Limit and Continuity

Transcription

1 Limit and Continuity Table of contents. Limit of Sequences Definitions and properties Definitions Proving existence and nonexistence of limits Properties Cauchy sequence Understanding convergence Subsequences and convergence Bolzano-Weierstrass Monotone sequences Proof of Bolzano-Weierstrass Classification of sequences Liminf and Limsup Infiinite Series Definition and Basic Properties Definition Arithmetics of infinite series Forbidden operations Convergence tests Convergence/divergence through comparison Typical non-negative series and their implications Ratio test and root test Other tests Limit of Functions Limit of functions at x 0 R Limit of functions at ± and ± as limits Left and right limits Properties of function limits Unifying Different Limits A Brief Introduction to Topology Continuity Definitions Arithmetics of continuous functions Continuity of composite functions Properties of Continuous Functions Intermediate value theorem Maximum and Minimum Continuity of Inverse Functions

2 2 Limit and Continuity.. Definitions and properties.... Definitions.. Limit of Sequences Definition. (Limit R) A sequence of real numbers {x n } is said to converge to a real number a R if and only if ε > 0 N N such that n > N, x n a <ε. () We denote this convergence by x n a, or lim n x n =a. Exercise. Suppose we have proved can we conclude x n a? What if we proved ε > 0 N N such that n > 5 N, x n a < 20 ε, (2) m N N N such that n >N 2, x n a < m, (3) can we conclude x n a? Definition 2. (Limit =± ) A sequence of real numbers {x n } is said to converge to + (also called diverge to + ) if and only if M >0 N N such that n >N, x n >M. (4) Denoted lim n x n = + or x n + as n. The definition for lim n x n = is similar and left as exercise. Remark 3. Often the + in + is omitted. Exercise 2. Give precise definition of lim n x n =, and find the working negation of it. Exercise 3. Suppose we have proved can we conclude x n +? What if we proved What if we proved ε > 0 N N such that n > 5 N, x n > 20 ε, (5) m N N N such that n > N 2, x n >m 2, (6) M R N N such that n >N, x n > M? (7)..2. Proving existence and nonexistence of limits. Existence. First we make it precise the meaning of {x n } has a limit in R. A sequence of real numbers {x n } has a limit in R if and only if there is a R such that ε >0 N N such that n > N, x n a <ε. (8) Exercise 4. Obtain the working negation of {x n } has a limit in R. Therefore, to show existence of limit for {x n }, we have to do two things: Guess the limit a.. Later we will see that the theory of Cauchy sequences simplifies the procedure.

3 3 Check that the conditions in the definitions hold. { } ( ) n Example 4. Find the limit of and justify. n Solution. First we guess that the limit is 0. Now for any ε > 0, let N > ε. Then for all n >N, ( )n 0 n = n < < ε. (9) N Exercise 5. What is wrong with the following proof: First we guess that the limit is 0. Now for any ε > 0, let N =ε. Then for all n > N, ( )n 0 n = n < = ε. (0) N Example 5. Find the limit of {log (n)} and justify. Solution. First we guess that the limit is +. Now for any M >0, take N >e M. Than for all n >N, log(n) > log(n) > log (e M ) =M. () Non-existence. To show non-existence, we need to show that. x n a for any a R; 2. x n + 3. x n. Example 6. (Nonexistence of limit) Show that {( ) n } has no limit. Proof. First we show that x n a for any a R. This is usually done through proof of contradiction. Assume the contrary, that is ( ) n a for some a. Then for any ε>0, there is N N such that for all n >N, ( ) n a <ε. We take ε =. There is N N such that for all n > N, ( ) n a <. Since this holds for all n>n, in particular it holds for all odd n>n, thus a <. On the other hand it also holds for all even n>n, which leads to a <. Combining the two together we get But on the other hand we have a + a <2. (2) a + a = +a + a +a+ a = 2 =2. (3) Therefore 2 < 2. Contradiction! Next we show x n +. We need to show the working negation of the definition: M > 0 N N n >N x n M. (4) For x n = ( ) n, we simply take M =. Now for any N N, we take n = N +. Then x n =( ) N+. The proof for x n is similar and left as exercise. Exercise 6. Let {x n } be as in the above example. Prove that x n.

4 4 Limit and Continuity.2. Properties. Proposition 7. (Arithmetics) Let a, b R x n a, y n b. Then a) x n ± y n a ±b; b) x n y n ab; c) If b 0, then x n /y n a/b. Proof. We only give detailed proof for b). Other cases are left as exercises. To show x n y n a b, all we need is for any given ε > 0, find N N such that for all n > N, x n y n ab <ε. First assume that a,b 0. Let N N be such that y n b < b. Now for any given ε>0, let N 2 be such that x n a < ε 4 b, and N 3 be such that y n b < ε 2 a. Take N = max {N, N 2, N 3 }. For any n > N, we have x n y n ab = (x n a) y n + a(y n b) (x n a) y n + a y n b < ε 4 b 2 b + a ε 2 a = ε. When either a or b is (or both are) 0, all we need to show is x n y n 0. The method is the same and the argument is simpler so omitted. Exercise 7. Prove a). Exercise 8. Find the mistake in the following proof of c): Let lim n xn = c. Then So c =a/b. yn a= lim x n = lim n n Then give a correct proof of c). ( xn y n ) ( y n = lim n ) ( x n lim y n )= c b. (5) y n n The above proposition can be readily extended to cover the cases of ± limits. Recall the following rules: x+ =, x =, x R (6) x =, x ( ) =, x >0 (7) x =, x ( )=, x <0 (8) + =, = (9) =( ) ( ) = ( )= ( ) ( ) = (20) Proposition 8. (Extension of Proposition 7) Let x n a, y n b where a,b are extended real numbers. Then a) x n ± y n a ±b; b) x n y n ab; c) If b 0, then x n /y n a/b. as long as the right hand sides are well-defined in the arithmetics of extended real numbers. Proof. We only prove the product case and leave other cases as exercises.

