《高数双语》课件section 1-2.pptx

1、Limits of Sequences of NumbersArea of a Circle2Do you ever remember how to obtain the area of a circle?RRThe area of a hexagon 6AThe area of a dodecagon 12A The area of a polygon with 162n sides 16 2nA .16126 2,nAAAS LLArea of a Circle3Informal Definition of Limit of a Sequence The area of the circl

2、e is theof the areas of the inscribed polygons.A sequence can be thought as a list of numbers written in a definite order:1234.,na a a aaIfnagets arbitrarily close to A as n becomes sufficiently large,or()limnnnaA naAnahas the limit A,we say the sequence and writeNote This definition is“informal”bec

3、ause phrases like and are imprecise;their meaning depends on the context.Limit of a SequenceSchool of Science,BUPT4nagets arbitrarily close to A as n becomes sufficiently large or()limnnnaA naAHow to get the precise definition of the limit of a sequence?5Limit of a Sequence A sequence of numbers 数列数

4、列 is just a special12:,nnaa aaLL,is called the general term 通项通项 denoted by:fNR,N function f,defined on the setwhere()naf n nN ,of the sequence.11 11:1,;2 3nnLL (1):1,1,1,(1),;nnLL 2:2,4,8,2,nnLL6Limit of a Sequence11 11:1,2 3nnLL11(1)1 1(1):1,2 3nnnn LL (1):1,1,1,(1),nnLL 2:2,4,8,2,nnLLApproach to

5、zero 趋于零趋于零Does not approachany valueApproachesThe limits are 0No limitsWhile n tends to infinity,7Limit of a Sequence5101520-1-0.50.51Since a sequence can be seen as a function of positive integer,we can plot a sequence on a coordinate plane.Limit of a Sequence8Observe the changing of a sequence wh

6、ile 11nn .n Limit of a Sequence9Observe the changing of a sequence while 11nn .n Limit of a Sequence10Observe the changing of a sequence while 11nn .n Limit of a Sequence11Observe the changing of a sequence while 11nn .n Limit of a Sequence12Observe the changing of a sequence while 11nn .n Limit of

7、a Sequence13Observe the changing of a sequence while 11nn .n Limit of a Sequence14Observe the changing of a sequence while 11nn .n Limit of a Sequence15Observe the changing of a sequence while 11nn .n Limit of a Sequence16Observe the changing of a sequence while 11nn .n Limit of a Sequence17Observe

8、the changing of a sequence while 11nn .n Limit of a Sequence18Observe the changing of a sequence while 11nn .n 19Limit of a SequenceObserve the changing of a sequence while 11nn .n 111()nnn 20Concept of Limit of a Sequence (Limit of a sequence of numbers)limnnaA or().naA n converges 收敛收敛 to the numb

9、er A if andnaThe sequence If there is no such number A,we say that an diverges 发散发散.|naA holds for all.nN only if to every positive number there exists a positive integer N such thatThen we say that the sequence naapproaches to A as n tends to infinity,denoted by And A is called the limit 极限极限 of th

10、e sequence .naConcept of Limit of a Sequence210,NN s.t.|naA holds for all.nN Another expression of the definition of limit for a sequence is:The definition can be used to judge whether A is the limit of an,but is not an efficient method used to find the limit or judge whether the limit exists.22Conc

11、ept of Limit of a SequenceLet us draw the sequence an on an axis to demonstrate the definition of limit by geometric graphxA(A )A 1a2a3a4a()nanN(,)AAnaQuestion:Isout of the neighborhoodfor all nN.Generally,N is depend on and denoted by()N to ensure and25Concept of Limit of a SequenceProof For any gi

12、ven 0 ,we want to find a()N,s.t.Prove that(1)lim1nnnn .Since(1)(1)11nnnnnn ,we need to find a N,s.t.1n or 1n for all.nN Choosing 1,N s.t.(1)1nnn ,for all.nN This means(1)lim1nnnn .This is the end.(1)1nnn ,for all nN.we have found a N,that 0,NN s.t.|naA for all.nN 26Concept of Limit of a SequenceProv

13、e that lim0,nnq|1q whereProof0,there is no harm in assuming that 01.We want to find a(),N such that forall|0|.nqnN|0|ln|ln.nnqqnqNotice that ln|q|N,naya.|nya.limnnya.Finish.30Concept of Limit of a SequenceFor example,the sequence 1111,1,1,1,1,23nLLdoes not approach a constant.|naA Note The requireme

