismethodoutsystematically.Andthat'sgoingtogiveustheapproachtowhat'sknownasL'Hôpital'sRule,what--mymainsubjectfortoday.Sohere'stheidea.Supposewe'reconsidering,ingeneral,alimitasxgoestosomenumberaoff(x)/g(x).Andsupposeit'sthebadcasewherewecan'tdecide.Soit'sindeterminate.f(a)=g(a)=0.Soitwouldbe0/0.Nowwe'rejustgoingtodoexactlythesamethingwedidoverhere.Namely,we'regoingtodividethe numeratoranddenominator,andwe'regoing torepeatthatargument.Sowehaveheref(x)/(x-a).Andg(x),dividedbyx-aalso.Ihaven'tchangedanythingyet.

AndnowI'mgoing towriteitinthissuggestiveform.Namely,I'mgoingtotakeseparatelythelimitinthenumeratorandthedenominator.AndI'mgoing tomakeonemoreshift.SoI'mgoingtotakethelimit,asxgoestoainthenumerator,butI'mgoing towriteitas((f(x) - f(a))/(x - a).Sothat'sthewayI'mgoing towritethenumerator,andI'vegottodraw amuchlongerlinehere.SowhyamIallowedtodothat?That'sbecausef(a)=0.SoIdidn'tchangethisnumeratorof thenumeratoranybysubtractingthat.f(a) =0.AndI'lldothesamethingto thedenominator.Again,g(a)=0,sothisisOK.Andloandbehold,Iknowwhattheselimitsare.Thisisf'(a)/g'(a).Sothat'sit.That'sthetechniqueandthisevaluatesthelimit.Andit'snotsodifficult.Theformula'sprettystraightforwardhere.Anditworks,providedthatg'(a)isnot0.Yeah,question.

STUDENT: [INAUDIBLE]

PROFESSOR: The questionis,isthere amoreintuitivewayofunderstandingthisprocedure.AndIthinktheshortansweristhatthere areother,similar,ways.Idon'tconsiderthemtobemoreintuitive.Iwillbementioningoneofthem,whichistheideaoflinearization,whichgoesbacktowhatwedidinUnit2.Ithinkit'sveryimportanttounderstandallofthese,moreorless, atonce.ButIwouldn'tclaimthatanyofthesemethods is amoreintuitiveonethantheother.Butbasicallywhat'shappeningis,we'relookingatthelinearapproximationtof, at a.Andthe linearapproximationto g ata.That'swhatunderliesthis.

SonowIgettoformulateforyouL'Hôpital'sRuleatleastinwhatIwouldcalltheeasyversionor,ifyoulike,Version1.Sohere'sL'Hôpital'sRule.Version1.It'snotgoingtobequitethesameaswhatwejustdid.It'sgoingtobemuch,muchbetter.Andmoreuseful.Andwhatisgoingtotakecareofisthisproblemthatthedenominatorisnot0.Sonowhere'swhatwe'regoing todo.We'regoingtosaythatitturnsoutthatthelimitasxgoestoaof f(x)/g(x)isequaltothelimitas xgoestoaoff'(x)/g'(x).Now,thatlookspracticallythesameaswhatwesaidbefore.And Ihavetomakesurethatyouunderstandwhenitworks.Soitworksprovidedthisisoneoftheseundefinedexpressions.Inotherwords,=g(a)=0.Sowehavea0/0expression,indeterminate.And,also,weneedonemoreassumption.Andtheright-handside,the right-handlimitexists.

Now,thisispracticallythesamethingaswhatIsaidoverhere.Namely,Itooktheratio ofthesefunctions,x^x^10-1andx^2-1.Itooktheirderivatives,whichiswhatIdidrighthere,right.I justdifferentiatedthemandItooktheratio.Thisiswayeasierthanthequotientrule,andisnothinglikethequotientrule.Don'tthinkquotientrule.Don'tthinkquotient rule.Sowedifferentiatethenumeratoranddenominatorseparately.AndthenItakethelimit asxgoesto1and Iget5.Sothat'swhatI'mclaimingoverhere.Itakethesefunctions,Ireplacethemwiththisratio ofderivatives,andthenItakethelimitinstead,overhere.AnditturnedoutthatthefunctionsgotmuchsimplerwhenIdifferentiatedthem.IstartedwiththismessyobjectandIgotthismucheasierobjectthatIcouldeasilyevaluate.Sothat'sthebiggamethat'shappeninghere.

Itworks,ifthislimitmakessenseandthislimitexists. Now,noticeIdidn'tclaimthatg,thatthedenominatorhadtobenonzero.Sothat'swhat'sgoing tohelpusalittlebitinafewexamples.Soletmegiveyouacoupleofexamplesandthenwe'llgofurther.Now,thisisonlyVersion1.Butfirstwehavetounderstandhowthisoneworks.Sohere'sanotherexample.Takethelimitas xgoesto0,ofsin(5x)/sin(2x).Thisisanotherkindofexampleofalimitthatwediscussedinthefirstpartofthecourse.Unfortunately,nowwe'rereviewingstuff.Sothisshouldreinforcewhatyoudidthere.Thiswillbeaneasierwayofthinkingaboutit.SobyL'Hôpital'sRule,sohere'sthestep.We're goingtotakeoneofthesesteps.Thisisthelimit,as xgoesto1,ofthederivativeshere.Sothat's5cos(5x)/(2 cos(2x)).