5 5 When both a, b R, the proof has been done in Proposition 7. Here we need to study the two new cases: a, b=± and a R, b=± or a=±,b R. Case. We only prove for a=b=. (The other three sub-cases can be proved similarly.) As =, we need to show that for every M R, there is N N such that for all n > N, x n y n >M. For every given M R, since x n, there is N N such that for all n > N, x n > M ; Similarly there is N2 N such that for all n > N 2, y n > M. Now take N = max {N,N 2 }. For any n >N, we have x n y n > M M = M M. Case 2. We only prove for a>0, b= (The other seven sub-cases can be proved similarly). We need to show that for every M R, there is N N such that for all n>n, x n y n >M. Since x n a>0, there is N N such that for all n>n, x n >a/2. On the other hand, as y n, there is N 2 N such that for all n>n 2, y n >2 M /a. Now take N = max{n,n 2 }, for every n >N we have x n y n > (a/2) (2 M /a)= M M. Exercise 9. Explain why in the following cases the limt is not determined by a, b only through examples for each cases. ; 0 ; / (2) Remark 9. These types of limit can be calculated using L Hospitale s rules. Theorem 0. (Comparison of limits) Suppose {x n } and {y n } are convergent sequences with limits a,b R ext. If there is N 0 N such that x n y n for all n N 0, then Proof. lim n x n lim n y n. (22) The case a, b R. Let x a, y b. We prove by contradiction. Assume a > b. Set ε = a b. Then there 2 is N N such that x n a < ε for all n > N, while N 2 N such that y n b < ε for all n >N 2. Take n > max {N, N 2 }. Then we have x n a x n a >a ε= a+b 2 ; (23) y n b+ y n b <b+ε= a +b 2. (24) But this combined with x n y n gives a+b < a + b 2 2, contradiction. The case a =. As b for every extended real number b, the proof ends. The case a =+. We need to show that b= or b R would lead to contradiction. If b=. By definition we have On the other hand, limx n = a=+ leads to Taking any n > max {N,N 2 }, we have contradiction. b R. Left as exercise. N N n > N y n < ; (25) N 2 N n >N 2 x n >. (26) x n >> > y n (27) The cases b=± can be proved similarly and are left as exercises.

6 6 Limit and Continuity Exercise 0. Prove the cases a = +, b R and b = ±. Exercise. Is the following claim true or false? Justify your answer. Suppose {x n }, {y n } are convergent sequences. If there is N 0 N such that x n < y n for all n N 0, then lim n x n < lim n y n. Lemma. A sequence can have at most one limit. Proof. We prove by contradiction. Assume x n a and x n b with a b. Without loss of generality we assume b > a. Take ε = b a 2. There is N N such that x n a < ε for all n > N ; There is N 2 N such that x n b <ε. Now take N = max {N,N 2 }. Now take any n>n, we have n>n,n>n 2 and therefore which combined to give Contradiction! x n a <ε, x n b <ε (28) b a= b a b x n + x n a = x n b + x n a <2ε=b a. (29) An important method of guessing/proving limit is the following theorem. Theorem 2. (Squeeze Theorem) Let a R ext, x n a and y n a. Let {w n } be a sequence. Assume that there is N 0 N such that for all n N 0, then a) w n converges. b) w n a. x n w n y n, (30) Remark 3. Of course proving w n a suffices. However we would like to emphasize that, the significance of this theorem is that the relation (29) forces w n to converge. Proof. We prove the case a R. For any given ε > 0, let N N be such that x n a < ε for all n > N ; Let N 2 N be such that y n a < ε for all n > N 2. Now let N = max {N, N 2 }. For all n >N, we have Thus ends the proof. w n a max { x n a, y n a } <ε. (3) Exercise 2. Prove the cases a= and a=. Do we really need both {x n }, {y n } in these two cases? Justify your answers. Corollary 4. Let {w n } be a sequence. If there is another sequence x n 0 such that w n x n, then w n 0. In particular, if w n 0, then w n 0. Exercise 3. Prove the above corollary. Example 5. Find lim n 2 n sin(n 8 ). We simply apply the above corollary: 2 n 0, therefore the limit is also 0. 2 n sin (n 8 ) 2 n (32)

7 7.3. Cauchy sequence. Definition 6. (Cauchy sequence) A sequence of real numbers {x n } is called a Cauchy sequence if and only if the following holds: ε > 0 N N such that m, n >N, x m x n <ε. (33) Often one simply says {x n } is Cauchy. Exercise 4. Obtain the working negation of {x n } is Cauchy. Exercise 5. Why is ε >0 N N such that n > N, a n+ a n < ε not enough? Recall that, to prove convergence for {x n } with formulas for x n given, we need to do two things:. Guess the limit a. 2. Prove that indeed x n a using either definition or properties (see Section? below) or both. On the other hand, there are many situations where we do not have explicit formulas for x n and thus cannot guess what the limit is. The following is the most useful result in those situations. Theorem 7. (Relation to Cauchy sequences) Let {x n } be a sequence. Then {x n } converges to some real number a R if and only if it is a Cauchy sequence. Proof. It s if and only if, so we need to show the only if : x n a for some a R {x n } is Cauchy (34) and the if : {x n } is Cauchyx n a for some a R. (35) Only if. Given any ε >0, we need to find N N such that for all n,m>n, x n x m <ε. We proceed as follows. For the given ε >0, since x n a, there is N N such that for all n > N. Now for any n, m > N, we have which combined to give x n a <ε/2 (36) x n a <ε/2, x m a <ε/2 (37) x n x m = (x n a)+ (a x m ) x n a + a x m = x n a + x m a <ε. (38) If. For this part we need to use the least upper bound property of R. Intuitively, {x n } is Cauchy means they cluster around something, but to show that something is a real number,

8 8 Limit and Continuity we need the property that R has no holes. Let the set A = {a R: N N, such that a < x n for all n > N }. We first show that A has an upper bound. Since {x n } is Cauchy, there is N N such that x n x m < for all n, m >N. In particular we have for all n > N. Now take x n < x N ++ (39) b=max {x +,x 2 +,,x N +, x N + +} (40) we have b > x n for all n N and clearly b > a for every a A. Using the same method one can show that A is not empty. Thanks to the least upper bound property, we have b min R such that b min a for every a A while for every b <b min there is a A with a >b. We prove the x n b min. Given any ε > 0. We know that there is a A such that a > b min ε. Therefore there is N N such that b min ε<x n for all n >N ; Now we show the existence of N 2 N such that b min +ε>x n for all n>n 2. We prove by contradiction. Assume that for every N N, there is n > N such that x n b min + ε. Since {x n } is a Cauchy sequence, there is N 3 N such that for all n,m >N 3, x n x m <ε/2. For this N 3 we can find l > N 3 with x l b min +ε. Consequently, for all n > N 3, we have x n x l x n x l >b min + ε/2. (4) But this means b min +ε/2 A, contradicting b min a for every a A. Finally take N = max {N, N 2 }. For every n >N, we have b min ε < x n < b min + ε x n b min <ε. (42) Therefore x n b min and the proof for if ends. Exercise 6. Can this theorem be extended to the cases a = ±? Why? Example 8. Let x 0 >2. Define x n iteratively by Prove that x n 2. x n+ = x n x n x n. (43) Proof. The idea is to show that it is Cauchy. Once this is done, we can take limits of both sides to reach (denote the limit by a): a =a a2 2 2 a a2 = 2 a = ± 2. (44)

