MATH 409 Advanced Calculus I Lecture 9: Limit supremum and infimum. Limits of functions.

Similar documents
MATH 409 Advanced Calculus I Lecture 10: Continuity. Properties of continuous functions.

MATH 409 Advanced Calculus I Lecture 7: Monotone sequences. The Bolzano-Weierstrass theorem.

1 Definition of the Riemann integral

MATH 409 Advanced Calculus I Lecture 12: Uniform continuity. Exponential functions.

Limit Theorems. MATH 464/506, Real Analysis. J. Robert Buchanan. Summer Department of Mathematics. J. Robert Buchanan Limit Theorems

The reference [Ho17] refers to the course lecture notes by Ilkka Holopainen.

MATH 409 Advanced Calculus I Lecture 11: More on continuous functions.

CHAPTER 3: CONTINUITY ON R 3.1 TWO SIDED LIMITS

Solutions Final Exam May. 14, 2014

h(x) lim H(x) = lim Since h is nondecreasing then h(x) 0 for all x, and if h is discontinuous at a point x then H(x) > 0. Denote

Exercise 2. Prove that [ 1, 1] is the set of all the limit points of ( 1, 1] = {x R : 1 <

MATH202 Introduction to Analysis (2007 Fall and 2008 Spring) Tutorial Note #7

REAL ANALYSIS II: PROBLEM SET 2

Advanced Calculus Math 127B, Winter 2005 Solutions: Final. nx2 1 + n 2 x, g n(x) = n2 x

FINAL REVIEW FOR MATH The limit. a n. This definition is useful is when evaluating the limits; for instance, to show

Iowa State University. Instructor: Alex Roitershtein Summer Homework #1. Solutions

Principle of Mathematical Induction

Mathematical Methods for Physics and Engineering

B553 Lecture 1: Calculus Review

Entrance Exam, Real Analysis September 1, 2017 Solve exactly 6 out of the 8 problems

REAL VARIABLES: PROBLEM SET 1. = x limsup E k

Advanced Calculus I Chapter 2 & 3 Homework Solutions October 30, Prove that f has a limit at 2 and x + 2 find it. f(x) = 2x2 + 3x 2 x + 2

Logical Connectives and Quantifiers

Existence of a Limit on a Dense Set, and. Construction of Continuous Functions on Special Sets

2.2 Some Consequences of the Completeness Axiom

Homework for MATH 4603 (Advanced Calculus I) Fall Homework 13: Due on Tuesday 15 December. Homework 12: Due on Tuesday 8 December

Econ Slides from Lecture 1

We say that the function f obtains a maximum value provided that there. We say that the function f obtains a minimum value provided that there

Solutions to Tutorial 8 (Week 9)

MA 123 (Calculus I) Lecture 3: September 12, 2017 Section A2. Professor Jennifer Balakrishnan,

Lemma 15.1 (Sign preservation Lemma). Suppose that f : E R is continuous at some a R.

Math 421, Homework #9 Solutions

Chapter 3 Continuous Functions

Induction, sequences, limits and continuity

,

Continuity. Chapter 4

This exam contains 5 pages (including this cover page) and 4 questions. The total number of points is 100. Grade Table

Continuity. Chapter 4

Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 7 Solutions Please write neatly, and in complete sentences when possible.

Sequences. Chapter 3. n + 1 3n + 2 sin n n. 3. lim (ln(n + 1) ln n) 1. lim. 2. lim. 4. lim (1 + n)1/n. Answers: 1. 1/3; 2. 0; 3. 0; 4. 1.

Structure of R. Chapter Algebraic and Order Properties of R

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Math 127C, Spring 2006 Final Exam Solutions. x 2 ), g(y 1, y 2 ) = ( y 1 y 2, y1 2 + y2) 2. (g f) (0) = g (f(0))f (0).

Appendix A. Sequences and series. A.1 Sequences. Definition A.1 A sequence is a function N R.

Math 117: Honours Calculus I Fall, 2012 List of Theorems. a n k b k. k. Theorem 2.1 (Convergent Bounded): A convergent sequence is bounded.

Contents Ordered Fields... 2 Ordered sets and fields... 2 Construction of the Reals 1: Dedekind Cuts... 2 Metric Spaces... 3

CH 2: Limits and Derivatives

Riemann Integrable Functions

REVIEW OF ESSENTIAL MATH 346 TOPICS

Continuous Functions on Metric Spaces

Analysis Finite and Infinite Sets The Real Numbers The Cantor Set

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

LIMITS AT INFINITY MR. VELAZQUEZ AP CALCULUS

In N we can do addition, but in order to do subtraction we need to extend N to the integers

1. Theorem. (Archimedean Property) Let x be any real number. There exists a positive integer n greater than x.

In N we can do addition, but in order to do subtraction we need to extend N to the integers

Solutions Final Exam May. 14, 2014

Selected solutions for Homework 9

MATH 1A, Complete Lecture Notes. Fedor Duzhin

Week 2: Sequences and Series

FIRST YEAR CALCULUS W W L CHEN

The Real Number System

Problems for Chapter 3.

