The Limit of a Sequence


 Cody Green
 2 years ago
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1 3 The Limit of a Sequece 3. Defiitio of limit. I Chapter we discussed the limit of sequeces that were mootoe; this restrictio allowed some shortcuts ad gave a quick itroductio to the cocept. But may importat sequeces are ot mootoe umerical methods, for istace, ofte lead to sequeces which approach the desired aswer alterately from above ad below. For such sequeces, the methods we used i Chapter wo t work. For istace, the sequece.,.9,.,.99,.,.999,... has as its limit, yet either the iteger part or ay of the decimal places of the umbers i the sequece evetually becomes costat. We eed a more geerally applicable defiitio of the limit. We abado therefore the decimal expasios, ad replace them by the approximatio viewpoit, i which the limit of {a } is L meas roughly a is a good approximatio to L, whe is large. The followig defiitio makes this precise. After the defiitio, most of the rest of the chapter will cosist of examples i which the limit of a sequece is calculated directly from this defiitio. There are limit theorems which help i determiig a limit; we will preset some i Chapter 5. Eve if you kow them, do t use them yet, sice the purpose here is to get familiar with the defiitio. Defiitio 3. The umber L is the limit of the sequece {a } if () give ǫ >, a ǫ L for. If such a L exists, we say {a } coverges, or is coverget; if ot, {a } diverges, or is diverget. The two otatios for the limit of a sequece are: lim {a } = L ; a L as. These are ofte abbreviated to: lim a = L or a L. Statemet () looks short, but it is actually fairly complicated, ad a few remarks about it may be helpful. We repeat the defiitio, the build it i three stages, listed i order of icreasig complexity; with each, we give its traslatio ito Eglish. 35
2 36 Itroductio to Aalysis Defiitio 3. lim a = L if: give ǫ >, a ǫ L for. Buildig this up i three succesive stages: (i) a L (a ǫ approximates L to withi ǫ); ( ) the approximatio holds for all a (ii) a L for ; ǫ far eough out i the sequece; (iii) give ǫ >, a L for ǫ (the approximatio ca be made as close as desired, provided we go far eough out i the sequece the smaller ǫ is, the farther out we must go, i geeral). The heart of the limit defiitio is the approximatio (i); the rest cosists of the if s, ad s, ad but s. First we give a example. Example 3.A Show lim + =, directly from defiitio 3.. Solutio. Accordig to defiitio 3., we must show: (2) give ǫ >, + ǫ for. We begi by examiig the size of the differece, ad simplifyig it: + = 2 + = 2 +. We wat to show this differece is small if. Use the iequality laws: 2 + < ǫ if + > 2 ǫ, i.e., if > N, where N = 2 ǫ ; this proves (2), i view of the defiitio (2.6) of for. The argumet ca be writte o oe lie (it s ugrammatical, but easier to write, prit, ad read this way): Solutio. Give ǫ >, + = 2 + < ǫ, if > 2 ǫ. Remarks o limit proofs.. The heart of a limit proof is i the approximatio statemet, i.e., i gettig a small upper estimate for a L. Ofte most of the work will cosist i showig how to rewrite this differece so that a good upper estimate ca be made. (The triagle iequality may or may ot be helpful here.) Note that i doig this, you must use ; you ca drop the absolute value sigs oly if it is clear that the quatity you are estimatig is oegative. 2. I givig the proof, you must exhibit a value for the N which is lurkig i the phrase for. You eed ot give the smallest possible N; i example 3.A, it was 2/ǫ, but ay bigger umber would do, for example N = 2/ǫ. Note that N depeds o ǫ: i geeral, the smaller ǫ is, the bigger N is, i.e., the further out you must go for the approximatio to be valid withi ǫ.