9 9 From this we see that we need to prove the following: First show that x n >0. We use mathematical induction. Let P(n) denote the statement x n > 0. Mathematical induction consists of two steps:. P(0) is true. 2 We have x 0 > 2 >0. 2. If P(m) is true then P(m +) is true. We have Therefore if x m > 0, x m+ > 0. x m+ =x m x m 2 2 = x m (45) 2 x m 2 x m Since a ratio is involved, we need to show x n 0 and a 0. We do this through showing x n 2 2 for all n N. This can be done directly as follows: ( x 2 n 2= x n x 2 n 2 2 x n ) 2 ( 2 xn 2= +2 2 x n ) 2 ( 2 xn 2 2 = 2 x n ) 2 0. (46) x n is Cauchy. The basic idea is to show that x n x n+ M r n for some 0<r < and some M > 0 (Why this is enough is left as exercise). Taking difference of x n+ = x n x n 2 2 = x n 2 x n 2 + and x n =x n x 2 n 2 = x n x n 2 x n 2 + x n (47) we have Therefore As x n > 2 [ ] x n+ x n = 2 (x n x n ). (48) x n x n x n+ x n = 2 x n x n x n x n. (49) we have xn x n 2 so 2 x n x. This leads to n 2 x n+ x n 2 x n x n 2 2 x n x n 2 ( ) n x x 0. (50) 2 Since x n is Cauchy, there is a R such that x n a. By (44) we have a = ± 2. Since xn > 0, we conclude a = 2 (using Comparison Theorem 7). Remark 9. The above is a very effective way to compute Because we start from x 0.

10 0 Limit and Continuity.4. Understanding convergence. In this section we understand what convergence means. Or more precisely, we understand what happens when a sequence is not convergent. Turns out, there are only two situations where a sequence does not converge:. The sequence is unbounded. 2. The sequence is oscillating and the amplitude does not tend to 0. Definition 20. (Boundedness) A sequence {x n } is said to be bounded above if there is a number M R such that x n M for all n, bounded below if there is a number M R such that x n M for all n, bounded if there is a number M R such that x n M for all n. Exercise 7. Obtain the working negation of {x n } is bounded. Lemma 2. (Relation between boundedness) {x n } is bounded if and only if it is bounded above and below. Proof. If. If {x n } is bounded above and below, then there are M,M 2 R such that M x n M 2 for all n N. Now take M = max { M, M 2 }. We have for all n N. M M M x n M 2 M 2 M (5) Only if. If {x n } is bounded, then there is M R such that M x n M. Thus {x n } is bounded both above and below. Oscillating is quantified through subsequences converging to different limits. Definition 22. (Subsequence) A subsequence of a sequence {x n } n N is a sequence of the form {x nk } k N, where each n k N with n <n 2 <n 3 <. Remark 23. It is important to understand that the n in the subsequence {x nk } is just a notation, it does not take any value. It is k that takes values, 2,3,. Remark 24. Recall that a real sequence is a function f: N R. Then its subsequence is the same function restricted to a subset E R. Exercise 8. A key fact for subsequences is n k k for all k N. (52) Prove it. (Hint: Use mathematical induction.) Exercise 9. Another way to understand subsequences is to view them as composite functions. Figure this out.

11 It is clear that a subsequence of a subsequence is a subsequence of the original sequence..4.. Subsequences and convergence. Lemma 25. Let a R ext. We have the following relations between convergence of {x n } and its subsequences. a) If x n a, then every of its subsequences also converge to a. b) If all its subsequences converge to the same a R ext, then x n a. Proof. We prove a) for a= and b) for a R and leave other cases as exercises. a) We have x n and need to prove x nk for any subsequence {x nk } of {x n }. More precisely, our hypothesis is and the conclusion to be proved is M R N N n >N x n <M (53) M R K N k >K x nk <M. (54) Given an arbitrary M R. Let N N be such that n>n, x n <M. Now take K =N. Then for every k > K, we have n k k > K = N which gives x nk <M. (55) b) We provce by contradiction. Assume {x n } does not converge to a. All we need to do is to find one subsequence which does not converge to a 3. As {x n } does not converge to a, we have ε 0 > 0 N N n >N, x n a >ε 0. (56) Now first take N =. There is n > such that x n a > ε 0. Next take N = n. There is n 2 >N =n such that x n2 a >ε 0. Next take N =n 2 and do the same and so on. This way we obtain a subsequence {x nk } satisfying x nk a >ε 0 for all k N. This subsequence does not converge to a. Exercise 20. Prove the above lemma for all other cases Bolzano-Weierstrass. Theorem 26. (Bolzano-Weierstrass) Let {x n } be bounded a sequence of real numbers. Then there is a converging subsequence. Remark 27. The theorem not only tell us something about simple cases, such as ( ) n for which we can easily get a convergent subsequence, it also tell us sequences like {sinn} also has convergent subsequence(s). 4 Monotone sequences. 3. Whether it converges to something else or does not converge at all does not matter here. 4. In fact, for every a, there is a subsequence of {sinn} convergent to a.

