NOTES ON INTRODUCTION TO ANALYSIS
ANDRZEJ ROSLANOWSKI
Abstract. These lecture notes are based on the textbook
Edward D. Gaughan. Introduction to Analysis. Fifth Edition. American
Mathematical Society. ISBN: 978–0–8218–4787–9.
URL: http://bookstore.ams.org/amstext-1/
Contents
0. Before we even start
0.1. Principles of Mathematical Writing
0.2. Sets, Relations and Functions
Exercises
1. Linear Orders, Sups and Infs
Exercises
2. Real numbers
Exercises
3. Sequences and Convergence
Exercises
4. Cauchy Sequences
Exercises
5. Arithmetic Operations on Sequences
Exercises
6. Subsequences and Monotone Sequences
Exercises
7. Limits of Functions
Exercises
8. Limits of Functions vs Sequences
Exercises
9. Algebra of Limits
Exercises
10. Limits of Monotone Functions
Exercises
11. Continuity of a Function at a Point
Exercises
12. Algebra of Continuous Functions
Exercises
13. Uniform Continuity
Exercises
14. Properties of Continuous Functions
Exercises
15. The Derivative of a Function
Date: January 2022.
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Exercises
16. The Algebra of Derivatives
Exercises
17. The Mean–Value Theorem
Exercises
18. L’Hospital’s Rule
Exercises
19. Solutions to selected problems
Problems for Section 1
Problems for Section 2
Problems for Section 3
Problems for Section 4
Problems for Section 5
Problems for Section 6
Problems for Section 7
Problems for Section 8
Problems for Section 9
Problems for Section 10
Problems for Section 11
Problems for Section 12
Problems for Section 13
Problems for Section 14
Problems for Section 15
Problems for Section 16
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NOTES ON INTRODUCTION TO ANALYSIS
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0. Before we even start
0.1. Principles of Mathematical Writing.
Reason Welcome to the world of mathematical writing. Before we get started,
you may be thinking, “Why are we writing in a math class?” It is a fair question,
but notice you would not think it is odd to write in a history or a philosophy class.
There is more to writing than just the words on a page - it is a tool that has been
used throughout the ages to communicate ideas to each other. With that in mind,
we realize there is not much di↵erence between math and literature, because they
are both about conveying ideas. Communicating ideas about math should not be
any di↵erent from communicating other ideas.
Language When writing in math, we will use two languages to convey ideas:
Math, and the language in which it is being conveyed - in our case, English. If we
were to think of English as an ocean, Math would be the fish that reside in it. Math
language, without English, would be like a fish without water, and would die, but
notice that the ocean still exists even in the absence of fish, as English still exists
without math. In fact, before the 16th century, we did not use formulas but rather
conveyed ideas in words. One way to see if you understand an idea in math is to
write it down. You can get correct answers by accident, but you cannot represent
the ideas in writing by accident. Perhaps you have noticed this when writing a
proof. When you make an error, it seems to pop out at you. That is because the
idea is incorrect. You can use all the right words, get the formatting just right, and
be flawless when it comes to the computation, but if the idea is wrong, it is all a
waste. The ideas are the core and the writing is the presentation.
Keep it Simple The English language has multiple tenses, but in math, there
is no need for them, we keep things as simple as possible. As far as grammar is
concerned, we reduce the language to its core. We do not want to distract the
reader from the important idea we are conveying. Therefore, we do not use fancy
language or complex sentence structures. Keep it simple, but retain importance.
Do not use a paragraph when a sentence will do, and try to use as few words as
possible.
You must control your use of English. Use only the basics, and try to keep your
sentences simple. More complex structures are rather saved for use with proofs by
contradiction. You might be tempted to use conditional structures and sometimes
it may be necessary, but be very careful with this. Read your paragraph twice,
three times if you have to. Make sure it will not confuse the reader.
In technical writing readers often just scan articles, they search for the information they need. This is a big mistake on their part, but you should be aware of
this fact. When people do not read carefully, they read/see what they expect to be
written. Try to follow traditions because the scanners in your audience will read
what they expect to see.
It helps to read your work out loud. If what you write can be broken into separate
sections, then break it up.
Always write the solution after you solve the problem. If there are natural
breaking points, such as a sequence of equalities, you should break it up. Do not
save ink, or paper. Granted, this is a matter of judgment. The general rule is that
it is better to do too much than not enough, if you are not sure.
4
ANDRZEJ ROSLANOWSKI
Audience Always remember your audience. This cannot be stressed enough.