3 Chapter 3. The Limit of a Sequece I Defiitio 3. of limit, the phrase give ǫ > has at least five equivalet forms; by covetio, all have the same meaig, ad ay of them ca be used. They are: for all ǫ >, for every ǫ >, for ay ǫ > ; give ǫ >, give ay ǫ >. The most stadard of these phrases is for all ǫ >, but we feel that if you are meetig () for the first time, the phrases i the secod lie more early capture the psychological meaig. Thik of a limit demo whose oly purpose i life is to make it hard for you to show that limits exist; it always picks upleasatly small values for ǫ. Your task is, give ay ǫ the limit demo hads you, to fid a correspodig N (depedig o ǫ) such that a ǫ L for > N. Remember: the limit demo supplies the ǫ; you caot choose it yourself. I writig up the proof, good mathematical grammar requires that you write give ǫ > (or oe of its equivalets) at the begiig; get i the habit ow of doig it. We will discuss this later i more detail; briefly, the reaso is that the N depeds o ǫ, which meas ǫ must be amed first. 4. It is ot hard to show (see Problem 33) that if a mootoe sequece {a } has the limit L i the sese of Chapter higher ad higher decimal place agreemet the L is also its limit i the sese of Defiitio 3.. (The coverse is also true, but more trouble to show because of the difficulties with decimal otatio.) Thus the limit results of Chapter, the Completeess Property i particular, are still valid whe our ew defiitio of limit is used. From ow o, limit will always refer to Defiitio 3.. Here is aother example of a limit proof, more tricky tha the first oe. Example 3.B Show lim ( + ) =. Solutio. We use the idetity A B = A2 B 2, which tells us that A + B (3) ( + ) = < ; give ǫ >, 2 < ǫ if 4 < ǫ2, i.e., if > 4ǫ 2. Note that here we eed ot use absolute values sice all the quatities are positive. It is ot at all clear how to estimate the size of + ; the triagle iequality is useless. Lie (3) is thus the key step i the argumet: the expressio must first be trasformed by usig the idetity. Eve after doig this, lie (3) gives a further simplifyig iequality to make fidig a N easier; just try gettig a N without this step! The simplificatio meas we do t get the smallest possible N; who cares?
4 38 Itroductio to Aalysis Questios 3.. Directly from the defiitio of limit (i.e., without usig theorems about limits you leared i calculus), prove that (a) + (b) cos a (a is a fixed umber) 2 + (c) 2 (d) 2 ( ) cf. Example 3.B: make 3 + a simplifyig iequality 2. Prove that, for ay sequece {a }, lim a = lim a =. (This is a simple but importat fact you ca use from ow o.) 3. Why does the defiitio of limit say ǫ >, rather tha ǫ? 3.2 The uiqueess of limits. The Kǫ priciple. Ca a sequece have more tha oe limit? Commo sese says o: if there were two differet limits L ad L, the a could ot be arbitrarily close to both, sice L ad L themselves are at a fixed distace from each other. This is the idea behid the proof of our first theorem about limits. The theorem shows that if {a } is coverget, the otatio lima makes sese; there s o ambiguity about the value of the limit. The proof is a good exercise i usig the defiitio of limit i a theoretical argumet. Try provig it yourself first. Theorem 3.2A Uiqueess theorem for limits. A sequece a has at most oe limit: a L ad a L L = L. Proof. By hypothesis, give ǫ >, a ǫ L for, ad a ǫ L for. Therefore, give ǫ >, we ca choose some large umber k such that L ǫ a k ǫ L. By the trasitive law of approximatio (2.5 (8)), it follows that (4) give ǫ >, L 2ǫ L. To coclude that L = L, we reaso idirectly (cf. Appedix A.2). Suppose L L ; choose ǫ = 2 L L. We the have Remarks. L L < 2ǫ, by (4); i.e., L L < L L, a cotradictio.. The lie (4) says that the two umbers L ad L are arbitrarily close. The rest of the argumet says that this is osese if L L, sice they caot be closer tha L L.