Thelimitwas1overthere,but nowit's0.a is0inthis case.Thisisthenumbera.Thankyou.Sothelimitasxgoesto0isthesameasthelimitofthederivatives.Andthat'seasytoevaluate.Cosineof0is1,right.Thisisequalto5 cos(5*0)--Andthat'samultiplicationsign.MaybeIshouldjustwritethisas0.Dividedby2cos0.Butyouknowthatthat's5/2.SothisishowL'Hopital'smethodworks.It'sprettypainless.

I'mgoing togiveyouanotherexample,whichshowsthatitworksalittlebetterthanthemethodthatIstartedoutwith.Here'swhathappensifweconsiderthefunction(cos x - 1)/x^2.Thatwasalittlehardertodealwith.Andagain,thisisoneofthese0/0thingsnear x=0.Asxtendsto0, thisgoestoanindeterminateformhere.Now,accordingtoourmethod,thisisequivalentto,nowI'mgoing tousethislittlewigglebecauseIdon'twant towritelimit,limit,limit,limitamilliontimes.SoI'mgoing tousealittlewigglehere.Soas xgoesto0,thisisgoing tobehavethesamewayasdifferentiatingnumeratoranddenominator.Soagainthisisgoing tobe-sin xin thenumerator.Inthedenominator,it'sgoingtobe2x.

Now,noticethatwestillhaven'twonyet.Becausethisisstillof0/0type.Whenyouplugin x=0youstillget0.Butthatdoesn'tdamagethemethod.Thatdoesn'tmakethemethodfail.This0/0,wecanapplyL'Hôpital'sRuleasecondtime.Andas xgoesto0thisisthesamethingas,again,differentiatingthe numerator anddenominator.SohereIget-cos xin thenumerator,andIget2inthedenominator.Againthisiswayeasierthandifferentiatingratiosoffunctions.We'reonlydifferentiatingthenumeratorandthedenominatorseparately.

Andnowthisistheend.Asxgoesto0,thisis--cos 0/2,whichis-1/2.Now,thejustificationforthiscomesonlywhenyouwinintheendandgetthelimit.Becausewhatthetheoremsaysisthatifoneoftheselimitsexists,thentheprecedingoneexists.Andoncetheprecedingoneexists,thentheonebeforeitexists.Soonceweknowthatthisoneexists,thatworksbackwards.Itappliestotheprecedinglimit,whichthenappliestotheveryfirstone.Andthelogicalstructurehereisalittlesubtle,whichisthatiftherightsideexists,thentheleftsidewillalsoexist.Yeah,question.

STUDENT: [INAUDIBLE]

PROFESSOR: Whydoestheright-handlimithavetoexist,isn'titjustthederivativethathastoexist?No.Thederivativeofthenumeratorhastoexist.Thederivativeofthedenominatorhastoexist.Andthislimithastoexist.Whatdoesn'thavetoexist,bytheway,Ineversaidthatfprimeofahas toexist.In fact,it'smuch,muchmoresubtle.I'mnotclaimingthatf'(a)exists,becauseinordertoevaluatethislimit,f'(a)neednotexist.Whathastohappenisthatnearby,forxnotequaltoa,thesethingsexist.Andthenthelimithas toexist.Sothere'snorequirementsthatthelimitsexist.Infact,that'sexactlygoing tobethepointwhen weevaluatethese limitshere.Iswedon'thavetoevaluate itrightattheend.

STUDENT: [INAUDIBLE]

PROFESSOR: Sothequestionthatyou'reaskingis,whyisthisthehypothesisofthetheorem?Inotherwords,whydoesthiswork?Well,theansweristhatthisisatheoremthat'strue.Ifyoudropthishypothesis,it'stotallyfalse.Andifyoudon'thavethishypothesis,youcan'tusethetheoremandyouwillgetthewronganswer.Imean,it'shardtoexpressitanyfurtherthanthat.Solook, inmanycaseswetellyouformulas.Andinmanycasesit'ssoobviouswhenthey'retruethatwedon'thavetoworryaboutwhatwesay.Andindeed,there'ssomethingimplicithere.I'msayingwell,youknow,ifIwrotethissymboldown,itmustmeanthatthethingexists.Sothat'sasubtlepoint.ButwhatI'memphasizingisthatyoudon'tneedtoknowinadvancethatthisoneexists.Youdoneedtoknowinadvancethat thatoneexists.Essentially,yeah.Sothat'sthedirectionthatitgoes.Youcan'tgetawaywithnothavingthisexist andstillhavethestatement betrue.

Alright,anotherquestion.Thankyou.

STUDENT: [INAUDIBLE]

PROFESSOR:SoI'm

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