You do not want to present Partial Di↵erential Equations to a class of toddlers,
nor do you want to present a popup book to your esteemed colleagues. Write to
your audience. Determine what you can assume they already know and what you
might need to clarify. You may be able to correct speech in person, but you cannot
correct or clarify your writing once it is in the reader’s hands. Remember, you are
never writing for only yourself, so keep your readers in mind.
Make a habit of using the royal “we.” This point of view connects you with
the reader, and is a matter of being polite. You are a guide, leading your reader
through your ideas. In mathematics, it is very unusual to see an author use first
person singular, “I.” It is actually quite irritating, and comes o↵ as hubris. Using
“we” gives the appearance to the reader that you are conversing with them, rather
than lecturing them.
Whatever you write, it should always be written under the assumption that
someone will read it. It does not make sense to write for no audience. Remember
to keep in mind who your audience is, even if your audience is you. First, knowing
your audience enables you to keep the complication level appropriate for the audience. You do not want to write beyond the reader. If you are writing something
for a masters thesis your audience is professional mathematicians. Do not bother
professors with basic algebra and addition, but if you are writing a paper for middle
school students, you do need to explain the basic algebra.
Often, an equation can be solved and simplified in various ways. The reader,
left on his own, will do it di↵erently, so do not force him to do it your way. Do
not leave out anything that is critical to the point of your paper. Saying what
is important and nothing more shows that you really understand the solution. If
you write too much or too little, your audience may question your understanding.
The same solution could be written in two di↵erent ways depending on the target
reader. If you are not sure, assume the audience is someone like you. When you
write for someone like you, you assume the person has your knowledge. You do
not want to o↵end your reader by oversimplifying. On the other hand, you want
to make sure you present all necessary steps. It is a balancing act, and you must
apply good judgment. Make sure everything you write is relevant to the problem.
Do not write something if it is not necessary. Do not discuss how the problem could
have been solved if you did not solve the problem that way. Everything must follow
from what was written before. If something is not connected, it is meaningless and
should be omitted.
Again, unless you are writing for kids, the reader will not be interested in calculations. Your algebraic manipulations or anything that is reduced to processing
should be done in your sketch papers not in your actual writing. Think about a
restaurant, where the wait sta↵ brings your food presented nicely on a plate. They
do not show you the kitchen where they prepare the food. You only want to see the
final dish, nicely presented and ready to eat, not the dirty kitchen. Likewise, do
not show the scratch calculations of a dirty math kitchen in your final presentation.
Rules Do not try to give new meaning to English words or redefine common
words. Trying to change words or put a weird spin on them will only confuse the
reader. Be aware of things like periods which could also be decimals. The way to
NOTES ON INTRODUCTION TO ANALYSIS
5
fix these types of problems is context - if there is nothing following the period, it is
obviously not a decimal.
Just as an English paper, every mathematical writing should have an introduction, body, and conclusion. The introduction should explain the idea you are trying
to prove, show, etc. The body should contain the method that shows your idea is
valid. The conclusion should mainly restate why your idea is true. More is to be
said on this later.
Keep things simple and be sure to follow the rules. The rules in English are the
same rules in math: everything you write should be a sequence of sentences and
be organized into paragraphs. The beginning starts with a capital, the end has a
period, etc. You may sometimes break the rules, but you have to know what you
are doing with the language and why you are doing it. You need have a purpose,
an intention for doing it.
Terms and Sentences Like English, every sentence in math has a noun and a
verb. Terms like x, y, ⇡ etc, act as the nouns of your math sentences, while relational
symbols like =, <, >, etc, act as your verbs. You can think of your terms as names
for objects. Some terms are specific, such as “Chelsea.” Other terms are general,
such as “student.”
5e2⇡ 10
specific
R
general
Sentences must always be complete. You would not write “The dog.” You would
say something more like “The dog ran.” or “The dog ate.” Without a verb, the
sentence doesn’t make sense. In the same way, your math sentence would not be,
5z 2 z.
It would be more like,
5x2 z = 10y.
Also, be sure to keep your languages, English and Math, separate. You would not
start a sentence in English and end it in Polish, would you? Look at this example:
5x2 z
something more than I want + Pi equals 2y.
This may be grammatically correct, but it is very bad style and will confuse the
reader.
Our math verbs, in the same way, need to be kept separate from the English
language. Never write:
The total revenue = pq where . . . .
“The total revenue is equal to pq where. . ..” is technically correct, but bad style.
It would be best to say,
Let R be total revenue such that R = pq where. . ..