12 2 Limit and Continuity To prove the Bolzano-Weierstrass theorem, we first study monotone sequences. Definition 28. A sequence {x n } is increasing if x n+ x n for all n (strictly increasing of x n+ >x n for all n N); A sequence is decreasing if x n+ x n for all n (strictly decreasing if x n+ < x n for all n N); A sequence is monotone if it is either increasing or decreasing. Convergence results when monotonicity meets boundedness. Lemma 29. Let {x n } be monotone increasing (decreasing) and bounded above (below). Then x n a for some a R. Proof. Since {x n } is bounded above. There is M R such that x n M for all n N. We now show that x n is Cauchy through proof by contradiction. Assume x n is not Cauchy, then there is ε 0 >0 such that for every N N, there are n,m >N with x n x m >ε 0. Now take N =. We find n 2 >n > with x n2 x n >ε 0. Next take N =n 2, we have n 4 >n 3 >n 2 with x n4 x n3 >ε 0. Doing this repeatedly, we obtain a (still increasing) subsequence x nk such that x n2k x n2k >ε 0. Since this subsequence is still increasing, x n2k (x n2k x n2k ) +(x n2k 2 x n2k 3 ) + + (x n2 x n ) +x n > kε 0 +x n. (57) Taking k >(M x n )/ε 0 we reach x n2k >M. Contradiction. Remark 30. An alternative proof is as follows: Let A = {a R: a < x n for some n N}. We can show that A has an upper bound and is nonempty. Applying the least upper bound property of R, there is b min R such that b min a for all a A, and for every b <b min, there is a A satisfying a >b. Then prove x n b min. This approach is left as exercise. Proof of Bolzano-Weierstrass. All we need to do is to find a monotone subsequence from {x n }. Proof. (of Bolzano-Weierstrass) Consider the following subsequences of {x n }: x m, x m+, x m+2,. There are only two cases: For every such sequence, there is a smallest element: x n = x, x nk : =min {x nk +,x nk +2, }. (58) In this case we know that {x nk } is a subsequence of {x n } and it is increasing and bounded above. Applying Lemma 29 to this subsequence we see that it converges to some a R. There is one m, such that there is no smallest element in {x nm +,x nm +2, }. Set x n =x nm +. Set n 2 to be the smallest number that x n2 <x n, that is x i x n for all n i<n 2. This can be done because there is no smallest number. Set n 3 to be the smallest number that x n3 <x n2, and so on. This way we obtained a decreasing subsequence which is bounded below. Applying Lemma 29 to this subsequence we see that it converges to some a R Classification of sequences. Lemma 3. We have the following relation between convergence and boundedness. a) If x n a R, then {x n } is bounded. b) If {x n } is bounded, then there is a subsequence {x nk } that converges to some a R.

13 3 Proof. a) Take ε=. Since x n a, there is N N such that for all n>n, we have x n a <ε. Thus for all n > N, we have Now set x n a + x n a < a +. (59) M = max { x,, x N, a +}. (60) We have x n M for all n N. Thus ends the proof. b) This follows immediately from Bolzano-Weierstrass. Exercise 2. Spot the mistake in the following proof of part a): Take ε = a. Since x n a,... (the remaining the same as that in the above proof) Now we almost have proved the following theorem. Theorem 32. (Classification of sequences) Let {x n } be a sequence of real numbers. Then one of the following is true: a) x n converges to a R; b) {x n } is not bounded. c) There are two convergent subsequences with different limits. Remark 33. a) and b),c) are clearly mutually exclusive. On the other hand both b) c) may be n =4k + n n =4k +2 true for a single sequence. For example let a n = for k N. n =4k +3 n n =4k +4 Proof. (of the Theorem) It is clear that a) and b) are mutually exclusive. The only thing left to show is that if {x n } is bounded and does not converge, then it has two converging subsequences with different limits. Since {x n } is bounded, by the Bolzano-Weierstrass theorem there is a converging subsequence. We denote its limit by a. Since x n a, by definition there must be ε 0 >0 such that for every N N there is n > N with x n a > ε 0. From this we can get a subsequence satisfying x nk a > ε 0 as follows: i. Take N =. There is x n a >ε 0 ; ii. Take N =n. Find x n2 a >ε 0 ; iii. Take N =n 2,... Now apply the Bolzano-Weierstrass theorem to this subsequence. We see that there is a converging subsequence (to this subsequence), called { } x nkl (l =, 2, 3, ). Let x nkl a as l. We show that a a. Assume the contrary. Then x nkl a. By definition there is N N such that xnkl a <ε0 for all l>n. But this contradicts the fact that x nkl are chosen from x nk which satisfies x nk a >ε 0 for all k N. Exercise 22. Prove the following:

14 4 Limit and Continuity If {x n } is unbounded, then one of the following is true: a) x n + ; b) x n ; c) There are two subsequences, one converges to + and one to Liminf and Limsup. Definition 34. Let {x n } be a real sequence. The limit supreme of {x n } is the extended real number ( limsupx n 8 lim sup x k ), (6) n n k n and the limit infimum of {x n } is the extended real number ( liminf x n 8 n lim n inf k n x k ). (62) Remark 35. Since sup k n x k is decreasing and inf k n x k is increasing, limsup and liminf always exist (in contrast to limits). Remark 36. One way to understand limsup, liminf is the following. Let {x n } be a sequence. Then we create a new sequence {y n } as follows. y 8 sup {x, x 2, }; (63) y 2 8 sup {x 2, x 3, }; (64) y n 8 sup {x n, x n+, }. (65) Now as y n is decreasing, its limit exists (either in R or is ). We call this limit the limit supreme of {x n } and denote it by limsup n x n. For example, if x n = ( ) n, then y 8 sup {,,, } = ; (66) y 2 8 sup {,,, } =; (67) y n 8 sup {( ) n, ( ) n+,( ) n+2, } =. (68) So in fact y n = for all n. We have lim n y n = which means limsup n x n =. Remark 37. Note that it is possible that limsup n x n = and liminf n =. Lemma 38. Let {x n } be a real sequence. Then limsupx n liminf x n (69) n n Proof. Follows from Comparison Theorem. Left as exercise. Exercise 23. Prove the above lemma.

15 5 Theorem 39. (Relation to limits and subsequences) Let {x n } be a real sequence. Then a) x n a if and only if limsup n x n = liminf n x n = a. 5 Proof. b) There are two subsequences converging to limsup n x n and liminf n x n, respectively. c) If {x nk } is a convergent subsequence, then liminf n x n lim k x nk limsup n x n. a) We only prove the case a R and left a = ± as exercise. if. Assume limsup n x n = liminf n x n =a. Then as supx k x n inf x k, (70) k n k n application of Squeeze Theorem gives the convergence of x n as well as lim n x n =a. only if. For any ε > 0, since x n a, there is N N such that x n a < ε. This implies sup n>n x n <a+ε and inf n>n x n > a ε. Therefore when n >N, Comparison Theorem now gives a ε < inf x n sup x n < a+ε. (7) k n k n a ε liminf x n limsup x n a+ε. (72) n n As this holds for all ε > 0, we must have limsup n x n = liminf n x n =a. b) We only show the existence of x nk limsup n x n. Thanks to Theorem?, there is n N such that sup k x k x n sup x k ; (73) k Now apply Theorem? to {x n +, }, we obtain n 2 N such that This way we obtain a subsequence satisfying sup x k k n x n 2 sup x k. (74) k n 2 + sup x k k n l + l x n l sup x k. (75) k n l + We take limit of both sides. As {sup k nl +x k } is a subsequence of {sup k n x k }, it converges to the same limit a. Using the fact that /l 0 as l, we apply Squeeze theorem to conclude x nl a. c) Let {x nk } be the subsequence. Then we have inf x l x nk supx l. (76) l n k l n k Taking limit of all three and applying Comparison Theorem we reach the conclusion. 5. We do not say {x n } converges to avoid dealing with a=± separately.

