Chapter 6 Limits of Sequences 6.1 Absolute Value 6.1 Definition (Absolute values.) Recall that if x is a real number, then the absolute value of x, denoted by |x|, is defined by 8 < x if x > 0, |x| = : 0 if x = 0, −x if x < 0. We will assume the following properties of absolute value, that follow easily from the definition: For all real numbers x, y, z with z 6= 0 |x| = | − x| |xy| = |x| · |y| ¯ ¯ ¯x¯ |x| ¯ ¯ = ¯ z ¯ |z| −|x| ≤ x ≤ |x|. For all real numbers x, and all a ∈ R+ (|x| < a) ⇐⇒ (−a < x < a) and (|x| ≤ a) ⇐⇒ (−a ≤ x ≤ a). (6.2) 116 6.1. ABSOLUTE VALUE 117 We also have |x| ∈ R≥0 for all x ∈ R, and |x| = 0 ⇐⇒ x = 0. 6.3 Theorem. Let a ∈ R and let p ∈ R+. Then for all x ∈ R we have |x − a| < p ⇐⇒ (a − p < x < a + p), and |x − a| ≤ p ⇐⇒ (a − p ≤ x ≤ a + p). Equivalently, we can say that {x ∈ R : |x − a| < p} = (a − p, a + p) and {x ∈ R : |x − a| ≤ p} = [a − p, a + p]. Proof: I will prove only the first statement. I have |x − a| < p ⇐⇒ −p < x − a < p ⇐⇒ a − p < a + (x − a) < a + p ⇐⇒ a − p < x < a + p. ||| 6.4 Definition (Distance.) The distance between two real numbers x and y is defined by dist(x, y) = |x − y|. Theorem 6.3 says that the set of numbers whose distance from a is smaller than p is the interval (a−p, a+p). Geometrically this is clear from the picture. p- p - a − p a a + p I remember the theorem by keeping the picture in mind. 6.5 Theorem (Triangle inequality.) For all real numbers x and y |x + y| ≤ |x| + |y|, (6.6) 118 CHAPTER 6. LIMITS OF SEQUENCES Proof For all x and y in R we have −|x| ≤ x ≤ |x| and −|y| ≤ y ≤ |y|, so −(|x| + |y|) ≤ x + y ≤ (|x| + |y|). Hence (Cf. (6.2)) |x + y| ≤ |x| + |y|. ³ ´ 6.7 Exercise. Can you prove that for all (x, y) ∈ R2 |x − y| ≤ |x| − |y| ? ³ ´ Can you prove that for all (x, y) ∈ R2 |x − y| ≤ |x| + |y| ? Remark: Let a, b, c, d be real numbers with a < c < b and a < d < b. a c d b Then |c − d| < |b − a| = b − a. This result should be clear from the picture. We can give an analytic proof as follows. (a < c < b and a < d < b) =⇒ (a < c < b and − b < −d < −a) =⇒ a − b < c − d < b − a =⇒ −(b − a) < c − d < (b − a) =⇒ −|b − a| < c − d < |b − a| =⇒ |c − d| < |b − a|. 6.8 Examples. Let A = {x ∈ R : |x − 2| < 5}, B = {x ∈ R : |x − 2| > 5}. 5- 5 - −3 2 7 6.1. ABSOLUTE VALUE 119 Then a number x is in A if and only if the distance from x to 2 is smaller than 5, and x is in B if and only if the distance from x to 2 is greater than 5. I can see by inspection that A = (−3, 7), and B = (−∞, −3) ∪ (7, ∞). Let ¯ ¯ ¯x − 1¯ C = {x ∈ R : ¯ ¯ < 1}. ¯ x + 1 ¯ If x ∈ R \ {−1}, then x is in C if and only if |x − 1| < |x + 1|, i.e. if and only if x is closer to 1 than to −1. c C - −1 0 1 I can see by inspection that the point equidistant from −1 and 1 is 0, and that the numbers that are closer to 1 than to −1 are the positive numbers, so C = (0, ∞). I can also do this analytically, (but in practice I wouldn’t) as follows. Since the alsolute values are all non-negative |x − 1| < |x + 1| ⇐⇒ |x − 1|2 < |x + 1|2 ⇐⇒ x2 − 2x + 1 < x2 + 2x + 1 ⇐⇒ 0 < 4x ⇐⇒ 0 < x. 6.9 Exercise. Express each of the four sets below as an interval or a union of intervals. (You can do this problem by inspection.) 1 3 A = {x ∈ R: |x − | < }, 1 2 2 1 3 A = {x ∈ R: |x + | ≤ }, 2 2 2 3 1 A = {x ∈ R: | − x| < }, 3 2 2 3 3 A = {x ∈ R: | + x| ≥ }. 4 2 2 120 CHAPTER 6. LIMITS OF SEQUENCES 6.10 Exercise. Sketch the graphs of the functions from R to R defined by the following equations: f1(x) = |x|, f2(x) = |x − 2|, f3(x) = |x| − |x − 2|, f4(x) = |x| + |x − 2|, 2 f5(x) = x − 1, 2 f6(x) = |x − 1|, 2 2 f7(x) = |x − 1| . (No explanations are expected for this problem.) 6.11 Exercise. Let f1, ··· , f7 be the functions described in the previous exercise. By looking at the graphs, express each of the following six sets in terms of intervals. S1 = {x ∈ R: f1(x) < 1} S2 = {x ∈ R: f2(x) < 1} S3 = {x ∈ R: f3(x) < 1} S4 = {x ∈ R: f4(x) < 3} S5 = {x ∈ R: f5(x) < 3} S6 = {x ∈ R: f6(x) < 3}. 