Section 2.5 : The Completeness Axiom in R

Math 341 Summer 2016 Midterm Exam 2 Solutions. 1. Complete the definitions of the following words or phrases:

Walker Ray Econ 204 Problem Set 3 Suggested Solutions August 6, 2015

HOMEWORK ASSIGNMENT 5

Exercises from other sources REAL NUMBERS 2,...,

(a) For an accumulation point a of S, the number l is the limit of f(x) as x approaches a, or lim x a f(x) = l, iff

Section 2.5. Evaluating Limits Algebraically

f ( x) = L ( the limit of f(x), as x approaches a,

LECTURE 6. CONTINUOUS FUNCTIONS AND BASIC TOPOLOGICAL NOTIONS

Calculus I. 1. Limits and Continuity

Solution. 1 Solution of Homework 7. Sangchul Lee. March 22, Problem 1.1

MATH 23b, SPRING 2005 THEORETICAL LINEAR ALGEBRA AND MULTIVARIABLE CALCULUS Midterm (part 1) Solutions March 21, 2005

MAS221 Analysis, Semester 1,

Introduction to Real Analysis

Copyright c 2007 Jason Underdown Some rights reserved. statement. sentential connectives. negation. conjunction. disjunction

Consequences of the Completeness Property

1. Supremum and Infimum Remark: In this sections, all the subsets of R are assumed to be nonempty.

Math 106 Fall 2014 Exam 2.1 October 31, ln(x) x 3 dx = 1. 2 x 2 ln(x) + = 1 2 x 2 ln(x) + 1. = 1 2 x 2 ln(x) 1 4 x 2 + C

The Dirichlet s P rinciple. In this lecture we discuss an alternative formulation of the Dirichlet problem for the Laplace equation:

2 2 + x =

Economics 204 Fall 2011 Problem Set 2 Suggested Solutions

Exercise 1. Let f be a nonnegative measurable function. Show that. where ϕ is taken over all simple functions with ϕ f. k 1.

Proof. We indicate by α, β (finite or not) the end-points of I and call

Problem Set 2: Solutions Math 201A: Fall 2016

d(x n, x) d(x n, x nk ) + d(x nk, x) where we chose any fixed k > N

Chapter 1 The Real Numbers

M17 MAT25-21 HOMEWORK 6

Section 4.3 Concavity and Curve Sketching 1.5 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Advanced Calculus II Unit 7.3: 7.3.1a, 7.3.3a, 7.3.6b, 7.3.6f, 7.3.6h Unit 7.4: 7.4.1b, 7.4.1c, 7.4.2b, 7.4.3, 7.4.6, 7.4.7

02. Measure and integral. 1. Borel-measurable functions and pointwise limits

Analysis III. Exam 1

Selected Solutions. where cl(ω) are the cluster points of Ω and Ω are the set of boundary points.

Limits at Infinity. Horizontal Asymptotes. Definition (Limits at Infinity) Horizontal Asymptotes

MATH NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS

G1CMIN Measure and Integration

1. For each statement, either state that it is True or else Give a Counterexample: (a) If a < b and c < d then a c < b d.

Some Background Material

Transcription:

MATH 409 Advanced Calculus I Lecture 9: Limit supremum and infimum. Limits of functions.

Limit points Definition. A limit point of a sequence {x n } is the limit of any convergent subsequence of {x n }. Properties of limit points. A convergent sequence has only one limit point, its limit. Any bounded sequence has at least one limit point. If a bounded sequence is not convergent, then it has at least two limit points. If all elements of a sequence belong to a closed interval [a, b], then all its limit points belong to [a, b] as well. If a sequence diverges to infinity, then it has no limit points. If a sequence does not diverge to infinity, then it has a bounded subsequence and hence it has a limit point.

Limit supremum and infimum Let {x n } be a bounded sequence of real numbers. For any n N let E n denote the set of all numbers of the form x k, where k n. The set E n is bounded, hence supe n and infe n exist. Observe that the sequence {supe n } is decreasing, the sequence {infe n } is increasing (since E 1,E 2,... are nested sets), and both are bounded. Therefore both sequences are convergent. Definition. The limit of {supe n } is called the limit supremum of the sequence {x n } and denoted lim sup x n. The limit of {infe n } is called the limit infimum of the sequence {x n } and denoted liminf x n.

Properties of lim sup and lim inf. liminf x n limsupx n. liminf x n and limsupx n are limit points of the sequence {x n }. All limit [ points of {x n } are contained in the interval lim inf x n,limsupx n ]. The sequence {x n } converges to a limit a if and only if liminf x n = limsupx n = a.