16 6 Limit and Continuity Remark 40. More discussions about liminf and limsup. Definition: limsup x n = lim n sup n k n x k = lim n sup {x n, x n+, }. (77) liminf x n= lim inf x k = lim inf {x n, x n+, }. (78) n n k n n In other words, we define a new sequence by and then define Understanding: y = sup {x, x 2,x 3, } (79) y 2 = sup {x 2, x 3,x 4, } (80) y n = sup {x n, x n+,x n+2, } (8) limsup x n = lim y n. (82) n n limsup x n = max {a R: subsequence x nk a}; (83) n liminf n x n=min {a R: subsequence x nk a}. (84) How to calculate: Evaluating exactly sup k n x k could be hard. There are two ways to overcome: Use Squeeze theorem: Instead of evaluating sup k n x n exactly, find an upper bound and an lower bound w n sup {x n, } z n (85) and try to show lim w n = lim z n. Note that the requirement for z n is z n x k for all k n while the requirement for w n is w n x k for some k n. Example 4. Consider x n = ( ) n + e n2. We have [ sup ( ) k ] +e k2 +e n 2. (86) k n Taking limit n we conclude limsup x n =. (87) Use limsup is the largest limit of convergent subsequences. To take this approach we need to first guess limsup n x n =. Then show. There is a subsequence converging to. Take n k =2 k then x nk =+e 4k2. 2. For every convergent subsequence x nk a, a. We do this through comparison theorem: x nk =( ) n k +e n k 2 +e k2 a= lim x ( ) n k lim +e k 2 =. (88) k k

17 7 2.. Definition and Basic Properties Definition. 2. Infiinite Series Definition 42. (Infinite series) Given a sequence {a n } of real numbers, the formal sum is called an infinite series. a n =a + a a n + (89) Remark 43. Up to now the summation of infinitely many real numbers a +a 2 + +a n + (90) is formal because it is not clear what it means to say a +a 2 + +a n + =s R ext. Remark 44. Note that a n is just another way of denoting the formal sum a +a a n +. Example 45. Some examples of infinite series: ( ) n = ( ) + +( ) ++ (9) sinn n = sin + sin (92) 2 n = (93) 2 n = (94) It is clear that the value of should be 2 while 2 n the first two sums corresponds to any value. 2 n should be. It s not clear whether Definition 46. (Partial sum and convergence) The nth partial sum of an infinite series a n is defined as s n = n m= a m. If the sequence {s n } converges to some real number s, then we say the infinite series converges, and say its sum is s, and simply write a n = s. (95) If s or, we say the infinite series diverges to or respectively and write a n = or. (96) Remark 47. There are many ways to assign a number s to an infinite sum a +a 2 +. The above is the most popular one. Recalling theorems for the convergence of sequences, we have

18 8 Limit and Continuity Theorem 48. a n = s if and only if for any ε > 0, there is N N such that for all n > N, s n a m < ε; (97) m= a n = if and only if for any M R, there is N N such that for all n >N, n a m >M; (98) m= a n = if and only if for any M R, there is N N such that for all n >N, Exercise 24. Prove the above theorem. n a m < M. (99) m= Theorem 49. (Cauchy) A infinite series a n converges to some s R if and only if for any ε > 0 there is N N such that for all m > n >N, m a k < ε. (00) Exercise 25. Prove the above theorem. k=n+ Corollary 50. If a n converges to s R then lim n a n = 0. Equivalently, if lim n a n does not exist, or exists but is not 0, then a n does not converge to any real number. Proof. For any ε >0, since all m >n>n, a n converges, it is Cauchy and there exists N N such that for m k=n+ a k < ε. (0) Now take N = N +. Then for any n >N, we have n > n >N which gives n a n 0 = a k <ε. (02) k=n Thus by definition of convergence of sequence lim n a n = 0. Remark 5. The above corollary is very useful, however we should keep in mind that:. The converse is not true. That is lim n a n =0 does not imply the convergence of a n. 2. It cannot be applied to conclude a n or. Exercise 26. Give a counterexample to the following claims: lim a n =0 n a n = s for some s R. (03) a n = s R ext lim a n =0. (04) n

19 9 Example 52. Let a n = r n for r R. Then Proof. a) If r <, then a n = +r +r 2 + = b) If r, then a n =. c) If r, then a) We have r. a n does not exist (as extended real number). n m= a m = + + r n = rn r. (05) For any ε > 0, take N N such that N log r [ε ( r)], then for any n > N, r n a m = r n r < r N <ε. (06) r m= b) For any M R. Take N N such that N > M. Then for every n >N we have n a m n =n>n > M M. (07) m= m= c) Since r, a n. Therefore by Corollary 42 a n does not converge to any real number. We still need to show that a n,. To do this, we show that s n = n m= a n satisfies s n 0 when n is odd and s n 0 when n is even. Clearly s = > 0, s 2 =+r 0. For n 3, calculate n s n = m= As r, r 2 which gives rn r r m + r n = rn r + r n 2 +r n. (08) r n r n. (09) Therefore, s n and r n cannot take opposite signs. Example 53. We have 2 n = = 2; 2 5 n = 5 4. (0) Example 54. Consider We show that it actually converges. Consider the partial sum sin n n = sin + sin2 2 + () S m = m k=n+ sin k k. (2)