1 Let S7 = {x ∈ R: f7(x) < 2 }. Represent S7 graphically on a number line. Remark: The notation |x| for absolute value of x was introduced by Weier- strass in 1841 [15][Vol 2,page 123]. It was first introduced in connection with complex numbers. It is surprising that analysis advanced so far without intro- ducing a special notation for this very important function. 6.2 Approximation 6.12 Definition (b approximates a.) Let ² be a positive number, and let a and b be arbitrary numbers. I will say that b approximates a with an error smaller than ² if and only if |b − a| < ². 6.2. APPROXIMATION 121 Remark: If b approximates a with an error smaller than ², then a approxi- mates b with an error smaller than ², since |a − b| = |b − a|. 6.13 Definition (Approximation to n decimals.) Let n ∈ Z+, and let a, b be real numbers. I will say that b approximates a with n decimal accuracy 1 if and only if b approximates a with an error smaller than · 10−n; i.e., if and 2 only if 1 |b − a| < 10−n. 2 6.14 Notation. If I write three dots (···) at the end of a number writ- ten in decimal notation, I assume that all of the digits before the three dots are correct. Thus since π = 3.141592653589 ··· , I have π = 3.1415 ···, and π = 3.1416 with 4 decimal accuracy. 6.15 Example. π = 3.141592653589793 ··· and 22 = 3.142857142857142 ··· . 7 Hence 22 3.1415 < π < < 3.1429, 7 and ¯ ¯ ¯22 ¯ 1 ¯ − π¯ < 3.1429 − 3.1415 = .0014 < .005 = · 10−2. ¯ 7 ¯ 2 22 22 Hence approximates π with an error smaller than .0014, and approxi- 7 7 mates π with 2 decimal accuracy. 6.16 Example. We see that .49 approximates .494999 with 2 decimal accuracy, and .50 approximates .495001 with 2 decimal accuracy, but there is no two digit decimal that approximates .495000 with 2 decimal accuracy. 122 CHAPTER 6. LIMITS OF SEQUENCES 6.17 Example. Since 1 |.49996 − .5| = .00004 < .00005 = · 10−4, 2 we see that .5 approximates .49996 with 4 decimal accuracy, even though the two numbers have no decimal digits in common. Since 1 |.49996 − .4999| = .00006 > · 10−4, 2 we see that .4999 does not approximate .49996 with 4 decimal accuracy, even though the two numbers have four decimal digits in common. 6.18 Theorem (Strong approximation theorem.) Let a and b be real numbers. Suppose that for every positive number ², b approximates a with an error smaller than ². Then b = a. Proof: Suppose that b approximates a with an error smaller than ² for every positive number ². Then |b − a| < ² for every ² in R+. Hence |b − a| 6= ² for every ² in R+, + i.e., |b − a| ∈/ R . But |b − a| ∈ R≥0, so it follows that |b − a| = 0, and consequently b − a = 0; i.e., b = a. ||| 6.3 Convergence of Sequences 6.19 Definition ({an} converges to L.) Let {an} be a sequence of real numbers, and let L be a real number. We say that {an} converges to L if for every positive number ² there is a number + N(²) in Z such that all of the numbers an for which n ≥ N(²) approximate L with an error smaller than ². We denote the fact that {an} converges to L by the notation {an} → L. Thus “{an} → L” means: 6.3. CONVERGENCE OF SEQUENCES 123 For every ² ∈ R+ there is a number N(²) in Z+ such that + |an − L| < ² for all n in Z with n ≥ N(²). Since ¯ ¯ ¯ ¯ |an − L| = |(an − L) − 0| = ¯|an − L| − 0¯, it follows immediately from the definition of convergence that ({an} → L) ⇐⇒ ({an − L} → 0) ⇐⇒ (|an − L| → 0). We will make frequent use of these equivalences. 6.20 Example. If a ∈ R+ then ½ µ 1 ¶ µ 1 ¶¾ a3 1 + 1 + → a3. n 2n Proof: Let ² be a generic element of R+. I must find a number N(²) such that ¯ µ ¶ µ ¶ ¯ ¯ 1 1 ¯ ¯a3 1 + 1 + − a3¯ < ² (6.21) n 2n whenever n ≥ N(²). Well, for all n in Z+ ¯ µ ¶ µ ¶ ¯ ¯ µ ¶ ¯ ¯ 1 1 ¯ ¯ 3 1 ¯ ¯a3 1 + 1 + − a3¯ = ¯a3 1 + + − a3¯ n 2n 2n 2n2 ¯ µ ¶ ¯ µ ¶ ¯ 3 1 ¯ 3 1 = ¯a3 + ¯ = a3 + 2n 2n2 2n 2n2 µ 3 1 ¶ 2a3 ≤ a3 + = . (6.22) 2n 2n n Now for every n in Z+ we have à ! à ! 2a3 2a3 < ² ⇐⇒ < n , n ² and by the Archimedean property of R there is some integer N(²) such that 2a3 < N(²). For all n ≥ N(²) we have ² à ! Ãà ! ! 2a3 2a3 (n ≥ N(²)) =⇒ < N(²) ≤ n =⇒ < ² , ² n 124 CHAPTER 6.