In last semester, we have seen some examples about it (See Tutorial Note #13). Try to have a look on that. Here we try to show more technique.

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1 MATH202 Introduction to Analysis (2007 Fall and 2008 Spring) Tutorial Note #4 Part I: Cauchy Sequence Definition (Cauchy Sequence): A sequence of real number { n } is Cauchy if and only if for any ε > 0, there is positive integer K such that m, n > K m n < ε Theorem: (Cauchy Theorem) The sequence is Cauchy if and only if { n } converges to some real number L. In last semester, we have seen some eamples about it (See Tutorial Note #3). Try to have a look on that. Here we try to show more technique. One useful technique is using mean value theorem, we state the theorem here (where the proof will be discussed in Chapter 8-Differentation) Theorem (Mean Value Theorem) Let f: a, b R be continuous on [a,b] and differentiable on (a,b). Then For some c (a, b) f b f a b a = f (c) Eample If a n 0 for n =,2,3, and {a n } is Cauchy Sequence. Show that the sequence {b n } defined by b n = ln ( + a n ) Cauchy by checking the definition. IDEA: Let f = ln + f = /( + ) b m b n = ln + a m ln + a n = + c a m a n a m a n < ε So we require a m a n < ε (this can be done as {a n } is Cauchy Sequence) For any ε > 0, since a n is Cauchy, then there eists K such that for m, n > K, we have a m a n < ε Pick K = K, then for m, n > K, from the previous work, we get b m b n < ε So {b n } is Cauchy

2 Eample 2 (Practice Eercise #47) Let a n 0 for n =,2,3, and {a n } is Cauchy. Show that { a n } is Cauchy by checking the definition. IDEA: If {a n } is Cauchy, then a n converges by Cauchy Theorem Case i) If lim n a n > 0, then say lim n a n = a. Consider f =, a m a n = 2 c a m a n < 2 a 2 a m a n = 2a a m a n < ε We require a n a < a 2 and a m a n < 2aε Case ii) If lim n a n = 0, then a m a n a m + a n < ε 2 + ε 2 = ε We require a n 0 < ε2 4 Case i) If lim n a n > 0, then say lim n a n = a, There eists K such that for n > K, a n a < a 2 There eists K 2 such that m, n > K 2, a m a n < 2aε Pick K = ma {K, K 2 }, then for m, n > K, from the arguments above, we get a m a n < ε Case ii) If lim n a n = 0 There eists K such that for n > K, a n 0 < ε2 Then for any m, n > K, we get a m a n a m + a n < ε 2 + ε 2 = ε Combining two cases, we complete the proof. 4 Eample 3 (Modified from Rudin P.82 #23) Suppose p n, {q n } are Cauchy Sequences in R, show that the distance d n = p n q n is a Cauchy Sequence. IDEA: Applying triangle inequality and assume d m d n we get d m d n = p m q m p n q n

3 = p m p n + p n q n + q n q m p n q n p m p n + p n q n + q n q m p n q n = p m p n + q m q n < ε 2 + ε 2 = ε We require p m p n < ε/2 and q m q n < ε/2 For any ε > 0, since p n, {q n } are both Cauchy Sequence, There eist K, such that m, n > K, p m p n < ε 2 There eist K 2, such that m, n > K 2, q m q n < ε 2 Pick K = ma {K, K 2 }, then for m, n > K, from the steps above, we get d m d n < ε Hence {d n } is Cauchy. Eample 4 Let { n } be the Cauchy such that n N for n =,2,3,.. (i.e. n is positive integers for every n). Show that there eists K such that for n > K, n is constant (i.e. n will become constant when n is large) IDEA: Since n is positive integer and two different integers must have distance at least (For eample:,2). Now n is Cauchy, for large n, the distance between m, n will be very close (distance less than one). It will force all n need to be same. Since n is Cauchy, pick ε = 0.5 (it can be any number less than ), there eists K such that for m, n > K, m n < 0.5..(*) Net we claim n is constant for n > K, we prove by contradiction, suppose there are m, n (m, n > K) such that m n. From (*), we get m n < 0.5, but since both m, n are both positive integers, then m n. Contradiction Hence n must be constant for n > K.