14、nt that not only for an infinite number of n.is for all nN,limnnaA or()naA n 0,NN s.t.|naA for all.nN 31Concept of Limit of a Sequence0 ,NN 00,s.t.|nnNaA Note The expression of the sequence na does not tend to A as n tends to infinity is:limnnaA or()naA n 0,NN s.t.|naA for all.nN 32Concept of Limit

15、of a SequenceTheorem(Uniqueness)The limit of any convergent sequence is unique.Assume that the sequence has two different limits A and,Bthat islimnnaA and limnnaB,AB.ProofLets suppose AB(If AB,we have the same conclusion)and choose a positive 02AB .Since limnnaA ,by the definition of limit,for any g

16、iven 0,there exists,such that 0|naA for all 1nN.It means thatSimilarly,since limnnaB ,we have This is impossible.The proof is completed.1+NN a 002nnABAaAa.002nnABBaBa for some N2 and for all n N2.BA2BA()0B0B()0A0Ax33Concept of Limit of a Sequence (Preservation of sign)Assume that limnnaA,which 0A .T

17、hen+NN such that na and A have the same sign for all nN;furthermore,if 0(0)A,then+NN,such that 0(0)nnaqaq for all nN.AO34Concept of Limit of a SequenceAssume that limnnaA and limnnbB,If+NN,such that nnab holds for all nN,then AB.()()ABnanb(Isotone)35 Conditions for Convergence of A Sequence(boundedn

18、ess of a sequence)If there exists a constant 0M (independent of n)naM holds for all+Nn.Then,Otherwise,na is said to be unbounded.for all+Nn,na is said to be bounded above(or below).Let na be a sequence.such thatna is said to be bounded.Moreover,if naA (or naB)5101520-1-0.50.51Bounded5101520-101020Un

19、bounded5101520-400-300-200-1005101520-0.5-0.4-0.3-0.2-0.1Bounded above510152010020030040051015200.10.20.30.40.5Bounded belowbounded:holds.Man|,0M,Nn,0M,Nnholds.Man|unbounded:36Conditions for Convergence of a SequenceTheorem(boundedness)Any convergent sequence must be bounded.na is a sequence andlim.

20、nnaA for given 1 ,there exists a+NN,such that Thus,whennN,we have|1nnaAaA or Proof By the definition of limit,naA holds for all nN|1|naA .Suppose that37Conditions for convergence of a sequenceThere are at most N points,12,Na aaL outside the interval(1|),(1|)AA .that we can find a number B,s.t.If we

21、choose max1|,MA B,then This is the end.121,max|,|,|NnNBaaa LL.na is bounded.Proof(continued)Since there are finite terms,it is obvious|1|naA .,nN Theorem(boundedness)Any convergent sequence must be bounded.38(monotonicity of a sequence)Conditions for convergence of a sequenceIf for every subscript n

22、,we always have then na is called to be monotone increasing;then na is called to be monotone decreasing.Let na be a sequence.1(1,2,)nnaa n L,if we always have A sequence which iseither monotone increasing or monotone decreasing is called monotone.510152010020030040051015200.10.20.30.40.5Monotone inc

23、reasingMonotone decreasing1(1,2,)nnaan L,39Conditions for convergence of a sequenceTheorem(monotone boundedness criterion)If a sequence is monotone increasing and bounded above or monotone decreasing and bounded below,then it must be convergent.5101520-0.5-0.4-0.3-0.2-0.1Monotone increasing and boun

24、ded above 单调增有上界单调增有上界M51015200.10.20.30.40.5Monotone decreasing and bounded below 单调减有下界单调减有下界M40Conditions for convergence of a sequence Note It is clear that changing any finite number of terms of a sequence does not affect its convergence and limit.For instance,let.51015200.20.40.60.81We had kno

25、wn that this sequence doesnot have limit or does not convergent.111,1,1,12nan LL.11,1,1,1,1,1,1n LL.If we change the first 2n numbers of this sequence to 1 and keep the others unchanged.We haveIt is clear that this sequence still does not have limit or does not convergent.51015200.20.40.60.8141Condi

