MATH 2710: NOTES FOR ANALYSIS

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1 MATH 270: NOTES FOR ANALYSIS The main ideas we will learn from analysis center around the idea of a limit. Limits occurs in several settings. We will start with finite limits of sequences, then cover infinite limits, and then look at limits of functions. We have not given a roer definition of R, so we can t really treat limits comletely because they may not exist just in Q. However, all the relevant definitions make sense in Q. I will note in the text where the situation is different between Q and R. Sequences and limits of sequences A sequence is just a function with domain N. We will mostly be interested in sequences with values in the real numbers. The usual function notation a : N R, which would list the values in the sequence as a(), a(2), a(3),... is often not the most convenient way to write a sequence. Instead we write the argument as a subscrit, so the foregoing would read: a, a 2, a 3,. Sometimes we write the sequence as {a n } or {a n }. Like functions, sequences are often written as rules. () The rule a n = n 2 defines a sequence with values, 4, 9, 6, 25,... (2) We could write { n } for the sequence, 2, 3, 4, 5,... (3) Often, a sequence is given by writing a few terms and leaving the reader to infer a rule, e.g. writing 2, 3, 5, 7,, 3,... to indicate the sequence of rimes. This is imrecise, but we will still do it from time to time. The rototye of a convergent sequence is something like { n }, where it seems that the values get arbitrarily close to zero and stay close to zero as n increases. The notion of closer is measured using the distance (in the real numbers). Definition. The distance between x, y R is x y, where the absolute value is defined to be { w if w 0, w = w if w < 0. Now we have a distance we can talk about x being closer to y than z is to y, meaning x y < z y. If ɛ > 0 we can also talk about x being closer than ɛ to y, meaning x y < ɛ. We also make a note of the most useful roerty of distance; we record it as a lemma. Lemma 2 (Triangle Inequality). For oints x, y, z R we have x y x z + z y. Intuitively this says the distance from x to y is no larger than the distance from x to y via z. Proof. This is a straightforward comutation using cases. Clearly the oints are arranged on a line. Swaing x and y does not change the exression, so we can assume x > y (the case x = y is obvious). The cases are then z x, x > z > y, y z. In all cases x y = x y. What changes is the right side. If z x then z y x y, so the inequality is true. If y z then also x z x y so the inequality is true. Finally, if x > z > y then x z = x z and z y = z y so x z + y z = x z = x z. This comletes the roof. A very useful version of the triangle inequality is as follows. Lemma 3. For oints a, b R we have a b a b.

2 2 MATH 270: NOTES FOR ANALYSIS Proof. The triangle inequality says x + y x + y, so x + y y x. Now ut x + y = a and y = b, so x = a b. We obtain a b a b. Reeating this for x + y = b, y = a so x = b a we have b a b a = a b. But a b is equal to one of a b or b a, both of which we showed are no larger than a b, comleting the roof. The idea of a sequence x n converging to a number L is suosed to cature that x n gets arbitrarily close to L as n gets large. The notion of arbitrarily close clearly requires a quantifier: there is no secific roximity we are looking for, but instead we are looking to get the x n closer to L than any ostive number ɛ. The notion of n being large also requires a quantifier. However, a little thought shows us that for a secfic tolerance ɛ > 0 the distance we have to go out along the sequence to get x n L < ɛ deends both on ɛ and on the sequence. For examle, with x n = n, L = 0 and ɛ = 0. we need n > 0 to get x n L < ɛ, while for ɛ = 0.0 we would need n > 00; if the sequence was instead y n = n 2 then to get y n L < 0. we d only need n > 3 and to get y n L < 0.0 we would need n > 0. The fact that the quantifier for how far out in the sequence we go can deend on ɛ means it should be determined after ɛ is determined. The above shows that for every ɛ > 0 we want that there is a notion of far enough out in the sequence, namely a N, ossibly deending on ɛ, such that going any further out, meaning taking n > N, gives x n L < ɛ. This is what we use to make the formal definition of a limit. Before we give this definition, one other imortant oint: if you don t understand the reasoning for defining the limit in this way, it is not essential to do so. Some eole use the definition for a long time before they really get it. If it doesn t make sense to you intuitively, don t worry about it just learn the definition and ractice using it Definition 4. The sequence a n converges to L if ɛ > 0 N such that n N = x n L < ɛ. This is often written as lim n a n = L or as a n L when n. From this we can make two other definitions. The first is that a sequence converges if there is some number to which it converges, while the second is that otherwise it diverges. Definition 5. A sequence a n is convergent if there is an L such that a n L when n. If a n is not convergent it is called divergent. Remark 6. The issue of whether a sequence converges is where Q and R are very different. Even if all the numbers we use in our definitions, excet L, are from Q, there are sequences that converge in R and do not converge in Q. A simle examle: in a homework roblem you saw how given a ositive number a you could construct a number ã so that ã 2 is closer to 2 than a 2. This can be used recursively to construct a sequence x j in Q such that x 2 j 2. Of course x j does not converge in Q, because the limit L would have to have L 2 = 2 and we also roved there is no such L Q. But it turns out one can rove that x j 2 in R. Now we are going to do some simle roofs that secific sequences converge to secified values. In each case the roof begins by saying that ɛ > 0 and then says how to choose N (deending on ɛ) so as to ensure that n > N = x n L < ɛ. However, the oint is to learn how to do these yourself, so each roof is receeded by a descrition of the thought rocess used to make the choice I gave. Note that the choice of N in each case is not unique any larger N would have worked so the reasoning I give is not unique either. Your reasoning for a roblem might be different to mine, which is fine so long as it works. Examle 7. Prove { n } converges to 0 as n. My reasoning: Given ɛ > 0 I want N so n > N will make x n 0 = n < ɛ. To get this I need n > ɛ, so I take N > ɛ. Proof: Given ɛ > 0 let N > ɛ. If n > N then x n 0 = n < N < ɛ, so n 0.