Never use the “=” sign to say “equals” in a line of text:
what is on the left = what is wanted.
You can use isolated “=” in your notes, otherwise only use in mathematical sentences. The same applies to quantifiers.
Thus, 8x we have ... .
6
ANDRZEJ ROSLANOWSKI
If you read relational symbols out loud you realize they are verbs. In 1 + 1 = 2,
the “=” acts as the verb. We use our symbols to create sentences with terms. The
inequality
3xy < 2
is a complete sentence, but the expression
5x2 z
10y
is not.
Keep symbols separate from English words and very seldom use relational symbols in isolation.
Never, ever, start a sentence with a symbol.
It should be clear where a math formula begins and ends. It is acceptable to keep
terms inline within the paragraph, but formulas and equations often look better on
separate line:
y 3 x + 22 + 2 = y 3 + x + 6.
Also, if a term or short equation is embedded in English statement, then it is good
to punctuate it:
Words, 2 + 2 = 4, words.
Symbols One of the best accomplishments in science was the advent of symbols.
You need to write with both symbols and words, and you must apply good judgement as to what should be written using symbols and what should be written using
words. A ten-page document of math in English with no symbols might be difficult
to read. When you use symbols you need to be in agreement with your reader, so
use widely used symbols. To verify whether something is widely used, check your
textbook. If your text uses a symbol, you may use it also. If a symbol is not in your
text, do not use that symbol. If you must use a symbol not in your text, define it
as soon as it appears. If two of your sources have conflicting notation, state which
source you are using. For calculus problems use your calculus text as a guide. If
you are a PhD student check with your advisor for questions about symbols. If you
are writing a thesis, refer to other published works if necessary.
When you use symbols in writing, your symbols are like names for objects. Those
names can be totally arbitrary, and we may name things any way we want, but there
are traditions and conventions to make our lives easier. You do not always know
who will be reading your writing, so play it safe.
We should always write what our symbols stand for. It is confusing if symbols
stand for more than one thing, so do not recycle your symbols.
Labels First of all, keep your labels uniform. You can keep the labels on the
left or the right, but stay consistent. It does not matter what you use for labels:
numbers, hearts, stars, roman numerals, etc. Once you label an equation you can
refer to that equation easily later. For example,
(1)
3z + 8 = y
can be referred to later as equation (1). You do not have to repeat the equation
anymore. The advantage is that you can avoid repetition, and you may phrase your
sentences in a more precise way. For example “By (⇤) and (⇤⇤) it follows that...”
is a great way to avoid repeating complicated equations. Labels help to guide the
NOTES ON INTRODUCTION TO ANALYSIS
7
reader, but keep in mind, you do not want to overdo it. You should only use labels
for equations you will refer to later. Write your draft and make your labels, then
when you proofread, erase the labels you did not end up referring to.
Hints Here is a list of what to do and what not to do, which will help get you
acquainted with the basics of mathematical writing. These are some of the most
common mistakes that new writers make, so if you follow these guidelines, you will
be o↵ to a good start.
DO: 3 + x < 0
DON’T: 3 plus x is less than zero.
When “math language” is clearer, use notation instead of English.
DO: the volume v of the solid s is...
DON’T: v of s is...
Always introduce new symbols.
DO: Because x > 3, we know x2 > 9 and x3 > 27.
DON’T: x > 3, x2 > 9, x3 > 27
Write in complete sentences.
DO: Since y is positive, it has a square root.
DON’T: y is positive, so it has a square root.
Never start a sentence with a symbol!
DO: Let p be a prime.
DON’T: Let m be a prime.
Your notation should match its context, so pick common sense variables.
DO: Let A be a set with two elements a1 and a2 .
DON’T: Let A be a set with two elements a and x.
Use caution when naming the elements of sets. They need to make sense as a
family.
DO: Be clear about variables’ definitions.
DON’T: Reuse variables or use them under multiple definitions.
Recycling variables (or labels) will only serve to confuse the reader.
DO: Let x be a negative integer.
DON’T: Let x < 0 2 Z.
Too much notation is confusing. This “don’t” has a di↵erent meaning than you
may expect!
DO: The conditions imply that a = 3 or there is no solution.
DON’T: The conditions imply that a = 3_ there is no solution.
Don’t use logical symbols when writing, except as a course on logic. Never use
them as shorthand!
DO: Let x be a positive integer.
DON’T: Let x be a Z+ .
Again, use proper English, not symbolic slang.
DO: By equation (6.2), we see...