4 Part 2: Limit of Function Definition: (Limit of Function) Given a function f(), we say lim 0 f = L if and only if for any ε > 0, there eists δ > 0 such that for 0 < δ, we have f L < ε Roughly speaking, the definition means if is sufficiently close to 0 ( 0 < δ), Then f() should be very close to L. Technically, when we apply the definition to show lim 0 f = L, similar as the one in sequence, we need to find the δ so that f L < ε Eample 5 Using the definition of limit, show that lim 2 = 2 IDEA: From the definition, we need to find the δ such that 2 = 2 2 < = 2 < ε (We hope 2 < 3ε) 3 Overall, we need 2 < min {0.5,3ε} 2 < ε We hope 2 < < < 2.5 For any ε > 0, pick δ = min {0.5,3ε}, then for 2 < δ, we get 2 < ε Eample 6 Using the definition of limit, show that lim = 4 IDEA: From the definition, we need to find the δ such that Note that = = < ε = =

5 = = < = 36 < ε We require 0 < 3 (so that 3 < < 3) and 0 < 36ε For any ε > 0, pick our δ = min {3,36ε}, then for 0 < δ, from the above steps, we get < ε. We completes the proof. Eample 7 Show by definition that if lim 0 f What is the case when a = 0? IDEA: Note that when 0, then a 0 f a al = a f a a f a al = a a = L R and a 0, then lim 0 f a L < ε = al So we need f a a L < ε a For any ε > 0, since lim y 0 f y y = L, then there eists δ > 0 such that for y 0 < δ, we get f y y L < ε a Pick δ = δ δ, then for 0 < δ = a a f a al = a f a a We complete the proof. f a al = a a a 0 < aδ = δ L = a ε a < ε Besides the definition, there is one useful theorem in limit. Theorem: (Sequential Limit Theorem) lim 0 f() = L if and only if for every sequence n 0 and n 0, we have lim n f( n ) = L

6 One application of this theorem is to show the limit of some functions DO NOT eist Eample 8 Show that lim [] does not eist (where [] denotes the greatest integer less than or equal to ) Consider two sequences which n = n+ and y n = + n+ Then lim n n = lim n n n + 0 = lim y n [y n ] = lim + n n n + = 0 Hence lim n n lim n y n, by sequential limit theorem, the limit does not eist Eample 9 Define f = For each 0, we pick if is rational 0 if is irrational, show that lim 0 f() does not eist. If 0 is rational, pick n = 0 n and y n = 0 2n We get lim n f( n ) = lim n = and lim n f(y n ) = lim n 0 = 0 If 0 is irrational, pick n = 0n 0 0 n and y n = 0 n, We get lim n f( n ) = lim n = and lim n f(y n ) = lim n 0 = 0 Hence the limit does not eists for every 0 Try to do the following eercises, you may submit your work to me so that I can give some comments to your work. Eercise (Practice Eercise #56, #70) Given { n } is a Cauchy Sequence, show that both e n and sin5 n are Cauchy Sequence by checking the definition. Eercise 2 Suppose {y n } is Cauchy Sequence. Show { y n 3 } is also Cauchy (Hint: The method is similar to Eample 2)

7 Eercise 3 (Practice Eercise #67) Let f: 0, R satisfy f f y sin 2 sin (y 2 ) for all, y > 0, show that the sequence, 2, 3,. defined by n = f n is a Cauchy Sequence. (Hint: From the mean value theorem, we get sina sinb cosc a b. Apply this result to R.H.S. of the inequality) Eercise 4 Let { n } and {y n } be two Cauchy Sequence in R. Show that the product { n y n } is also Cauchy. (Hint: Apply the similar trick from Eample 3 on m y m n y n ) and use the fact that if { n } is Cauchy { n } converges { n } is bounded (i.e. n M)) Eercise 5 Show by definition of limit that a) lim + = 2 (Practice Eercise #57) b) lim = 5 (Practice Eercise #09c) c) lim = 4 d) lim a tan = tan a for a 0 ( Difficult!!) Eercise 6 Consider f = if 3 if >, show that lim f does not eist. (Hint: Try to plot the graph and get the idea, then prove it property) Eercise 7 Consider f = 2 if is rational. Determine with proof whether 2 if is irrational lim f() and lim f() eist or not. 2 4 Eercise 8 Suppose lim a f() = A > 0, show that there eists δ > 0 such that for 0 < a < δ, we have f > 0. (Hint: If f() has a positive limit at a, then it implies that if is close enough to a, f() will be very close to a, then f() will be eventually positive.)