3 Examle 8. Prove { 4n n+3 } converges to 4 as n. My reasoning: With x n = 4n n+3 we have x n 4 = 4n 4(n+3) n+3 MATH 270: NOTES FOR ANALYSIS 3 = 2 n+3 which will be less than ɛ if n + 3 > 2 ɛ. It is temting to take N > 2 ɛ 3, so n > N imlies n + > 2 ɛ and 2 n+3 < ɛ, but it is just as simle and erhas a little neater to take N > 2 ɛ and say n > N imlies 2 n+3 < 2 N < ɛ. Proof: Given ɛ > 0 take N > 2 ɛ. Then n > N imlies x n 4 = 4n 4(n+3) n+3 = 2 n+3 < 2 N < ɛ. So x n 4. Examle 9. Prove 3 n 3 converges to 0 as n. Reasoning: We need N so n > N imlies < ɛ. Evidently N > ( ) 3 /3 n 3 ɛ will work, so we can use that. An alternative would be to say that if n 2 > 3, for which we only need n > N > 2 we would have n 3 > 3N so 3 < n 3 N < ɛ if N > ɛ. I refer this latter way, which requires N > max{2, ɛ } only because the method is more flexible. Proof: Given ɛ > 0 take N > max{2, ɛ }. Then N 2 > 4 so n 3 > 3n and 3 < n 3 n < ɛ which shows the result. Examle 0. Prove { n+ } converges to 0. n 2 3 Reasoning: We need n+ < ɛ. It it a bit messy to figure out what the best choice of N so n > N n 2 3 would make this haen we d have to solve a quadratic equation but we don t need the best N we just need one that works. Accordingly we make some crude inequalities that bound the numerator and denominator just like we did in the revious roblem. Notice n+ < 2n for n >. Also, lets look for where n 2 3 > n2 two inequalities and also 2. This occurs when n2 > 6, so when n > 2. If we have those, then we get the inequality we want. So we 2n (n 2 /2) < ɛ, meaning n > 4 ɛ take N > max{2, 4 ɛ }. Proof: Given ɛ > 0 let N > max{2, 4 ɛ }. If n > N then n > 2 which imlies n + < 2n and n2 > 6, hence n 2 3 > n2 n+ 2. Then n 2 3 < 4n = 4 n 2 n < ɛ, showing the result. At this oint it might be worth summarizing the structure that aeared in all these roofs. When giving a roof of a limit of a sequence from the definition, you can always use the following structure: Given ɛ > 0 here is a rule for getting N (usually as a function of ɛ) and an argument showing that this choice of N ensures that a n L < ɛ. It is now about time to rove that some sequences are divergent. The basic idea is to use contradiction: assume that the sequence is convergent and derive a contradiction. This works well on sequences that diverge to infinity, but also on sequences that oscillate around without settling down in the real numbers. One way to have the latter occur is if the sequence aears to have arts that are converging to different values. We will see examles of both tyes. In both cases the key is that if the sequence did converge then all x n values are eventually close together, so if this is not true the sequence must diverge. Examle. Prove that {( ) n } does not converge as n. Reasoning: No matter how far you go out in the sequence,,,,,... you have oints and that are searated by distance 2. If x n is within ɛ of L for all n then L is within ɛ of both and. But the triangle inequality allows us to say that the distance between and is at most the distance via L, which is L + L so 2 < 2ɛ which gives ɛ >. If we take ɛ = we should get a contradiction. Proof: If the sequence converged to L then for ɛ = there is N so n > N imlies x n L <. But then 2 = 2ɛ > x n L + x n+ L = L + L = 2 by the triangle inequality, a contradiction.