DON’T: By the equation 11 lines up, we see...
If something is important enough to be referred to later, give it a label and use
that label to direct the reader back to it.
8
ANDRZEJ ROSLANOWSKI
DO:
(x + 2)3
=
=
=
(x + 2)2 (x + 2)
(x2 + 4x + 4)(x + 2)
x3 + 6x2 + 12x + 8.
or
(x + 2)3 = (x + 2)2 (x + 2) = (x2 + 4x + 4)(x + 2) = x3 + 6x2 + 12x + 8.
DON’T:
(x + 2)3 =
(x + 2)3 =
(x + 2)3 =
Do not be repetitious by ’stacking’
(x + 2)2 (x + 2)
(x2 + 4x + 4)(x + 2)
x3 + 6x2 + 12x + 8.
computations vertically.
DO: (x + y)2 = x2 + 2xy + y 2 .
DON’T: (x + y)2 = x2 + (2xy + y 2 ) .
Use parentheses only when necessary. If you’re unsure, err on the side of excess,
but be smart.
DO: Write professionally.
DON’T: Use contractions.
This list excluded, of course!
DO: Use correct spelling and proper grammar.
DON’T: Make words be spelled incorectly and let you’re grammer get sloppy
If you’re not sure, and a computer can’t help, ask someone to proofread it. This
is not a matter of compromise!
DO: Proofread your work multiple times to make sure it is perfect.
DON’T: Just scan and submit.
A good way to improve your math writing is by reading your work out loud.
Doing so forces you to concentrate and actually read what is written.
NOTES ON INTRODUCTION TO ANALYSIS
9
LATEX TEX is a typesetting system that was created by Donald Knuth and released in 1978. Using it may be a very di↵erent experience from writing in you
typical word processing program.
Text editors let you type what you want and choose formulas. What you see is
what you get with this program. You type something and then you see immediately
what you get.
In a programming language you write a program then you compile it, in e↵ect
creating a binary executable file. What you write in TEX is a programof a sort.
Programs are text files, using ASCII codes only. You then use the compiler to
compile it. Filename.tex is the default for naming a TEX file.
LATEX is a set of macros for TEX. You do not have to worry how to do some
of the programming; it is done for you with the macros. LATEX from programming
point of view is a bunch of libraries. LATEX is not in place of TEX it is around TEX
. You can change things not only in LATEX, but also in TEX . TEX overrides LATEX,
so be careful when writing in TEX . It is not necessary to mess with TEX in LATEX
for our purposes.
Here are some latex symbol codes to help get you started.
\alpha
\zeta
\lambda
⇢ \rho
\chi
⇥ \Theta
⌥ \Upsilon
1 \infty
\leftarrow
⇠
R \sim
\int
↵
⇣
\beta
\eta
\mu
\sigma
\psi
⇤ \Lambda
\Psi
 \le
( \Leftarrow
H9 \exists
\oint
⌘
µ
✓
⌫
⌧
!
⌅
6=
8
()
\gamma
\theta
\nu
\tau
\omega
\Xi
\Phi
\ge
\ne
\forall
\i↵
◆
⇠
⇧
⌦
!
⌘
P
¬
\delta
\iota
\xi
\upsilon
\Gamma
\Pi
\Omega
\rightarrow
\equiv
\sum
\neg
\epsilon
\kappa
\pi
\phi
\Delta
⌃ \Sigma
± \pm
) \Rightarrow
⇡ \approx
Q
\prod
^ \land
✏

⇡
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ANDRZEJ ROSLANOWSKI
0.2. Sets, Relations and Functions. Symbols and notation used in this Notes
and lectures should be compatible with those you learned and used in other classes.
However, to make sure that we understand each other, let us review the basic
terminology.
Logical symbols
symbols.
We will use the following representations of classical logical
Notation 0.1.
• _ stands for logical connective or, and/or, it is inclusive
“or”;
• ^ and occasionally & stand for logical connective and ;
• ) stands for logical connective if . . . then i.e., it denotes the implication;
• , stands for logical connective if and only if i.e., it denotes the biconditional;
• ¬ stands for logical connective not i.e., it denotes the negation;
• 8x is the universal quantifier, i.e., for every x;
• 8x 2 A is the universal quantifier bounded/restricted to set A, i.e., for
every x in a given set A;
• 9x is the existential quantifier, i.e., there exists x;
• 9x 2 A is the existential quantifier bounded/restricted to set A, i.e., there
is x in a given set A.