20 20 Limit and Continuity Denote A k = k l= m k=n+ sink k sin l, and B k =. Then we have k = m k=n+ (A k A k ) B k = [A n+ B n+ A n B n+ ] +[A n+2 B n+2 A n+ B n+2 ] + +[A m B m A m B m ] = [A m B m A n B n+ ]+ +[A n+ (B n+ B n+2 ) +A n+2 (B n+2 B n+3 )+ +A m (B m B m )] m = [A m B m A n B n+ ]+ [A k (B k B k+ )] = [ Am m A ] n n + k=n+ m [ + k=n+ A k ( k k + Now notice that {A n } is in fact a bounded sequence: )]. (3) A n = sin + sin sinn sin [sin + sin sinn] = sin cos( ) cos( + )+cos(2 ) cos(2 +) + + cos(n ) cos(n +) = 2 sin [cos0 + cos+ + cos(n )] [cos2+cos3+ + cos(n +)] = 2 sin cos0+cos cosn cos(n +) =. (4) 2 sin Now it is clear that A n 2 for all n N. sin Back to (3): m sink k A m m m + A n n + + ( A k k ) k + k=n+ [ k=n+ 2 sin m + m n + + ( k ) ] k + k=n+ = 2 [ ] 2 4 = sin n + (n +) sin. (5) Now we are ready to show the series is Cauchy: For any ε >0, take N N such that N + 4, then for any m >n>n, we have m k=n+ Therefore the series converges. sink k ε sin 4 (n + ) sin < 4 ε. (6) (N +) sin Exercise 27. Prove the following Abel summation formula: Let A k = k l= m k=n+ a k B k = [A m B m A n B n+ ] m k=n+ Draw analogy to the formula of integration by parts: b f(x) G(x) dx = [F(b) G(b) F(a)G(a)] a where F(x) = x f(t) dt, G(x)= x g(t) dt. a a a b a l, B k = k l= b l, then A k b k+ (7) F(x) g(x) dx (8)

21 Arithmetics of infinite series. Theorem 55. (Arithmetics) If a n = s, b n =t with s,t R ext. Then For any c R, if c s is defined, (c a n )= c s. If s+t is defined, (a n + b n ) =s+t. Exercise 28. What happens if c=±? Exercise 29. Let c = 0, s = or. We know that c s is not defined. However the sum (c a n ) is still well-defined. Explain why and find this sum. Example 56. Let a n = c r n for r R and c R. Then a) If r <, then a n = +r +r 2 + = c r. b) If r and c 0, then c > 0 a n =c ={ c < 0. c) If r and c 0, then a n does not exist (as extended real number). d) If c =0 then a n =0 no matter what value r takes. a),b),d) clearly follow from Theorem 47. We prove c) by contradiction: If a n =s R {, }, then since c 0, we have Contradiction. Example 57. We have r n = (c a n ) =c s R {, }. (9) 5 n = 5 5 n = = 4. (20) Remark 58. In general there is no relation between a n b n and a n, b n. On the other hand, with some extra assumption we can define the product ( ) ( ) a n b n = [ ] n a k b n+ k = (a b ) + (a b 2 + a 2 b ) + (a b 3 + a 2 b 2 + k= a 3 b ) + (2) This will be discussed in Math Forbidden operations. Example 59. (Grouping) Unless a n 0 (or all 0) and a n converges, the order of summation cannot be changed. For example let a n = ( ) n+. If we are allowed to group terms together and sum them first, we would have both a n =+( )+ + = +[( ) + ] +[( )+ ]+ =+0+0+ =; (22) a n = +( ) ++ =[ + ( )] +[ +( )]+ = = 0. (23) Definition 60. (Rearrangement) A rearrangement of an infinite series infinite series m= a n(m) where m: n(m) is a bijection from N to N. a n is another Example 6. An example of rearrangement of a +a 2 +a 3 + is a 2 +a 4 +a 7 +a +a 5 +a 3 +a 6 +.

22 22 Limit and Continuity Example 62. (Rearrangement) Consider the sequence a n with a n = ( )n+. If we are n allowed to freely rearrange (that is choose the order of summation), then for any s R {, }, there is a rearrangement such that it converges to s. Proof. Consider the case s R. The cases s=, are left as exercises. Consider the rearrangement b n defined as follows: Let k 0 be such that k 0 s but <s. Set 2 k 0 3 b =, b 2 = 3,,b k 0 = 2 k 0 ; (24) The case k 0 = is when s. Then we just set b = and turn to the next step. Let k be such that and set k 0 b k k= Let k 2 be such that k 0 +k b k +( s k= and set And so on. Now set ( k 2 b k0 + = 2, ) s, 2 k k 0 +2k 2 b k0 +k + = k 0 k= ( b k ) <s (25) 2 k b k 0 +k = 2 (k + ). (26) ) k s, 0 +k b k +( k= 2 k k k 2 3 ) < (27) 2 k 0 +,,b k 0 +k +k 2 = 2 k 0 +2k 2 +, (28) S l = l b k. (29) k 0 +k + +k k= Then we see that if n [k 0 + +k l, k 0 + +k l+ ], then n s n = b m (30) m= is always between S l and S l+. Finally notice that by construction, S l s < l. Thus for any ε>0, take L N such that L>ε. Now set N = k k L. For any n > N, there is l L such that n [k k l, k k l+ ]. Therefore we have s n s max { S l s, S l+ s } l < ε. (3) L That is b n s by definition. Remark 63. Note that the above proof depends on the fact that k= Exercise 30. Prove the above two facts. ( 2 k =, k= 2 k ) =. (32)

23 Convergence tests Convergence/divergence through comparison. In most situations, it is very hard or impossible to explicitly calculate the partial sum S n 8 n m= a m and is therefore not possible to establish convergence/find the sum based on definition. Similar to the idea of Cauchy sequences, we need a way to determine the convergence of a series without obtaining explicitly the partial sums. This is possible thanks to the following theorem. Theorem 64. Let a n and b n be two infinite series. Assume that there are c > 0 and N 0 N such that a n cb n for all n >N 0. Then a) b n converges a n converges. b) a n does not converge to some real number b n =. Proof. Note that a) and b) are equivalent logical statements, so we only need to prove a). We show that a n is Cauchy. For any ε >0, since b n converges, it is Cauchy and there is N N such that for all m >n>n, m b k < ε c. (33) k=n+ Take N = max {N,N 0 }. Then for any m > n >N, m k=n+ a k m k=n+ a k c m k=n+ b k < ε. (34) So a n is Cauchy and therefore converges. Example 65. It is clear by Theorem 56 that if a n converges, so does a n. The converse is not true, as can be seen from the following example: Take a n = ( )n+. Then we clearly see that n S 2n = n 2 n = (2 n ) (2 n) (35) converges. On the other hand, we have S 2n+ S 2n 0 so S 2n+ converges to the same limit. From here it is easy to prove by definition that S n to the same limit, which turns out to be ln2. Remark 66. A sequence a n that converges but with a n = is called conditionally convergent. It turns out that the phenomenon we have seen in Example 54 is quite generic for conditionally convergent series. More specifically, if a n is conditionally convergent, then it can be re-arranged to converge to any extended real number. On the other hand, a sequence a n such that a n converges is said to be absolutely convergent. Absolutely convergent sequences can undergo any re-arrangement and still converge to the same sum. Exercise 3. Prove that if a n does not converge to some s R, then a n =.