4 4 MATH 270: NOTES FOR ANALYSIS In the revious examle it looks like this sequence has two ieces: the odd terms are all so are converging to and the even terms are all so are converging to. The issue is that the sequence can t converge to two different laces. The idea of the sequence having ieces that do these two different things can be made recise using the idea of subsequences and limit oints. We won t do that right now, but we will reeat the above argument to show that, in fact, it is always imossible for a sequence to converge to two laces. Lemma 2. A convergent sequence has a unique limit. Reasoning: we use a small modifiation of the argument used in the revious exercise. The idea is that the sequence can t simultaneously converge to two different oints L and L because these oints are a ositive distance aart, so it is not ossible to be simultaneously arbitrarily close to both of the oints. The issue is to make this clear with some sort of ɛ and N. It may hel to draw a number line showing L and L. When you do so you will see that it is not ossible to be within 2 L L of both L and L, so this is what we take for ɛ in order to get a contradiction. Proof. Suose a n L and also a n L as n. If L L take ɛ = 2 L L > 0 and get N so n > N imlies x n L < ɛ and N 2 so n > N 2 imlies x L < ɛ. Then for n > N = max{n, N 2 } we have by the triangle inequality L L L x n + L x n < ɛ + ɛ = L L which is a contradiction. We conclude L = L, so the limit is unique. Examle 3. Prove that {n 2 } does not converge as n. Reasoning: If it converged to L then for any tolerance ɛ it would eventually be within ɛ of L. If this haened then n > N would mean n 2 and (n + ) 2 would be within 2ɛ of each other, but the difference between them is 2n + 2 > 2. So the condition for convergence will fail if ɛ <. Proof: If the sequence converged to L then for ɛ = 2 we could take N so n > N would imly both n 2 L < ɛ and (n+) 2 L < ɛ. But then using the triangle inequality n 2 (n+) 2 + < 2ɛ =, so 2n + < when n > N, which is a contradiction. For the revious examle, if you were given the roblem in a calculus class you would have been exected to say that the limit is. Obviously we can t rove this using the definition we have for sequences with finite limits, because the notion of x n < ɛ does not make sense the symbol does not refer to a number, and the distance we are using is not defined if one of the quantities is infinite. Instead, what x n is suosed to mean is that the values get arbitrarily large, arbitrarily meaning larger than any fixed M. The exression x n has a similar meaning. Definition 4. We say x n as n or lim n x n = if M N such that n > N = x n > M. We say x n or lim n x n = if M N such that n > N = x n < M. Examle 5. Prove that {n 2 } converges to as n. Reasoning: Given M we need to show that n 2 is bigger than M. To do so we need to say n 2 is larger than something. One way is to say n 2 2n > n if n > 2, and thus n 2 > M if N > M. So take N = max{2, M} to get the right inequalities. Proof: Given M let N = max{2, M}. If n > N then n > 2 so n 2 2n > n > N M. Algebraic roerties of limits In calculus you would have been taught certain basic rules about limits that were used to comute limits of comlicated limits from simler limits. They are established in the following theorems, which are collectively referred to as the limit laws. Theorem 6. If a n a and b n b then (a n + b n ) (a + b).

5 MATH 270: NOTES FOR ANALYSIS 5 Reasoning: We know (a n + b n ) (a + b) a n a + b n + b. If we want this latter to be less than a given ɛ > 0 we could make both a n a < ɛ 2 by making n > N and b n b < ɛ 2 by making n > N 2. Then n > max{n, N 2 } would make both true and give us the result. Proof. Given ɛ > 0 use the assumed limits to take N so n > N imlies a n a < ɛ 2 and N 2 so n > N 2 so b n b < ɛ 2. Let N = max{n, N 2 } so n > N imlies both n > N and n > N 2, whence a n a < ɛ 2 and b n b < ɛ 2 so (a n + b n ) (a + b) a n a + b n b < ɛ 2 + ɛ 2 = ɛ. Theorem 7. If a n a and b n b then a n b n ab. Reasoning: The key to this comutation is that a n b n ab = a n (b n b) + b(a n a) a n b n b + b a n a. We can take N so n > N imlies a n a < ɛ 2 b and therefore the second term less than ɛ 2. We d like to deal with the first term in a similar way but in that case both factors deend on n, so we d want to a n to be bounded by something. However, since a n a we can see that a n a + if n is large enough. Secifically, if ɛ = then there is N 2 so if n > N 2 then a n a < and ɛ thus a a n a +. Moreover, there is N 3 so if n > N 3 then b n b < 2( a +). If n > N = max{n, N 2, N 3 } then we should get the result. Along the way it seems we roved the following result which will be useful later as well. Lemma 8. If a n a then there is N so n > N imlies a a n a +. Proof. Take ɛ = and N so n > N imlies a n a <. Then a < a n < a +. Proof of theorem. Given ɛ > 0 use the assumed limit of a n to take N so n > N imlies a n a < ɛ 2 b. Using the lemma and a n a take N 2 so n > N 2 so a n < a +. Finally use the limit b n b ɛ to take N 3 so n > N 3 imlies b n b < 2( a +). Then ɛ a n b n ab = a n (b n b) + b(a n a) a n b n b + b a n a ( a + ) 2( a + ) + b ɛ 2 b = ɛ. A secial case of the receding is that if b n b then b n b, because it is easy to rove that the constant sequence converges to. Then the fact that one can add convergent sequences term by term and add the limits means one can do the same with differences. Secifically, a n a and b n b imlies, as we just said, b n b, and using the result about sums, a n b n a b. To do the result corresonding to division we have to work a little harder. Theorem 9. If a n a and a 0 then a n a. Reasoning: When we comute the difference between the sequence elements and the utative limit we find = a n a a a n a a n which looks romising because a a n can be made small by making n large. However, to get this less than ɛ > 0 we need to know a lower bound for a a n. There is no roblem with a because it is a non-zero constant. For a n we could use the lemma already established and find N so n > N imlies a n a, but this would not be useful unless we knew a > 0. We d need something better, like a n > a a 2. But we can get this: just take ɛ = 2 > 0 and then N so n > N imlies a n a < ɛ = a 2, from which a n > a 2. Then a a n > a 2 2. We could then take N 2 so n > N 2 imlies a n a < ɛ a 2 2.