Remark 0.2. Remember, the proposition p ) q is logically equivalent to ¬q ) ¬p
and to ¬p _ q, and the proposition ¬(p ) q) is logically equivalent to p ^ ¬q.
Example 0.3.
(1) (9x 2 R)(x2 = 2) is a true statement, while (9x 2 Q)(x2 =
2) is false.
(2) The order of quantifiers is very important. For instance, let P (x, y) means
“y makes x happy”. Then
(8x)(9y)P (x, y) stands for a reasonably true statement
for every x there is a y making x happy,
but (9y)(8x)P (x, y) is a (probably) false sentence
there is a y making everybody happy.
(3) However, the order of quantifiers of the same sort does not matter. Therefore we write 8x, y instead of 8x8y (“for every x, y”), and 9x, y instead of
9x 9y (“there are x, y”).
Sets The most basic notation concerning sets is a 2 A which signifies the assertion
that “an element a belongs to a set A”, “a is a member of a set A”.
We will very often introduce/define sets using one of the following methods.
• Listing elements as in {a, b, c} — a set consisting of 3 points.
• Showing a pattern as in {1, 2, 3, 4, . . .} — the set of positive integers. But
be careful! If you write {3, 5, 7, . . .} you may mean the set of all odd
numbers or the set of all prime numbers greater than 2, {3, 5, 7, 9, . . .}
or {3, 5, 7, 11, . . .}.
• We may use word description, as in “the set of all integers”.
• We may use set builder notation as in {x 2 A : P (x)}, where P (x) is a
property of x.
Notation for some sets is well established, for instance:
• R denotes the set of all real numbers,
NOTES ON INTRODUCTION TO ANALYSIS
11
• Q denotes the set of all rational numbers,
• Z denotes the set of all integers,
• J denotes the set of all positive integers, sometimes it may be denoted by
N,
Operations on sets
Definition 0.4. For sets A and B we define their union, intersection and di↵erence
as follows.
Union
Intersection
Di↵erence
A [ B = {x : x 2 A or x 2 B},
A \ B = {x : x 2 A and x 2 B}
A\B = {x : x 2 A and x 2
/ B}
Proposition 0.5. For any sets A, B, and C we have:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
A\B =B\A
A[B =B[A
A \ (B \ C) = (A \ B) \ C
A [ (B [ C) = (A [ B) [ C
A \ (B [ C) = (A \ B) [ (A \ C)
A [ (B \ C) = (A [ B) \ (A [ C)
A \ (B [ C) = (A \ B) \ (A \ C)
A \ (B \ C) = (A \ B) [ (A \ C)
A \ B = A \ (A \ B)
Proof. See your notes from MATH 2230.
⇤
Definition 0.6.
(1) An indexed family of sets is a function whose values are
sets.
An indexed family of sets is denoted by an expression of the form {A :
2 ⇤} which should not be mistaken for a set. It is a function with domain
⇤ and the value at 2 ⇤ being A .
(2) A family of sets is a set whose elements are sets.
Remark 0.7. While a family of sets and an indexed family of sets are di↵erent
notions, for our purposes you do not have to di↵erentiate between them. Pick the
one that you are more comfortable with and always use this one. Note: if A is a
family of sets, that it can be thought of as an indexed family {B : 2 ⇤}, where
⇤ = A and BA = A. On the other hand, the range of an indexed family of sets is
a family of sets.
Definition 0.8.
(1) If A is a family of sets then we define its union as
[
A = {x : (9A 2 A)(x 2 A)}
and if {Ai : i 2 I} is an indexed family of sets then
[
[
{A : 2 ⇤} =
A = {x : (9 2 ⇤)(x 2 A )}.
2⇤
(2) If A is a family of sets then its intersection is
\
A = {x : (8A 2 A)(x 2 A)}
12
ANDRZEJ ROSLANOWSKI
and if {A : 2 ⇤} is an indexed family of sets then
\
\
{A : 2 ⇤} =
A = {x : (8 2 ⇤)(x 2 A )}.
2⇤
Example 0.9. For a real number r 2 R let Ar = [3 r2 , 5 + r2 ]. Then {Ar : r 2 R}
is an indexed family of sets and
[
\
Ar = R and
Ar = [3, 5].
r2R
Theorem 0.10. Let {A :
Then
(1)
r2R
2 ⇤} be an indexed family of sets and B be a set.
[
\
B\
A =
(B \ A )
2⇤
and
(2)
B\
\
2⇤
Proof. See your notes from MATH 2230.
2⇤
A =
[
2⇤
(B \ A ).