5 Chapter 3. The Limit of a Sequece Before, we emphasized that the limit demo chooses the ǫ; you caot choose it yourself. Yet i the proof we chose ǫ = 2 L L. Are we blowig hot ad cold? The differece is this. Earlier, we were tryig to prove a limit existed, i.e., were tryig to prove a statemet of the form: give ǫ >, some statemet ivolvig ǫ is true. To do this, you must be able to prove the truth o matter what ǫ you are give. Here o the other had, we do t have to prove (4) we already deduced it from the hypothesis. It s a true statemet. That meas we re allowed to use it, ad sice it says somethig is true for every ǫ >, we ca choose a particular value of ǫ ad make use of its truth for that particular value. To reiforce these ideas ad give more practice, here is a secod theorem which makes use of the same priciple, also i a idirect proof. The theorem is obvious usig the defiitio of limit we started with i Chapter, but we are committed ow ad for the rest of the book to usig the ewer Defiitio 3. of limit, ad therefore the theorem requires proof. Theorem 3.2B {a } icreasig, L = lim a a L for all ; {a } decreasig, L = lim a a L for all. Proof. Both cases are hadled similarly; we do the first. Reasoig idirectly, suppose there were a term a N of the sequece such that a N > L. Choose ǫ = 2 (a N L). The sice {a } is icreasig, a L a N L > ǫ, for all N, cotradictig the Defiitio 3. of L = lim a. The Kǫ priciple. I the proof of Theorem 3.2A, ote the appearace of 2ǫ i lie (4). It ofte happes i aalysis that argumets tur out to ivolve ot just ǫ but a costat multiple of it. This may occur for istace whe the limit ivolves a sum or several arithmetic processes. Here is a typical example. Example 3.2 Let a = + si +. Show a, from the defiitio. Solutio To show a is small i size, use the triagle iequality: + si + + si +. At this poit, the atural thig to do is to make the separate estimatios < ǫ, for > ǫ ; si + < ǫ, for > ǫ ; so that, give ǫ >, + si + < 2ǫ, for > ǫ. This is close, but we were supposed to show a < ǫ. Is 2ǫ just as good?
6 4 Itroductio to Aalysis The usual way of hadlig this would be to start with the give ǫ, the put ǫ = ǫ/2, ad give the same proof, but workig always with ǫ istead of ǫ. At the ed, the proof shows + si + < 2ǫ, for > ǫ ; ad sice 2ǫ = ǫ, the limit defiitio is satisfied. Istead of doig this, let s oce ad for all agree that if you come out i the ed with 2ǫ, or 22ǫ, that s just as good as comig out with ǫ. If ǫ is a arbitrary small umber, so is 22ǫ. Therefore, if you ca prove somethig is less tha 22ǫ, you have show that it ca be made as small as desired. We formulate this as a geeral priciple, the Kǫ priciple. This is t a stadard term i aalysis, so do t use it whe you go to your ext mathematics cogress, but it is useful to ame a idea that will recur ofte. Priciple 3.2 The Kǫ priciple. Suppose that {a } is a give sequece, ad you ca prove that (5) give ay ǫ >, a Kǫ L for, where K > is a fixed costat, i.e., a umber ot depedig o or ǫ. The lim a = L. The Kǫ priciple is here formulated for sequeces, but we will use it for a variety of other limits as well. I all of these uses, the essetial poit is that K must truly be a costat, ad ot deped o ay of the variables or parameters. Questios 3.2. I the last (idirect) part of the proof of the Uiqueess Theorem, where did we use the hypothesis L L? 2. Show from the defiitio of limit that if a L, the ca cl, where c is a fixed ozero costat. Do it both with ad without the Kǫ priciple. ( 3. Show from the defiitio of limit that lim + 2 ) =. 3.3 Ifiite limits. Eve though is ot a umber, it is coveiet to allow it as a sort of limit i describig sequeces which become ad remai arbitrarily large as icreases. The defiitio is like the oe for the ordiary limit. Defiitio 3.3 We say the sequece {a } teds to ifiity if (6) give ay M, a > M for. I symbols: lim {a } =, or a as.