Limit of a function Let I R be an open interval and a I. Suppose f : E R is a function defined on a set E I \{a}. Definition. We say that the function f converges to a limit L R at the point a if for every ε > 0 there exists δ = δ(ε) > 0 such that 0 < x a < δ implies f(x) L < ε. Notation: L = lim f(x) or f(x) L as x a. Remark. The set (a δ,a) (a,a+δ) is called the punctured δ-neighborhood of a. Convergence to L means that, given ε > 0, the image of this set under the map f is contained in the ε-neighborhood (L ε,l+ε) of L provided that δ is small enough.

Limits of functions vs. limits of sequences Theorem Let I be an open interval containing a point a R and f be a function defined on I \{a}. Then f(x) L as x a if and only if for any sequence {x n } of elements of I \{a}, lim x n = a implies lim f(x n ) = L. Remark. Using this sequential characterization of limits, we can derive limit theorems for convergence of functions from analogous theorems dealing with convergence of sequences.

Limits of functions vs. limits of sequences Proof of the theorem: Suppose that f(x) L as x a. Consider an arbitrary sequence {x n } of elements of the set I \{a} converging to a. For any ε > 0 there exists δ > 0 such that 0 < x a < δ implies f(x) L < ε for all x R. Further, there exists N N such that x n a < δ for all n N. Then f(x n ) L < ε for all n N. We conclude that f(x n ) L as n. Conversely, suppose that f(x) L as x a. Then there exists ε > 0 such that for any δ > 0 the image of the punctured neighborhood (a δ,a) (a,a+δ) of the point a under the map f is not contained in (L ε,l+ε). In particular, for any n N there exists a point x n (a 1/n,a) (a,a+1/n) such that x n I and f(x n ) L ε. We have that the sequence {x n } converges to a and x n I \{a}. However f(x n ) L as n.

Limit theorems Squeeze Theorem If lim f(x) = lim g(x) = L and f(x) h(x) g(x) for all x in a punctured neighborhood of the point a, then lim h(x) = L. Comparison Theorem If limf(x) = L, g(x) = M, and f(x) g(x) for all x in lim a punctured neighborhood of the point a, then L M.

Limit theorems Theorem If limf(x) = L and lim g(x) = M, then lim(f +g)(x) = L+M, (f g)(x) = L M, lim lim (fg)(x) = LM. If, additionally, M 0 then lim (f/g)(x) = L/M.

Divergence to infinity Let I R be an open interval and a I. Suppose f : E R is a function defined on a set E I \{a}. Definition. We say that the function f diverges to + at the point a if for every C R there exists δ = δ(c) > 0 such that 0 < x a < δ implies f(x) > C. Notation: lim f(x) = + or f(x) + as x a. Similarly, we define the divergence to at the point a.

One-sided limits Let f : E R be a function defined on a set E R. Definition. We say that f converges to a right-hand limit L R at a point a R if the domain E contains an interval (a, b) and for every ε > 0 there exists δ = δ(ε) > 0 such that a < x < a+δ implies f(x) L < ε. Notation: L = lim + f(x). Similarly, we define the left-hand limit lim f(x). Theorem f(x) L as x a if and only if lim f(x) = lim f(x) = L. +

Limits at infinity Let f : E R be a function defined on a set E R. Definition. We say that f converges to a limit L R as x + if the domain E contains an interval (a, + ) and for every ε > 0 there exists C = C(ε) R such that Notation: L = lim x +. x > C implies f(x) L < ε. x + f(x) or f(x) L as Similarly, we define the limit lim x f(x).

Examples Constant function: f(x) = c for all x R and some c R. lim f(x) = c for all a R. Also, lim f(x) = c. x ± Identity function: f(x) = x, x R. lim f(x) = a for all a R. Also, lim f(x) = + and x + f(x) =. lim x Step function: f(x) = lim f(x) = 1, lim f(x) = 0. x 0+ x 0 { 1 if x > 0, 0 if x 0.

Examples f : R\{0} R, f(x) = 1 x. lim f(x) = 1/a for all a 0, lim f(x) = +, x 0+ f(x) = 0. lim x 0 f(x) =. Also, lim x ± f : R\{0} R, f(x) = sin 1 x. lim f(x) does not exist since f((0,δ)) = [ 1,1] for any x 0+ δ > 0. f : R\{0} R, f(x) = x sin 1 x. lim x 0 f(x) = 0, which follows from the Squeeze Theorem since x f(x) x.

Examples Dirichlet function: f(x) = { 1 if x Q, 0 if x R\Q. lim f(x) does not exist since f((c,d)) = {0,1} for any interval (c, d). In other words, both rational and irrational points are dense in R. Riemann function: { 1/q if x = p/q, a reduced fraction, f(x) = 0 if x R\Q. lim f(x) = 0 for all a R. Indeed, for any n N and a bounded interval (c, d), there are only finitely many points x (c,d) such that f(x) 1/n. On the other hand, lim x + f(x) and lim x f(x) do not exist.