6 6 MATH 270: NOTES FOR ANALYSIS Proof. First take ɛ = a 2 and use a n a to get N so n > N imlies a n a < a 2 so that a n > a 2. Then take N 2 so n > N 2 imlies a n a < ɛ a 2 2. For n > N = max{n, N 2 } comute = a n a a a n a a n < 2 ɛ a 2 a 2 = ɛ 2 which roves the result. Obviously we can combine this with the revious theorem to obtain that a n a 0 and b n b imlies bn a n b a. At this oint we can use these theorems and the fact that n 0 to obtain the results of the other exercises we have done so far which established convergence, as follows. We have 3 = 3 n 3 n n n is a roduct of a constant sequence and three sequences that converge to zero. By our roduct result alied three times we find 3 0. n 3 Rewrite n+ n 2 3 = (/n)+(/n)(/n) 3(/n)(/n). Then the numerator converges to zero (by alying the roduct result and the sum result) and the denominator converges to by the same argument. Using the ratio result we find that the sequence converges to 0. In fact it is not hard to see that the limit of any rational function of n can be comuted in this manner as n, if the limit exists. This does not mean that all limits can be done in this way. Here is an imortant one that cannot be done in this manner. Lemma 20. If r < then r n 0 as n. Reasoning: Given ɛ > 0 we need N so n > N imlies r n < ɛ. We don t know how to do this for general r because we don t know anything about the owers of r, but there is one situation in which we do know about owers the situation in which we can aly the binomial theorem. Some thought suggests that this will let us rove the result in the secial case where r = then ( + ) n = n ) 0 n j n n. So ( n j ( ) n n + n n = 0 as n. n +, because What is more, if r < + this still roves r n 0 as n, and since the distance from r to is ositive (because r < ) and + as we can find a so + is in the interval ( r, ), which makes r <. This is the idea of the roof. + Proof. We begin with the fact that as, which is easily established using the limit laws. Use this to find a for which + < r. Since + < we conclude that r < + <. Now r n 0 = r n < ( ) n. + Given ɛ > 0 take N > ɛ (notice that this is ok because we already decided on the value for ). Then comute ( + ) n n n using the binomial theorem and comlete the roof by deducing ( ) n + n N < ɛ. One can also determine that r > imlies r n as n by a similar method. Try it for yourself!

7 MATH 270: NOTES FOR ANALYSIS 7 Series We know how to add finitely many terms. Suose we have a sequence a, a 2, a 3,.... We can make a new sequence as follows: s = a s 2 = a + a 2 s 3 = a + a 2 + a 3. s n = n j= If the sequence s, s 2, s 3,... converged to a limit L, it would be natural to think of L as the result of the infinite sum a + a 2 + a 3 +. We define this to be the case. Definition 2. If the sequence { n j= a } j n= converges to L as n then we set a j = L and call this a convergent series. There are some basic examles of convergent series, as follows. Examle 22. Prove Reasoning: We consider n k(k+) = k k+. Then n a j j(j+) =.. Several aroaches are ossible. One is to recognize that j(j+) j(j + ) = ( ) ( ) ( ) ( ) Internal cancellation then suggests that the sum is n+. We should be able to rove it by induction. When n = we get 2. If the result was true for n then adding the next term we would get n + + (n + )(n + 2) = ( ) = n + n + 2 n + 2, which verifies the inductive ste. Using this and the fact that n+ 0 as n gives the result. Proof. We first rove by induction that n sides are 2. If it is true for n then n+ so the formula is true for all n by induction. Then use that j(j+) = n+. This is true for n = where both j(j + ) = n + + (n + )(n + 2) = n + 2, n+ 0 as n, which may be roved directly (given ɛ > 0 take N > ɛ and n + > N > (/n) ɛ imlies the result) or by writing it as +(/n) and using the limit laws. This imlies as n, establishing the convergence of the series and that its sum is. n j(j+) Examle 23. If r < rove that rn = r. Reasoning: This is done using a trick. Comute n n n ( r) r j = r j r j+ = Thus n rj = rn+ r n n+ r j r j = r n+.. This and the fact that r n 0 as n roves the result. 2