⇤
Relations and functions The ordered pair (a, b) is a creature which knows a is
first and b is second, it knows that a is followed by b. We can define it in a formal
as follows.
Definition 0.11. For any objects a, b, the ordered pair is defined by
(a, b) = {{a}, {a, b}}.
For sets A and B, the Cartesian product A ⇥ B is the set
A ⇥ B = {(a, b) : a 2 A and b 2 B}
If A and B are the same, write A2 instead of A ⇥ A.
Definition 0.12.
(1) A binary relation is a set of ordered pairs. Thus, R is a
binary relation if, for some sets A, B we have R ✓ A ⇥ B. Often we write
a R b instead of (a, b) 2 R.
(2) For a binary relation R:
the domain of R is the set of all first coordinates used in R, that is,
dom(R) = {x : (9 y)(x R y)};
the range of R is the set of all second coordinates of pairs from R, that is,
rng(R) = {y : (9 x)(x R y)}.
(3) A binary relation R on a set X is called a function if and only if for every
element x in dom(R) there is exactly one y in rng(R) such that xRy.
(4) We write f : A ! B to signify that f is a function with domain A and
the range included in the set B. Also then, for x 2 A the unique y 2 B
such that (x, y) 2 f is denoted by f (x).
Definition 0.13.
(1) A function f : X ! Y is one-to-one (or injection) if
(8x, y 2 X)(x 6= y ) f (x) 6= f (y)). It is onto Y (or surjection) if for every
y 2 Y we may find x 2 X such that f (x) = y.
(2) A function f : X ! Y is a bijection if it is both onto and one-to-one.
Proposition 0.14. If f : X ! Y is a bijection, then there is (unique) g : Y ! X
such that g f (x) = x for all x 2 X and f g(y) = y for all y 2 Y . This unique g
is called the inverse of f and it is denoted by f 1 .
NOTES ON INTRODUCTION TO ANALYSIS
Proof. See your notes from MATH 2230.
13
⇤
Exercises.
Problem 0.1. Show Theorem 0.10.
Problem 0.2. Show Proposition 0.14.
Problem 0.3. Using binary relational symbols <,  and =, 6= and binary functional
symbols +, , ·, exponentiation and constants 0, 1, 2, 3, . . . express the following
statements. (Do not pay attention to the truth value of these sentences.) The range
of all quantifiers and/or variables is assumed to be the set of all real numbers.
(1) There is no real number whose square is smaller than zero.
(2) Between any two distinct real numbers there is another real number.
(3) There is no smallest real number.
(4) The equation x7 x4 2x = 2021 has at least one real solution.
(5) The equation x7 x4 2x = 2021 has at least two real solutions.
(6) The equation x8 + x4 + 1 = 0 has no real solutions.
(7) Every quadratic equation has at most two real solutions.
(8) Every cubic equation has at least one real solution.
Problem 0.4. A formula is in prenex form if it is written as a string of quantifiers
(referred to as the prefix ) followed by a quantifier–free part (referred to as the
matrix). For each of the following formulas find a formula in the prenex form
equivalent to the given formula and the matrix simplified as much as possible. (Do
not pay attention to the truth value of these sentences.)
(1) ¬(8x)(8y)(x + y = y + x)
(2) ¬(8x)(9y)(x + y = 2021)
(3) ¬(9x)(8y)(2x > sin(y))
(4) ¬(8" > 0)(9 > 0)(8x)(8z)(|x z| < ) | sin(x) sin(z)| < ").
(5) ¬(8x)(8" > 0)(9 > 0)(8z)(|x z| < ) |2x 2z | < ").
(6) (8x) x 0 ) (9y)(y 2 = x) .
(7) ¬(8x) (9z)(z 2 = x) ) (9y)(y 4 = x) .
14
ANDRZEJ ROSLANOWSKI
1. Linear Orders, Sups and Infs
You may have seen some of the material presented here in your Intro to Abstract
Math class. To play it safe, however, let us use at the concepts of linear orders and
least upper bounds again.
Definition 1.1. A binary relation R on a set X is
(1) weakly antisymmetric if for every x, y 2 X we have
xRy^y Rx
)
x = y,
(2) antisymmetric if for every x, y 2 X we have
xRy
)
¬(y R x),
(3) reflexive if for every x 2 X we have x R x,
(4) irreflexive if for every x 2 X we have ¬(x R x),
(5) transitive if for all x, y, z 2 X we have
xRy ^ yRz
)
x R z,
(6) a partial order if R is transitive, reflexive, and weakly antisymmetric,
(7) a strict partial order if R is transitive, irreflexive and antisymmetric.