24 24 Limit and Continuity In light of the above theorem, it is important to study non-negative series, that is infinite series a n satisfying a n 0 for all n N. Once a non-negative sequence b n is shown to be convergent, we know that any a n satisfying a n c b n for some constant c is also convergent. It is further possible to make this comparison intrinsic, that is design some criterion involving a n only and guarantees the relation a n c b n. Such criteria are usually called tests. We will study the simpliest tests in the following section. Exercise 32. Let a n be a non-negative series. Then it converges it is bounded above Typical non-negative series and their implications. Ratio test and root test. Both tests are based on the convergence/divergence of the Geometric series. Example 67. (Geometric series) We have seen that a consequence, if another series a n satisfies for some c > 0 and for all n > some N 0 N, then r n converges when 0 r <. As a n cr n (36) a n converges. The following two intrinsic convergence tests based on comparison with geometric series are the simplest and most popular tests for convergence/divergence. Theorem 68. (Ratio test) Let a n a infinite series. Further assume that a n 0 for all n N. Then If limsup n a n+ a <, the series converges. n If liminf n a n+ a >, the series diverges. n Proof. Assume limsup n a n+ L + a = L <. Set r = n limsup n a n+ = lim a n n 2 { and ε 0 = L. By definition 2 a } k+ sup k n a k (37) therefore there is N N such that for all n >N, sup a k+ k n a k L < ε 0 (38) which gives sup a n+ < L+ε 0 =r < 0 < a n+ < r <. (39) n>n a n This gives, for all n >N +, a n a n < a N+ r n N = a N+ r N rn. (40)

25 25 Note that since N is fixed, we have a n <cr n (4) for all n > N and consequently a n converges. Assume liminf n a n+ a =L>. Set ε0 =L. Then by definition, similar to the limsup n case above, there is N N such that for all n >N, a n+ a n > (42) which means for all n N + a n a N+ (43) As a consequence a n 0. By Corollary 42 we know that a n diverges. Example 69. Prove that Proof. We have therefore So the series converges. n! converges. a n+ a n = n + (44) limsup a n+ n a n =0<. (45) Exercise 33. Let a n be a infinite series. Let b n be the series obtained by dropping all 0 s from {a n }. Prove that a n = s b n = s (46) for s R ext. Exercise 34. Can the ratio test be modified to apply to a n without assuming a n 0? Theorem 70. (Root test) Let a n be a infinite series. Then If limsup n a n /n <, then the series converges. If liminf n a n /n >, then the series diverges. Proof. Assume limsup n a n /n = L <. Set r = L + and ε 2 0 = L. Then by definition, as in 2 the proof of the above ratio test, there is N N such that for all n > N, Therefore a n converges. a n /n <r < a n <r n. (47) Assume liminf n a n /n >. The proof is left as exercise. Remark 7. In fact, for x n >0 x liminf n+ liminf n x n n (x n) /n limsup n (x n ) /n limsup n x n+ x n. (48)

26 26 Limit and Continuity Therefore the root test is sharper than the ratio test, in the sense that any series that passes the ratio test for convergence will also pass the root test. { Example 72. Consider the infinite series = a n where a n = 2 k n = 2 k 3 k. It can be easily verified that n = 2 k liminf n a n+ a n =0; limsup n so the ratio test does not apply. On the other hand so the root test tells us that the series converges. Other tests. a n+ a n = (49) limsup (a n ) /n = (50) n 2 Example 73. (Generalized harmonic series) The series converges when a > and n diverges when a. a Proof. When a, we have n a n therefore it suffices to show the divergence of the following trick: 6 Therefore we have (recall s n is the partial sum) n. We use > 2, (5) > = 2, (52) > = 2, (53) s nk > k + 2 (54) where n k = k =2 k+. Therefore s n is not bounded and the series diverges to. When a >, we use the following trick: ; (55) 2 a +3 a < 2 2 a = 2 a ; (56) 4 a +5 a +6 a +7 a < 4 4 a = 2 2( a) ; (57) We see that N n a < ( 2 ( a) ) n = <+. (58) 2 a 6. This proof is attributed to Nicole Oresme (c ).

27 27 Thus this non-negative series is bounded from above and therefore converges. Exercise 35. Let {x n } be a sequence. Prove: {x n } is bounded Every subsequence of {x n } is bounded. Remark 74. The standard proof for a >, usually presented after introduction of integration, is as follows: n n n n a = n n a dx dx x a. (59) Thus N n a = + N n=2 n a + N n=2 n n dx x a =+ N dx N a xa =+ a a a. (60) Remark 75. Recall that weeks ago we proved the convergence of the a=2 case through the trick n(n ) n2, however for more exotic a such tricks are not available anymore. Remark 76. Also note that neither the ratio test nor the root test works for the generalized harmonic series. Remark 77. It is also possible to design convergence/divergence tests using generalized harmonic series as the gauge. Since n a converges to 0 slower than r n (in the sense that lim n n a r n = 0 if r <), these tests will be more refined than either the ratio test or the root test. One of such test is the following (Raabe s test) a n >0. Then a n converges if a n diverges if ( ) liminf n an > (6) n a n+ ( ) an limsup n <. (62) n a n+ Note that the convergent part of Raabe s test can be turned into a convergence test for general series (without requiring a n >0) as ( ) liminf n an n a n+ > a n converges (63) but the divergent part cannot, that is ( ) an limsup n n a n+ < a n diverges (64) is not true due to the existence of convergent sequences such as If we consider even slower convergent series, such as ( ) n+ n. n (log n) α, n (log n) (log (log n))(log ( log n)) α (65)