7 Chapter 3. The Limit of a Sequece 4 As for regular limits, to establish that lim{a } =, what you have to do is give a explicit value for the N cocealed i for, ad prove that it does the job, i.e., prove that a > M whe N. I geeral, this N will deped o M: the bigger the M, the further out i the sequece you will have to go for the iequality a > M to hold. As before, it is ot you who chooses the M; the limit demo does that, ad you have to prove the iequality i (6) for whatever positive M it gives you. Note also that eve though we are dealig with size, we do ot eed absolute values, sice a > M meas the a are all positive for. Oe should ot thik that ifiite limits are associated oly with icreasig sequeces. Cosider these examples, either of which is a icreasig sequece. Examples 3.3A Do the followig sequeces ted to? Give reasoig. (i) {a } =,, 2, 2, 3, 3, 4, 4,..., k, k,..., (k ); (ii) {a } =, 2,, 3,...,, k,..., (k ). { 5, eve; Solutio. (i) A formula for the th term is a = ( + )/2, odd. This shows the sequece teds to sice (6) is satisfied: give M >, a > M if ( + )/2 > M; i.e., if > 2M. (ii) The secod sequece does ot ted to, sice (6) is ot satisfied for every give M: if we take M =, for example, it is ot true that after some poit i the sequece all a >, sice the term occurs at every odd positio i the sequece. Example 3.3B Show that {l }. Solutio. We use the fact that lx is a icreasig fuctio, that is, therefore, la > lb if a > b; give M >, l > l(e M ) = M if > e M. Questios 3.3. (a) Formulate a defiitio for lim a = : a teds to. (b) Prove l(/). 2. Which of these sequeces ted to? For those that do, prove it. (a) ( ) (b) si π/2 (c) (d) + cos 3. Prove: if a, the a is positive for large.
8 42 Itroductio to Aalysis 3.4 A importat limit. As a good opportuity to practice with iequalities ad the limit defiitio, we prove a importat limit that will be used costatly later o. Theorem 3.4 The limit of a. (7) lim a = Proof., if a > ;, if a = ;, if a <. We cosider the case a > first. Sice a >, we ca write a = + k, k >. Thus a = ( + k), which by the biomial theorem ( ) = + k + k 2 ( )( 2) + k k. 2! 3! Sice all the terms o the right are positive, (5) a > + k ; > M, for ay give M >, if > M/k, say. This proves that lim a = if a >, accordig to Defiitio 3.3. The secod case a = is obvious. For the third, i outlie the proof is: a < ( ) a > a. a Here the middle implicatio follows from the first case of the theorem. The last implicatio uses the defiitio of limit; amely, by hypothesis, ( ) give ǫ >, > for large; a ǫ by the reciprocal law of iequalities (2.) ad the multiplicatio law for, a < ǫ for large. Why did we begi by writig a = + k? Experimetally, you ca see that whe a >, but very close to (like a =.), a icreases very slowly at first whe raised to powers. This is the worst case, therefore, ad it suggests writig a i a form which shows how far it deviates from. The case a is ot icluded i the theorem; here the a alterate i sig without gettig smaller, ad the sequece has o limit. A formal proof of this directly from the defiitio of limit is awkward; istead we will prove it at the ed of Chapter 5, whe we have more techique. Questios 3.4. Fid (a) lim cos a; (b) lim l a, for a. 2. Suppose oe tries to prove the theorem for the case < a < directly, by writig a = k, where < k < ad imitatig the argumet give for the first
9 Chapter 3. The Limit of a Sequece 43 case a >. Where does the argumet break dow? Ca oe prove a is small by droppig terms ad estimatig? Could oe use the triagle iequality? 3.5 Writig limit proofs. Get i the habit of writig your limit proofs usig correct mathematical grammar. The proofs i the body of the text ad some of the aswers to questios (those aswers which are t just brief idicatios) are meat to serve as models. But it may also help to poit out some commo errors. Oe frequetly sees the followig usages ivolvig for large o studet papers. Your teacher may kow what you mea, but the mathematical grammar is wrog, ad techically, they make o sese; avoid them. Wrog Right a for ; a as ; lim 2 = for ; lim 2 = ; { lim(/) = ; lim(/) = if > /ǫ; (/) if > /ǫ. ǫ I the first two, the limit statemet applies to the sequece as a whole, whereas for > some N ca oly apply to idividual terms of the sequece. The third is just a geeral mess; two alteratives are offered, depedig o what was origially meat. As we said earlier, give ǫ > or give M > must come first: Poor: / < ǫ, for (what is ǫ, ad who picked it?) Wrog: For, give ǫ >, / < ǫ. This latter statemet is wrog, because accordig to mathematical covetios, it would mea that the N cocealed i for should ot deped o ǫ. This poit is more fully explaied i Appedix B; rather tha try to study it there at this poit, you will be better off for ow just rememberig to first preset ǫ or M, ad the write the rest of the statemet. Aother poit: write up your argumets usig plety of space o your paper (sorry, clarity is worth a tree). Ofte i the book s examples ad proofs, the iequality ad equality sigs are lied up below each other, rather tha strug out o oe lie; it is like properlywritte computer code. See how it makes the argumet clearer, ad imitate it i your ow work. If equalities ad iequalities both occur, the covetio we will follow is: A < ( + ) A < ( + ) = 2 rather tha +, < 2 +. The form o the right does t tell you explicitly where the secod lie came from; i the form o the left, the desired coclusio A < 2 + is t explicitly stated, but it is easily iferred.