8 8 MATH 270: NOTES FOR ANALYSIS Proof. Comute ( r) n r j = n r j n r j+ = n n+ r j r j = r n+ 2 and deduce n rj = rn+ r. Use Lemma 20 from the end of the last section and the given fact r r r < to get r n 0 as n and multily by the constant tells us that lim n n rj = r, which is what we wanted to rove. to get rn+ r 0 as well. This There are some other basic facts one can rove about convergence of series. One of the most imortant is an easy test for divergence. Theorem 24. If r j converges then r j 0. Reasoning: The oint here is that convergence of the series means the sequence n r j converges, so the terms of this sequence of artial sums are all close to a limiting value when n is large, which in turn means that all the terms must be close to one another. However the difference between consecutive terms for n and n + is recisely r n+, so r n+ must also be small, in fact (by the triangle inequality) less than twice the distance from the artial sum to its limit value. What remains is to quantify this with ɛ and N. Proof. The series converges to a limit L, so given ɛ > 0 there is N so n > N imlies n r j L < ɛ 2. In articular, using n > N and thus n + > N, and the triangle inequality: r n+ = r j L ) ( n r j L ) n+ r j L + n r j L < ɛ (n+ so for n > N we have r n+ < ɛ which roves the result. It is a very imortant fact that you learned in calculus class that the converse of this result is false: there are sequences for which the terms go to zero but the series diverges. A basic examle is the harmonic series j, for several reasons, one of which is that the method of roof is useful elsewhere. Theorem 25. The series j diverges to. Reasoning: The roof contains an imortant idea, which is that one can break the sum u into blocks of different lengths such that each block contributes at least a constant amount to the sum. The decomosition into blocks is as follows: j = ( 3 + ) ( ) ( 4 + ) ( ) What we see is that adding the terms in the block from 2 n to 2 n we have 2 n terms each of size at least 2 n, so the sum of this block contributes at least 2 to the series. We can make this recise using induction, because this will show 2 n + n 2. From here it is easy to show the series diverges as stated.

9 Proof. We rove by induction that 2 n inductive ste is 2 n+ 2n j = 2n+ j + 2 n+ MATH 270: NOTES FOR ANALYSIS 9 j + n 2 j + n 2n n+ for n. This is true for n =, and the 2 (n+) + n 2 + 2n 2 (n+) = + n + 2 so it is true for all n. Now given M take N > 2 2M. Then n > N imlies N j 2 2M + M > M, so the artial sums diverge to. Limits of functions The limits you most frequently encounter in calculus class are limits of functions not sequences. In these notes we will limit ourselves to considering limits of the form lim x c f(x) = L where f : (a, b) R is a function from an oen interval (a, b) R and c (a, b). The idea is that we can make the values of f(x) arbitrarily close to the limiting value L by making x close enough to c, though not equal to c. As with limits of sequences we want f(x) arbitrarily close to L, so we want for all ɛ > 0 that we can make f(x) L < ɛ. In order to get this to haen we need to take x close enough to c, which requires quantification: we need a distance δ > 0, deending on f and ɛ, such that making x c < δ but x c will ensure f(x) L < ɛ. Definition 26. For a function f : (a, b) R and c (a, b) we say lim x c f(x) = L if ɛ > 0 δ > 0 such that 0 < x c < δ = f(x) L < ɛ. Remark 27. The condition 0 < x c is what ensures x c. Our choice of c (a, b) ensures that if δ < min{ a c, b c } then 0 < x c < δ imlies x (a, b) and therefore f(x) makes sense. Following the same notation as was used for limits of sequences we make the following further deifnitions. As was the case for sequences, the question of whether a limit exists can have different answers in Q and R, for the same reason. Definition 28. The limit lim x c f(x) exists if there is L such that lim x c f(x) = L. If there is no such L then we say the limit does not exist. As with the definition of a limit of a sequence, do not worry unduly if you do not feel any intuition for this definition; ractice using the definition will eventually hel you to make sense of it. It is, however, critical that you know the definition and get comfortable with the structure of arguments using it. As with limits of sequences, the structure of a roof is often different than the intuition needed to figure it out, but once you have it you can ut the roof in fairly standard form like the following: Given ɛ > 0 define δ > 0 deending on ɛ and f, assume 0 < x c < δ and then give an argument which reasons from this assumtion to the conclusion that f(x) L < ɛ. We illustrate with some examles. Examle 29. Problem: Show lim x 2x 3 =. Reasoning: For f(x) = 2x 3 we need to get ɛ > f(x) ( ) > 2x 3 + = 2(x ) by making 0 < x < δ. If we ut δ = ɛ/2 then x < δ = ɛ/2 so 2 x < ɛ, which works. Proof: Given ɛ > 0 let δ = ɛ/2 and suose 0 < x < δ. Then f(x) ( ) = 2(x ) < 2δ = ɛ In the revious roblem we did not really need 0 < x because at x = c we had 2x 3 ( ) = 0. However it is not so difficult to give a similar examle where the 0 < x c condition is essential. x 2 + 5x + 6 Examle 30. Problem: Show that lim =. x 2 x + 2 Reasoning: Factoring the numerator we have x 2 + 5x + 6 = (x + 2)(x + 3). For x 2 we can