Example 1.2.
(1) Let  be the standard weak inequality on the set R of all
real numbers. Then  is a partial order.
(2) The strict inequality < on R is a strict partial order.
(3) Define a relation on R ⇥ R by
(x1 , x2 )
Then
(y1 , y2 ) , x1  y1 ^ x2  y2 .
is a partial order on R ⇥ R.
Definition 1.3.
(1) Let v be a partial order on X. We say that elements
x, y 2 X are v-comparable if x v y or y v x. Elements which are not
comparable are said to be incomparable.
(2) If any two elements of X are v-comparable, then the partial order v is
called a linear (total) order on X.
(3) We say that a strict partial order @ on X is a strict linear (total) order if
for all x, y 2 X:
either x = y, or x @ y, or y @ x
Example 1.4. The standard inequality  on R is a linear order and < is a linear
strict order. The partial order on R ⇥ R (defined in Example 1.2(3)) is not linear.
We will be interested in the properties of (R, <) and/or (R, ), but many of the
relevant concepts can be considered in general linear orders or even partial orders.
Remark 1.5. Suppose v is a linear order on a set X and for x, y 2 X let x @ y if
and only if x v y and x 6= y. Then @ is a strict linear order on X.
Conversely, if @ is a strict linear order on a set X and for x, y 2 X we define
x v y if and only if x @ y or x = y, then v is a linear order on X.
Definition 1.6. Let v be a linear order on a set X and ; =
6 S ✓ X.
NOTES ON INTRODUCTION TO ANALYSIS
15
(1) Let x 2 X. We say that x is the largest element of S if
x2S
and
x2S
and
(8y 2 S)(y v x),
and x is the smallest element of S if
(8y 2 S)(x v y).
(2) We say that M 2 X is an upper (lower, respectively) bound to S if and
only if a v M (M v a, respectively) for all a 2 S.
(3) The set S is bounded from above if it has an upper bound. It is bounded if
it has both upper and lower bounds.
Example 1.7. Let J, Z, Q and R be equipped with the natural ordering .
(1) Every nonempty subset of J has the smallest element.
(2) Z has neither smallest nor largest elements.
(3) The interval [0, 1) is bounded in R from above by, e.g., 1, 2, 3, ⇡ and it
is bounded from below by, e.g., 2, 1, .5, .25, 0. It has the smallest
element (it is 0) and it has no largest element.
(4) The set [0, 1) \ Q is bounded in Q from above by, e.g., 1, 2, 3 and it is
bounded from below by, e.g., 2, 1, .5, .25, 0. It has the smallest element (it is 0) and
p it
p has no largest element.
p p
(5) The interval ( 2, 3) is bounded in R from abovep by, e.g., 3, 3 +
.002,
2, 1, 1.4142, and
p 3, ⇡ and it is bounded from below by, e.g.,
2. It has
no
smallest
element
and
it
has
no
largest
element.
p p
(6) The set ( 2, 3)\Q is bounded in Q from above by, e.g., 3, 2, 1.75, 1.74, 1.733
and it is bounded from below by, e.g., 1, 0, 1.4, 1.41, and 1.4142.
Observation 1.8. Let v be a linear order on X and ; =
6 S ✓ X.
(1) If both x1 and x2 are the largest elements of S, then x1 = x2 .
(2) If both x1 and x2 are the smallest elements of S, then x1 = x2 .
(3) If M is an upper bound of S and M v M 0 , then M 0 is also an upper bound
of S.
(4) If m is a lower bound of S and m0 v m, then m0 is also a lower bound of
S.
Proof. Left for Exercises.
⇤
Definition 1.9. Let v be a linear order on a set X and ; =
6 S ✓ X. If we write
M = max(S) we mean that M is the (unique) largest element of S. Similarly,
m = min(S) means that m is the (unique) smallest element of S.
Definition 1.10.
(1) We say that M is a least upper bound of S if and only if
• M is an upper bound to S, and
• no x strictly @–smaller than M is an upper bound to S.
A least upper bound of S may be also called a supremum of S.
(2) The element m is a greatest lower bound of S if and only if
• m is a lower bound to S, and
• no x strictly @–greater than m is a lower bound to S.
A greatest lower bound of S may be also called an infimum of S.