10 44 Itroductio to Aalysis 3.6 Some limits ivolvig itegrals. To broade the rage of applicatios ad get you thikig i some ew directios, we look at a differet type of limit which ivolves defiite itegrals. Example 3.6A Let a = (x 2 + 2) dx. Show that lim a =. The way ot to do this is to try to evaluate the itegral, which would just produce a uwieldy expressio i that would be hard to iterpret ad estimate. To show that the itegral teds to ifiity, all we have to do is get a lower estimate for it that teds to ifiity. Solutio. therefore, Thus We estimate the itegral by estimatig the itegrad. x for all x; (x 2 + 2) 2 for all x ad all. (x 2 + 2) dx 2 dx = 2. Sice lim 2 = by Theorem 3.4, the defiite itegral must ted to also: give M >, Example 3.6B Show lim (x 2 + 2) dx 2 M, for > log 2 M. (x 2 + ) dx =. Solutio. Oce agai, we eed a lower estimate for the itegral that is large. The previous argumet gives the estimate (x 2 + ) =, which is useless. However, it may be modified as follows. Sice x 2 + is a icreasig fuctio which has the value A = 5/4 at the poit x =.5 (ay other poit o (,) would do just as well), we ca say therefore, x 2 + A > for.5 x ; (x 2 + ) A for.5 x ; sice lim A = by Theorem 3.4, the defiite itegral must ted to also: give M >, (x 2 + ) dx.5 A dx = A 2 M, for large. Questios 3.6. By estimatig the itegrad, show that: 2. Show without itegratig that lim 3 x 2 + x dx. x ( x) dx =.
11 Chapter 3. The Limit of a Sequece Aother limit ivolvig a itegral. We give a third example, this oe a little more sophisticated tha the other two. You ca skip it, but it is worth studyig ad uderstadig. Example 3.7 Let a = π 2 si x dx. Determie lim a. Remarks. As before, attemptig to evaluate the itegral will lead to a expressio i whose limit is ot so easy to determie. The itegral represets a area, so it helps to have some idea of how the curves si x look. Sice six < si x o the iterval x < π 2, by Theorem 3.4 the powers = si 2 x. Thus as icreases, the successive curves get closer ad closer to the xaxis, except that the righthad ed always passes through the poit ( π 2,). x π/2 The area uder the curve seems to get small, as icreases, so the limit should be. But how do we prove this? For a big value of, the area really is composed of two parts, both of which are small, but for differet reasos. The vertical strip o the right is small i area si x because it is thi. The horizotal strip below is small i si a area because the curve is ear the xaxis. ε This suggests the followig procedure. a π/2 Solutio. Give ǫ >, we show: area uder si x ad over [, π 2 ] < 2ǫ, for. (This suffices, by the Kǫ priciple of sectio 3.2.) Mark off the the poit a show, where a = π 2 ǫ. Divide up the area uder the curve ito two pieces as show, ad draw i the two rectagles. righthad area < area of righthad rectagle = ǫ ; lefthad area < area of horizotal rectagle = asi a < ǫ, for, sice si a < (cf. Theorem 3.4). Therefore, π 2 si x dx = total area uder curve < area of the two rectagles < 2ǫ, for. It would be easy to get rid of the pictures ad just use itegral sigs everywhere, but it would t make the argumet more rigorous, just more obscure. x
12 46 Itroductio to Aalysis Exercises 3.. Show that the followig sequeces have the idicated limits, directly from the defiitio of limit. si cos (a) lim (c) lim (e) = 3.2 lim 2 = (b) lim = (d) lim + 2 = 2 3 = 2. Prove that if a is a oegative sequece, lim a = lim a =.. Prove that if a L ad b M, the a + b L + M. Do this directly from the defiitio 3. of limit. 2. Suppose {a } is a coverget icreasig sequece, ad lim a = L. Let {b } be aother sequece iterwove with the first, i.e., such that a < b < a + for all. Prove from the defiitio of limit that lim b = L also. 3. (a) Prove the sequece a = (b) Criticize the followig proof that its limit is : Give ǫ >, the for i =,2,3,..., we have < ǫ, if < ǫ, i.e., if > /ǫ. + i Addig up these iequalities for i =,..., gives therefore, < a < ǫ, for > /ǫ ; a ǫ, for. By the defiitio of limit ad the Kǫ priciple, lim a =. ( 4. Prove that lim ) 2 =. + (Modify the icorrect argumet i the precedig exercise.) has a limit. 5. Let {a } be a coverget sequece of itegers, havig the limit L. Prove that it is evetually costat, that is, a = L for large. (Apply the limit defiitio, takig ǫ = /4, say. Why is it legal for you to choose the ǫ i this case?)