26、tions for convergence of a sequencethen the new sequence becomes convergent or has limit.But if we change terms with subscript as 2(1,)k k L to 212(1)kk k ,51015200.20.40.60.81111,1,1,12nan LL.51015200.20.40.60.81By the theorem of monotone boundedness criterion,this new sequence is convergent3 151,4

27、 2 12na L42Conditions for convergence of a sequence Note If a sequence is bounded but monotone only starting form term,then it is still convergent.Note It should be recognized that the theorem is only a sufficient condition for convergence.A convergent sequence is not necessarily monotone.For instan

28、ce,letnann)(1an is a convergent but is not monotone.43Conditions for convergence of a sequence Let 11a and 111nnnaaa Proof Since 1111111(1)(1)nnnnnnnnnnaaaaaaaaaa and 21aa,by mathematical induction method,we have 1121nnnaaa Then,by monotone boundedness criterion,we can finish the proof.,prove that l

29、imnna exists.10nnaa for all n.That is the sequence na is monotoneincreasing.Moreover,it is clear that for all n.44Conditions for convergence of a sequence(number e)Prove that sequence 11nn is convergent.10203040502.452.552.62.6511nn 45Conditions for convergence of a sequence(number e)Prove that sequ

30、ence 11nn is convergent.Therefore,for any n,we have 1nnaa .So,If we can prove na is bounded above,then we will finish the proof.11nnan,thenProof Let nais monotone increasing.46Conditions for convergence of a sequenceProof (continued)11nnanActually,term can be expended asThen na is bounded above.Fini

31、sh.10203040502.452.552.62.65(number e)Prove that sequence 11nn is convergent.4710203040502.452.552.62.65Conditions for convergence of a sequenceRemark This example shows that sequence 11nn has limit and.That isIt can be proved that e is an irrational number2.7182818284590452354.e ethis limit is deno

32、ted by 1lim 1nnen.ande(number e)Prove that sequence 11nn is convergent.48Conditions for convergence of a sequence11lim 1lim 1nnnnenn 1()lim 1().f nnf neIf lim()0nf n,where f(n)is an integer variable function,then we have 1lim 1nnen 1()()lim 1nf nfenFor example,Remarkuse the following conclusion:1lim

33、 1nnenThe limitwill be used often.Generally,we often 55lim 1,nnen49Conditions for convergence of a sequenceTheorem(Cauchys convergence principle)The necessary and sufficient condition for a sequence to be convergent is that 0,+()NN,such thatRemarkaccumulate in an arbitrarily small neighborhood of so

34、me point.is arbitrarily small for any nN and mN,then all terms will accumulate in an arbitrarily small is large enough|,nmaanNmN .nThis theorem means that while is sufficient large,all the terms willConversely,nama and if the distance between two points nneighborhood of some point when converges to

35、this point.and the sequence50Conditions for Convergence of a Sequence Prove that the sequence na is convergent,where222111123nanL.0 ,+()NN,s.t.|,nmaanNmN .convergenceProof We will use Cauchys theorem.+Nn and+Nm is equivalent to taking any+Nn,+Np and considering n and np.ThenIn fact,taking any.51Cond

36、itions for convergence of a sequence 0 ,+()NN,s.t.|,nmaanNmN .convergenceProof (continued)0,we want to judge whether there exists a N,s.t.Since 1|npnaan,then we can choose Finish.|npnaa .1N .Hence,such that0 1N ,there exists|npnaa for all nN.Prove that the sequence na is convergent,where222111123nan

37、L.52Conditions for convergence of a sequence is divergent,whereProof00,+NN,m nN,such that0|.mnaa We now take 012 ,2mn.SinceFinish.naProve that the sequence 111123nanL.We only need to show that the condition of Cauchys theorem can not be hold.That is,0 ,+()NN,s.t.|,nmaanNmN .convergence53Conditions f

38、or convergence of a sequenceDefinition(subsequence)Any infinite number of terms of a sequence na,arranged in increasing order of its subscripts,is called a subsequence子列子列 of the sequence na,denoted by 12:,kknnnnaaaaLL where 12knnnLL.k expresses the fact that kna is the thk term of the subsequence k

39、na;kn expresses the fact that kna is the thkn term of the original sequence na.Then,knk and jknn while jk.54Conditions for convergence of a sequenceTheorem(aggregation principle)The necessary and sufficient condition that limnnaA is that for any subsequence kna of na,we have limknkaA.Determine wheth