10 0 MATH 270: NOTES FOR ANALYSIS then cancel the common factor of x + 2 from the denominator and the numerator, so f(x) = x + 3 = x + 2, and we must rove this is less than ɛ when 0 < x + 2 < δ. Clearly it suffices to set δ = ɛ. Proof: Given ɛ > 0 let δ = ɛ and suose 0 < x + 2 < δ. Factoring the function in the limit we obtain x2 + 5x + 6 (x + 2)(x + 3) = = x + 3 = x + 2 < δ = ɛ x + 2 x + 2 where the cancellation is justified by the fact that x + 2 > 0. This roves the result. Examle 3. Problem: Show lim x 2 x 2 = 4. Reasoning: We need to show x 2 4 < ɛ if x 2 is small enough. The essential oint is that x 2 4 contains a factor of x 2. Secifically, x 2 4 = (x 2)(x + 2), which imlies x 2 4 = x 2 x + 2. If we knew x 2 < δ then 2 δ < x < 2 + δ so x + 2 < 4 + δ, which would make x 2 4 < (4 + δ) x 2 < (4 + δ)δ. We d want (4 + δ)δ ɛ to comlete the argument. It is a bit messy to choose δ using the quadratic formula, so instead we just require that δ < so (4 + δ)δ < 5δ and choose δ so 5δ ɛ. To get our two conditions to be true we take δ = min{, ɛ/5}. Proof: Given ɛ > 0 let δ = min{, ɛ/5}. Then x 2 4 = x + 2 x 2 < (4 + δ) x 2 5δ ɛ. The idea that if f(x) is a olynomial then f(x) L should contain a factor of x c is alicable in general. Here is an examle that looks a little different. Examle 32. Problem: Show lim x x 4 3x 3 + x 2 = 2. Reasoning: We need to show x 4 3x 3 + x 2 ( 2) is small when x is small, and have just said that this should occur because the first exression contains a factor x. Let s try to find the factor by long division of olynomials. (I may not use your favorite format for such long division, but you can always do it yourself. however you refer.) x 4 3x 3 + x 2 ( 2) = x 4 3x 3 + x 2 + = x 3 (x ) 2x 3 + x 2 + = x 3 (x ) 2x 2 (x ) x 2 + = (x 3 2x 2 )(x ) x(x ) x + = (x 3 2x 2 x)(x ) (x ) = (x 3 2x 2 x )(x ) Note that although we had to do this working ste-by-ste to get the factorizations, in the roof we only need to state what the factorization is (we don t need to include all stes) because the reader could verify it quickly by multilying out the factors. Now we want to make sure the factor multilying (x ) is not too big. As in the revious roblem it hels to insist that δ <. When x < δ < we have 0 < x < 2, so x 3 < 8, 2 x 2 < 8 and x < 2 from which we have the following (admittedly not very good) bound using the triangle inequality: x 3 2x 2 x x x 2 + x + < = 9. The reason we did not try to get a better bound in this exression is that it does not matter. If we take δ 9ɛ then we will still get the result we seek. To fit our two conditions on δ we use a minimum as usual. Proof: Given ɛ > 0 let δ = min{, ɛ/9}. If 0 < x < δ then x (0, 2) so x < 2 and we can comute x 4 3x 3 + x 2 ( 2) = x 3 2x 2 x x ( x x 2 + x + ) x < 9δ ɛ.