13 Chapter 3. The Limit of a Sequece For each sequece {a }, tell whether or ot a ; if so, prove it directly from Defiitio 3.3 of ifiite limit. (a) a = 2 (b) a = 2 cos π (c) a = (d) a =, 2, 3, 2, 3, 4, 3, 4, 5,... (e) (f) l l() 2. Prove: if a < b for, ad a, the b. Base the proof o Defiitio Prove: if {a }, the {a } is ot bouded above. (This is obvious ; the poit is to get practice i usig the defiitios to costruct argumets. Give a idirect proof; see Appedix A.2.) 3.4. Defie a recursively by a + = ca, where c <. Prove lim a =, usig Theorem 3.4 ad Defiitio Let a = r /!. Prove that for all r R, we have a. (If r, this is easy. If r >, it is more subtle. Compare two successive terms of the sequece, ad show that if, the a + is less tha half of a. The complete the argumet.) 3. Prove that if a >, the a /. (Hit: imitate the proof i theorem 3.4, but use a differet term i the biomial expasio.) 4. Prove that a if < a <, usig the result i the precedig exercise. 5. Prove that lima / =, if a >. (Here a / meas the real positive th root of a. Nothig is said about a <, sice the a / is ot a real umber if is eve. Note how {a / } ad {a } have opposite behavior as sequeces. As we take successive th roots, all positive umbers approach ; i cotrast, as we take successive th powers, all positive umbers recede from.) (a) Cosider first the case a >. For each, put a / = + h, ad show that h, by reasoig like that i Theorem 3.4. (b) Now cosider the case a <. If a <, the /a > ; use this ad the defiitio of limit to deduce this case from the previous oe. (Use oly the defiitio of limit i this chapter, ot other obvious facts about limits.) 3.6. Modelig your argumets o the two examples give i this sectio, prove the followig without attemptig to evaluate the itegrals explicitly. (a) lim 2 l x dx = (b) lim 3 2 l x dx =
14 48 Itroductio to Aalysis 3.7. Show that lim the itegral explicitly. ( x 2 ) dx =, without attemptig to evaluate Problems 3 Let {a } be a sequece ad {b } be its sequece of averages: b = (a a )/ (cf. Problem 2). (a) Prove that if a, the b. (Hit: this uses the same ideas as example 3.7. Give ǫ >, show how to break up the expressio for b ito two pieces, both of which are small, but for differet reasos.) (b) Deduce from part (a) i a few lies without repeatig the reasoig that if a L, the also b L. 32 To prove a was large if a >, we used Beroulli s iequality : ( + h) + h, if h. We deduced it from the biomial theorem. This iequality is actually valid for other values of h however. A sketch of the proof starts: ( + h) 2 = + 2h + h 2 + 2h, sice h 2 for all h; ( + h) 3 = ( + h) 2 ( + h) ( + 2h)( + h), by the previous case, = + 3h + 2h 2, + 3h. (a) Show i the same way that the truth of the iequality for the case implies its truth for the case +. (This proves the iequality for all by mathematical iductio, sice it is trivially true for =.) (b) For what h is the iequality valid? (Try it whe h = 3, = 5.) Recocile this with part (a). 33 Prove that if a is a bouded icreasig sequece ad lima = L i the sese of Defiitio.3A, the lima = L i the sese of Defiitio Prove that a coverget sequece {a } is bouded. 35 Give ay c R, prove there is a strictly icreasig sequece {a } ad a strictly decreasig sequece {b }, both of which coverge to c, ad such that all the a ad b are (i) ratioal umbers; (Theorem 2.5 is helpful.) (ii) irratioal umbers.