40、er the sequence converges or diverges.sin4n Solution Taking n=4k,we get a subsequence sin044k,which converges to 0.Taking n=8k+2,we get a subsequence which converges to 1.According to the aggregation principle,the sequence sin4n is divergent.(82)sin1,4k 55Rules of Operations on Convergent Sequence(R

41、ules of rational operations),thenlim()limlimnnnnnnnabABab(2)lim()limlimnnnnnnnabABab(3)limlimlimnnnnnnnaaAbBb,if 0B .limnnaA and limnnbB Assume that(1)Rules of Operations on Convergent Sequence56Solution,which will be solved later.we have then we have Finish.limnna Let 11a and 111nnnaaa ,find.and we

42、 know that That is Since 111nnnaaa ,That is 11 ,(the negative value should be has limitnaWe had proven that sequence We can suppose the limit is if a sequence has limit,then the limit must unique.limnna .So,251nnalim152 .omitted).57If k is a constant and then(2)lim()limmmmnnnnaAa,+Nm.Rules of Operat

43、ions on Convergent SequenceCorollary(1)lim()limnnnnkakAka,limnnaA,NoteThe rules(1)and(2)rational operations theorem may be extended to anyif the limit of each one exists,but they are not necessarilytrue for an infinite number of sequences.finite number of sequence58Rules of Operations on Convergent

44、Sequence22222233331212limlimlimlimnnnnnnnnnnLL Find 2223.12limnnnL2223312(1)(21)1111266nn nnnnnnL,Solution Since?infinite number0000L L59Rules of Operations on Convergent SequenceSolution we obtain Using rules of rational operations theorem and its corollary we have Find 52534265lim.324nnnnnnn5nIf w

45、e divide numerator and denominator by,60lim.liml mlnnmanb 0.lmlmalmblm Discuss the convergence of the following sequence and find its limit if it exists.11101110nnn,(0,0).nnnllllnlmmmmmaaaaxabbbbb LLSolution11101110nnnlimlimnnnllllnmmnnmmaaaaxbbbb LL11011101111n()lim111n()nnnlllllnmmmmmaaaannnbbbb L

46、LdivergenceRules of Operations on Convergent Sequence61Rules of Operations on Convergent SequenceSolution Sincewe have Find 111lim.1 22 3(1)nn nL,62 Find 11lim.12nnkk Land Then,it is clear that SinceSolution.Rules of Operations on Convergent sequence63(Sandwich Theorem 夹逼准则),such thatnnnacb holds fo

47、r all nN and limlim,nnnnabAthen the sequence nc is also convergentSandwich Theorem+NNIf limnncA.and 51015201.011.021.031.041.05nanbnc64Sandwich Theorem Find 1lim1mpnn Solution:and by the Sandwich theorem we known1lim 11.mpnn,where,m+N.p Since?using the sandwich theorem1lim 11pnn,65Sandwich TheoremSo

48、lution and Then,by the Sandwich theorem we known222111lim1.12nnnnnL Find 222.111lim12nnnnnLSince,.66Sandwich Theorem1,0,1,2,nnnab bnL L Find lim,nnawhere a 0 is a constant.Solution When a=1,it is easy to obtain.lim1nnaWhen a 1,letBy the binomial formula,we havennba)(1nnb1So thatnabnbann11e.i.and.nab

49、ann1111Then by the Sandwich theorem we known.lim1nnaWhen 0 a 1,then,11a.lim11nnaso thatHence.limlim111nnnnaa.lim1nnnPlease prove that2(1)12!nnnnn nnbbb L L67Examples12lim11332lim13nnnn 1123lim.23nnnnn Find 1.3 1111231323limlim232313nnnnnnnnnn Solution12113lim3213nnn 68Examples10001000 1000,nnnxglim

50、1231000.nnnnnnL Find lim1000.nnx lim 1000 1000lim1000 10001000.nnnnngSolution1231000,nnnnnnx LLetsinceandThen,by the Sandwich theorem we known1212Find lim(,0).nnnnmmnaaaa aaLL?69Examples(1)!(1)!2!nnnn 2221111lim 111lim232nnnnn L222111lim 111.23nnL Find 11lim(1)2nn(1),2nn 1.2 222111(21)(21)(31)(31)(1