28 28 Limit and Continuity for α > (the proof of convergence of these series is similar to that of the generalized harmonic series), we will obtain even sharper tests (Gauss test from the former, Bertrand s test from the latter), but the formulas become quite baroque. Finally let s look at a few fun examples. Example 78. We know that the harmonic series n obtained by dropping all terms involving 9: diverges to. Now consider the sequence (66) Does this series converge? As it is a non-negative sequence, we only need to check whether it s bounded above (Theorem 59). We have, + + <8, (67) 8 ( ) ( ) < In ( general, ) there are 8 9 k terms between 0 k and 9 k. 8 0 Overall the sum is bounded above by ( ) n 9 8 = 8 0 n=0 Therefore the new series converges. 9 0 (68), so their sum is bounded above by 0 k+ = 80. (69) Remark 79. Obviously we can try to study the sequence resulted from deleting all terms involving other digits, or sequence of numbers. For example we can delete all terms involving the combination 43, that is 4352 is deleted while is not. We can even play some silly games such as deleting all 4537 terms involving 222, or someone s birthday. The resulting sequences are all convergent. Example 80. (Fermat s Last Theorem) This is adapted from the blog of Terence Tao of UCLA 7. We all know that Fermat s Last Theorem claims that x n + y n = z n, x, y, z N (70) does not have any solution when n 3. On the other hand, it is well-known that when n=2, there are infinitely many solutions. But why? What s the difference between n=2 and n>2? It turns out that we can reveal some difference through knowledge of convergence/divergence of infinite series. Let s consider the chance of three numbers a,b,a+b are all the nth power of a natural number. If we treat a as a typical number of size a, then it s chance of being an nth power is roughly a /n /a. Ignoring the relation between a,b,a+b, we have the following probability for a,b,a+b solving the equation (70): a n b n (a+b) n. (7) 7. The probabilistic heuristic justification of the ABC conjecture, link at

29 29 Now consider all numbers a, b, we sum up the probabilities: I 8 [ a= b= a n b n (a+b) n ] (72) and apply the following intuition based on the so-called Borel-Cantelli Lemma in probability: If I <, then the chance of (70) having a solution is very low, while if I =, the chance is very high. We notice that I has perfect symmetry between a and b, which means I = 2 a= a [ b= a n b n (a +b) n ] + a= a 2 n 2 (2 a) n. (73) It is clear that the second series converges for all n 2 so can be ignored for our purpose. Now consider J = a [ a n b n (a +b) ] n. (74) a= b= When b a, we have a <a+b<2a therefore which gives a a n b n (2a) n <J < a a n b n (a) n (75) a= a= b= b= 2 n a a 2 n 2 b n < J < a a 2 n 2 b n. (76) a= b= a= b= So finally all we need to study is the convergence/divergence of We have Therefore K = a a 2 n 2 b n = [ a 2 n 2 a b n ]. (77) a= b= a= b= a b= b n a x n dx an. (78) K a= a 3 n 2 (79) which is convergent when n 4 while divergent when n = 2. Remark 8. The case n = 3 is a bit tricky here. The series a= a 3 n 2 becomes the Harmonic series a= a which is the borderline between convergence and divergence. Our argument does not provide any insight on why x 3 + y 3 = z 3 should not have solutions.

30 30 Limit and Continuity 3.. Limit of functions at x 0 R. 3. Limit of Functions Definition 82. We say that a real number L is the limit of f at x 0, denoted lim x x0 f(x) = L (or f(x) L as x x 0, or lim f(x) = L as x x 0 ), if ε >0 δ > 0 such that for all 0 < x x 0 <δ f(x) L <ε. (80) Remark 83. Note that the requirement 0< x x 0 is important. The limit of f at x 0 has nothing to do with whether f(x 0 ) is defined or not, not to say its value. This is very reasonable: Consider f(x) =0 everywhere except f(x) = at x=0. Then clearly we should have lim x 0f(x) =0. Exercise 36. If we use 0 x x 0 < δ instead, then what can we say about lim x 0f(x) for the above f(x)? Example 84. The following hold: a) lim x x 0 a =a for constant function a. b) lim x x 2 =. Proof. We prove b). Given any ε > 0, we need to find δ >0 such that for all 0 < x <δ, x 2 <ε. Since x 2 = x x+ <δ x+ <δ (2 +δ). (8) we see that we need to choose a δ such that δ (2 +δ) <ε. Notice that δ (2+δ)<ε δ δ +<ε+ δ + < ε + Thus for any ε > 0, if we take δ = ε Limit of functions at ± and ± as limits. Definition ε + <δ < ε +. (82), then for 0 < x <δ, x 2 <ε. We say lim x f(x) = L R if for every ε > 0 there is some M R such that x > M f(x) L <ε. The limit lim x f(x) =L is defined similarly. Let x 0 R. We say lim x x 0 f(x)= + iff for every M R there is δ >0 such that lim x x 0 f(x) = can be defined similarly. x x 0 <δ f(x) >M. (83) We say lim x + f(x) =+ iff for every M R there is K R such that The other three cases are similarly defined. Example 86. Prove that ln(/ x ) as x 0. x >K f(x) > M. (84) Proof. For any M R, we need to find δ >0 such that when 0< x <δ, ln(/ x )>M. It is clear that we can take δ =e M.

31 3 Example 87. Find and prove the limit (only discuss x > 0) lim x ( x x x ). (85) Solution. The function is quite complicated and the limit is not obvious. Thus we first try to simplify: ( x x x x )( x x +x ) +2x x= ( x x +x ) = 2 x x x +x = x (86) All we need to do is to find the limits of the numerator and denominator. Clearly lim 2 =2. (87) x On the other hand we show lim x +2/x =. For any ε > 0, take M = /ε. Then for all x >M, we have 2/x <2ε and consequently 0 < +2/x < +2ε < ( +ε) 2 =ε. (88) Therefore +2/x ( < ε for all x >M. We have proved limx +2/x + ) =2 0. Now we can apply Theorem? once more to conclude ( lim x 2 +2x x ) 2 = lim =. (89) x x x 3.3. Left and right limits. Definition 88. (Left and right limits) Let f(x) be a real function and x 0 R. We say L is the left-limit of f at x 0 if We denote it by ε > 0 δ > 0 such that for all δ <x x 0 <0, f(x) L <ε. (90) We say L is a right-limit of f at x 0 if We denote it by lim f(x) = L; (9) x x 0 ε > 0 δ >0 such that for all 0 < x x 0 < δ, f(x) L <ε. (92) Exercise 37. Prove that lim x x0 lim f(x) =L. (93) x x 0 + f(x) = L R ext lim f(x)=l and lim f(x) =L. (94) x x0 x x Properties of function limits. First we establish relation between function limits and sequence limits. Theorem 89. Let x 0,L R ext. Then lim x x 0 f(x)=l if and only if for every sequence x n x 0 with x n x 0 for all n N, lim n f(x n )= L.