11 MATH 270: NOTES FOR ANALYSIS Sometimes one has a rational function where the factors in the denominator do not cancel with terms from the numerator. In this case we can still factor out a coy of x c but will be left bounding a rational function rather than a olynomial. The imortant thing is that the bounds for the terms must say the terms in the denominator stay away from 0 and that the terms in the numerator stay away from. x 3 x 2 Examle 33. Problem: Show that lim x 0 x 2 + x + 3 = 0. x 3 x 2 Reasoning: We need to show that x 2 < ɛ when x < δ. Clearly we should factor out one + x + 3 or more owers of x, but one will suffice. So if we wrote x 3 x 2 x 2 x 2 x + + x + 3 x 2 + x + 3 we see it is enough to have an uer bound for the factor with the fraction when x < δ. An uer bound for a fraction uses a lower bound for the denominator, so we want to find a number d, so x 2 +x+3 > d. We d exect this exression to be close to 3 if x is very small. One otion to show this occurs is to use the form of the triangle inequality in Lemma 3, because with a = x 2 + x + 3 and b = 3 we obtain (also using the usual triangle inequality and x < δ at the end: x 2 + x x 2 + x δ 2 + δ If we knew δ then this is no more than 2δ 2, which gives x 2 + x + 3. Then we d get x 3 x 2 x 2 x 2 x + + x + 3 x 2 + x x 2 < 2δ 2 2δ where at the end we again used δ. Taking 2δ < ɛ will finish the roof. Proof: Given ɛ > 0 let δ = min{, ɛ/2}. If 0 < x < δ then x 2 + x x 2 + x δ 2 + δ 2 so that x 2 + x + 3. Then we may comute x 3 x 2 x 2 x 2 x + + x + 3 x 2 + x x 2 < 2δ 2 < 2δ < ɛ to comlete the roof. Proving a limit does not exist frequently roceeds in the way we did for sequences. Aly contradiction, use the hyothesis of the existence of the limit to show that the function gets within ɛ of two values that are searated by distance more than 2ɛ and obtain a contradiction by the triangle inequality. All that is different is that in the sequence case the function values were oints arbitrarily far out in the sequence, whereas for a limit of a function they are values f(x) that occur for x very close to c. Here are two examles. x Examle 34. Problem: Show that lim x 0 x does not exist. Reasoning: If x > 0 then x /x = x/x =. If x < 0 then x /x = x/x =, so we should be able to use the same aroach as in Examle. Suose the limit exists and is L, take ɛ = and use the limit definition to find δ so 0 < x < δ imlies f(x) L < ɛ. Then for any x, x 2 with x < δ, x 2 < δ we must have f(x ) f(x 2 ) = f(x ) L (f(x 2 ) L) < 2ɛ = 2 by the triangle inequality. But if x > 0 then f(x ) = and if x 2 < 0 then f(x 2 ) =, so if we can find such numbers with size less than δ we get the contradiction 2 = ( ) = f(x ) f(x 2 ) < 2. One way to get these numbers is just to take x = δ/2 and x 2 = δ/2. Proof: Suose the limit exists and is equal to L. Let f(x) = x /x when x 0. Given ɛ = take

12 2 MATH 270: NOTES FOR ANALYSIS δ > 0 so 0 < x < δ imlies f(x) L < ɛ =. Let x = δ/2 and x 2 = δ/2. Observe f(x ) = and f(x 2 ) =. Both x < δ and x 2 < δ, so 2 = ( ) = f(x ) f(x 2 ) = f(x ) L (f(x 2 ) L) f(x ) L + f(x 2 ) L < ɛ + ɛ = 2 which gives a contradiction. Examle 35. Problem: Show that lim x 0 x does not exist. Reasoning: We see that x is large when x is small. It seems likely that we can get it to outut values that have arbitrary searation, just by taking, eg, inuts x and x/2. This gives f(x/2) f(x) = (2/x) (/x) = / x. If we ut x < then this difference is larger than. The usual way to get the contradiction is to take ɛ so 2ɛ is the searation of the function values, so lets try ɛ = /2. Proof: In order to obtain a contradiction, suose there is L so lim x 0 x = L. Write f(x) = x. With ɛ = 2 take δ > 0 so 0 < x < δ imlies f(x) L < ɛ. If 0 < x < min{, δ} then also 0 < x 2 < δ, so that x = 2 x = f( x x 2 ) f(x) f( x 2 ) L + f(x) L < 2ɛ =. This gives a contradiction because 0 < x <.

13 Exercises for Chater 2 37 EXERCISES FOR CHAPTER 2 Section 2.: Limits of Sequences 2.. Give an examle of a sequence that is not exressed in terms of trigonometry but whose terms are exactly those of the sequence of {cos(nπ)} Give an examle of two sequences different from the sequence {n 2 n! + n 2 } whose first three terms are the same as those of {n 2 n! + n 2 } Prove that the sequence { } 2n converges to Prove that the sequence { } n 2 + converges to Prove that the sequence { } + 2 converges to. n 2.6. Prove that the sequence { } n+2 2n+3 converges to By definition, lim n a n = L if for every ɛ>0, there exists a ositive integer N such that if n is an integer with n > N, then a n L <ɛ. By taking the negation of this definition, write out the meaning of lim n a n L using quantifiers. Then write out the meaning of {a n } diverges using quantifiers Show that the sequence { n 4} diverges to infinity. { } n 2.9. Show that the sequence 5 +2n diverges to infinity. n (a) Prove that n < 2 n for every ositive integer n. (b) Let s n = n + 2n + 3n + + for each n N. Prove that the sequence {s n 2 n } converges to Prove that if a sequence {s n } converges to L, then the sequence {s n 2} also converges to L. Section 2.2: Infinite Series 2.2. Prove that the series k= converges and determine its sum by (3k 2)(3k+) (a) comuting the first few terms of the sequence {s n } of artial sums and conjecturing a formula for s n ; (b) using mathematical induction to verify that your conjecture in (a) is correct; (c) comleting the roof Prove that the series k= converges and determine its sum by 2 k (a) comuting the first few terms of the sequence {s n } of artial sums and conjecturing a formula for s n ; (b) using mathematical induction to verify that your conjecture in (a) is correct; (c) comleting the roof The terms a, a 2, a 3, of the series k= a k are defined recursively by a = 6 and 2 a n = a n n(n + )(n + 2) for n 2. Prove that k= a k converges and determine its value Prove that the series k= k+3 (k+) 2 diverges to infinity (a) Prove that if k= a k is a convergent series, then lim n a n = 0. (b) Show that the converse of the result in (a) is false Let k= a k be an infinite series whose sequence of artial sums is {s n } where s n = 3n 4n+2. (a) What is the series k= a k? (b) Determine the sum s of k= a k and rove that k= a k = s.