15 Chapter 3. The Limit of a Sequece 49 Aswers 3.. We write these up i four slightly differet styles; take your pick. (a) Give ǫ >, + = +, (b) Give ǫ >, (c) Give ǫ >, which is < ǫ if > /ǫ, or > /ǫ. cos a, sice cos x for all x; ad / < ǫ if > /ǫ = 2 2, 2 < ǫ if 2 > 2/ǫ, or > 2/ǫ +. (d) Give ǫ >, 3 + < < ǫ if > /ǫ. 2. lim a = meas: give ǫ >, a < ǫ for. lim a = meas: give ǫ >, a < ǫ for. But these two statemets are the same, sice a = a = a = a. 3. If the limit demo were allowed to give you ǫ =, the sice is the same as equality =, it would have to be true that a = L for ; i other words, ay sequece which had the limit L would from some poit o have to be costat ad equal to L. This would be too restricted a otio of limit. (The sequeces which do behave this way are said to be evetually costat ; cf. Exercise 3.2/5.) 3.2. L L ǫ = L L /2 >, which is essetial (cf. Questio 3./3). 2. (a) Give ǫ >, a L < ǫ for > N, say. Therefore, so we re doe by the Kǫ priciple. ca cl = c a L < c ǫ for > N, (b) Give ǫ >, a L < ǫ/ c for > N, say. Therefore, ca cl = c a L < c ǫ/ c = ǫ for > N. 3. Give ǫ >, (triagle iequality) < ǫ + ǫ, if + > /ǫ, ad > 2/ǫ; so we re doe by the Kǫ priciple. < 2ǫ, if > + 2/ǫ;
16 5 Itroductio to Aalysis (a) Give M <, a < M for. (The sigs ca be omitted.) (b) l(/) = l < M if l > M, i.e., if > e M. 2. (a) o; alterate terms are egative; (b) o; alterate terms are ; (c) yes; give M >, > M if > M 2 ; (d) yes; give M >, + cos > > M, if > M Give M, a > M for. Take M = : a > for.. (a) limit is, except: limit is if a = 2π, o limit if a = (2 + )π. {, if a < e; (b) limit is:, if a = e;, if a > e. 2. Sice the terms alterate i sig, oe caot get a iequality after droppig most of the terms to simplify the expressio. Basically, a is small ot because the idividual terms are small, but because they cacel each other out. Thus the triagle iequality caot help either. So this approach does t lead to a usable estimatio that would show a is small Over the iterval x, x 2 + 2; 2 x By stadard reasoig (see Example 2.2B for istace), 3 x2 + x ; therefore its itegral over the iterval [,] of legth also lies betwee these bouds. 2. O [,], the maximum of x( x) occurs at x = /2, therefore give ǫ >, x( x) /4 o [,]; therefore, deotig the itegral by I, we see that which proves that limi =. x ( x) dx (/4) < ǫ, if 4 > /ǫ ; give ǫ >, I < ǫ if > l(/ǫ)/l 4,
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