14 38 Chater 2 Proofs in Calculus Section 2.3: Limits of Functions ( 2.8. Give an ɛ δ roof that lim 3 x 2 2 x + ) = Give an ɛ δ roof that lim x (3x 5) = Give an ɛ δ roof that lim x 2 (2x 2 x 5) = Give an ɛ δ roof that lim x 2 x 3 = Determine lim x and verify that your answer is correct with an ɛ δ roof. 5x 4 3x Give an ɛ δ roof that lim x 3 4x+3 = 2 3. x Determine lim 2 2x 3 x 3 and verify that your answer is correct with an ɛ δ roof. x 2 8x Show that lim x 0 does not exist. x The function f : R R is defined by x < 3 f (x) =.5 x = 3 2 x > 3. (a) Determine whether lim x 3 f (x) exists and verify your answer. (b) Determine whether lim x π f (x) exists and verify your answer A function g : R R is bounded if there exists a ositive real number B such that g(x) < B for each x R. (a) Let g : R R be a bounded function and suose that f : R R and a R such that lim x a f (x) = 0. Prove that lim x a f (x)g(x) = 0. (b) Use the result in (a) to determine lim x 0 x 2 sin ( x 2 ) Suose that lim x a f (x) = L, where L > 0. Prove that lim x a f (x) = L Suose that f : R R is a function such that lim x 0 f (x) = L. (a) Let c R. Prove that lim x c f (x c) = L. (b) Suose that f also has the roerty that f (a + b) = f (a) + f (b) for all a, b R. Use the result in (a) to rove that lim x c f (x) exists for all c R Let f : R R be a function. (a) Prove that if lim x a f (x) = L, then lim x a f (x) = L. (b) Prove or disrove: If lim x a f (x) = L, then lim x a f (x) exists. Section 2.4: Fundamental Proerties of Limits of Functions 2.3. Use limit theorems to determine the following: (a) lim x (x 3 2x 2 5x + 8) (b) lim x (4x + 7)(3x 2 2) (c) lim x 2 2x 2 3x Use induction to rove that for every integer n 2 and every n functions f, f 2,, f n such that lim x a f i(x) = L i for i n, lim x a ( f (x) + f 2 (x) + + f n (x)) = L + L 2 + +L n.

15 Additional Exercises for Chater Use Exercise 2.32 to rove that lim x a (x) = (a) for every olynomial (x) = c n x n + c n x n + +c x + c Prove that if f, f 2,..., f n are any n 2 functions such that lim x a f i (x) = L i for i n, then Section 2.5: Continuity lim x a ( f (x) f 2 (x) f n (x)) = L L 2 L n The function f : R {0, 2} Ris defined by f (x) = x2 4. Use limit theorems to determine whether f x 3 2x 2 can be defined at 2 such that f is continuous at The function f defined by f (x) = x2 9 is not defined at 3. Is it ossible to define f at 3 such that f is x 2 3x continuous there? Verify your answer with an ɛ δ roof Let f : R Z be the ceiling function defined by f (x) = x. Giveanɛ δ roof that if a is a real number that is not an integer, then f is continuous at a Show that Exercise 2.33 imlies that every olynomial is continuous at every real number Prove that the function f :[, ) [0, ) defined by f (x) = x is continuous at x = (a) Let f : R R be defined by f (x) = { 0 if x is rational if x is irrational. In articular, f (0) = 0. Prove or disrove: f is continuous at x = 0. (b) The roblem in (a) should suggest another roblem to you. State and solve such a roblem. Section 2.6: Differentiability 2.4. The function f : R R is defined by f (x) = x 2. Determine f (3) and verify that your answer is correct with an ɛ δ roof The function f : R { 2} R is defined by f (x) = x+2. Determine f () and verify that your answer is correct with an ɛ δ roof The function f : R R is defined by f (x) = x 3. Determine f (a) for a R + and verify that your answer is correct with an ɛ δ roof The function f : R R is defined by { x 2 sin if x 0 x f (x) = 0 if x = 0. Determine f (0) and verify that your answer is correct with an ɛ δ roof. ADDITIONAL EXERCISES FOR CHAPTER Prove that the sequence { } n+ 3n converges to 3. 2n Prove that lim 2 n 4n 2 + = Prove that the sequence { + ( 2) n } diverges Prove that lim n ( n 2 + n) = 0.