Remark 1.11. Let v be a linear order on a set X and ; =
6 S ✓ X.
(1) An M 2 X is a least upper bound of the set S if and only if
(a) (8a 2 S)(a v M ), and
16
ANDRZEJ ROSLANOWSKI
(b) (8M 0 @ M )(9a 2 S)(M 0 @ a).
(2) Similarly, m 2 X is a greatest lower bound of the set S if and only if
(a) (8a 2 S)(m v a), and
(b) (8m0 A m)(9a 2 S)(a @ m0 ).
Observation 1.12. Let v be a linear order on X and ; =
6 S ✓ X.
(1) If both x1 and x2 are least upper bounds of S, then x1 = x2 .
(2) If both x1 and x2 are greatest lower bounds of S, then x1 = x2 .
(3) If M = max(S), then M is a least upper bound of S.
(4) If m = min(S), then m is greatest lower bound of S.
Proof. Left for Exercises.
⇤
Definition 1.13. Let v be a linear order on a set X and ; 6= S ✓ X. If we write
M = sup(S) we mean that M is the (unique) least upper bound of S. In particular,
this means that the least upper bound exists. Similarly, m = inf(S) means that m
is the (unique) greatest lower bound of S.
Example 1.14. Let Q and R be equipped with the natural ordering .
⇣
⌘
⇣
⌘
(1) In R, sup [0, 1) = 1 and inf [0, 1) = 0.
⇣
⌘
⇣
⌘
(2) In Q, sup [0, 1) \ Q = 1 and inf [0, 1) \ Q = 0.
⇣p p ⌘ p
⇣p p ⌘ p
(3) In R, sup ( 2, 3) = 3 and inf ( 2, 3) = 2.
p p
(4) In Q, the set ( 2, 3) \ Q is bounded but it has no least upper bound nor
greatest lower bound.
(5) Let A = [0, 1) [ 1 + n1 : n 2 J . Then (in R) inf A = min A = 0 and
sup A = max A = 2.
(6) Assume ; 6= A ✓ R and let B = { x : x 2 A}. If ↵ = sup A , then
↵ = inf B , and if = inf A , then
= sup B .
NOTES ON INTRODUCTION TO ANALYSIS
17
Exercises.
Problem 1.1. Let v be a linear order on X and ; =
6 S ✓ X. Prove that:
(1) If both x1 and x2 are the largest elements of S, then x1 = x2 .
(2) If both x1 and x2 are the smallest elements of S, then x1 = x2 .
Problem 1.2. Let v be a linear order on X and ; =
6 S ✓ X. Prove that:
0
(1) If M is an upper bound of S and M v M , then M 0 is also an upper bound
of S.
(2) If m is a lower bound of S and m0 v m, then m0 is also a lower bound of S.
Problem 1.3. Let v be a linear order on X and ; =
6 S ✓ X. Demonstrate that:
(1) If both x1 and x2 are least upper bounds of S, then x1 = x2 .
(2) If both x1 and x2 are greatest lower bounds of S, then x1 = x2 .
Problem 1.4. Let v be a linear order on X and ; =
6 S ✓ X. Demonstrate that:
(1) If M = max(S), then M is a least upper bound of S.
(2) If m = min(S), then m is greatest lower bound of S.
Problem 1.5. Let A be a nonempty subset of R. Define
A = { a : a 2 A}.
(a) Prove that if A is bounded from below, then A is bounded from above.
(b) Prove that if A has an infimum in R, then A has a supremum and inf(A) =
sup( A).
Problem 1.6. Let A be a nonempty subset of R and ↵ > 0. Define
↵A = {↵a : a 2 A}.
(a) Prove that if A is bounded from above, then ↵A is bounded from above.
(b) Prove that if A has a supremum in R, then ↵A has a supremum and
sup(↵A) = ↵ sup(A).
Problem 1.7. Let B = {⇡ +
1
2k
: k 2 J} ✓ R. Show that
1
= sup(B).
2
Does the set B have the largest and/or the smallest elements?
⇡ = inf(B)
and
⇡+
Problem 1.8. Let C = {r 2 Q : 0 < r < 1} and B = {⇡ + r : r 2 C} ✓ R. Show
that
⇡ = inf(B) and ⇡ + 1 = sup(B).
Does the set B have the largest and/or the smallest elements?
Problem 1.9. For each subset of R below, determine if it is bounded above,
bounded below, or both, and if it has the supremum (infimum) and what they
are. Justify all your conclusions.
(a) {1, 3, 5, 7, 11}
(b) [ 2, 5)
(c) {2 + 2 n : n 2 J}
(d) {2 + 2n : n 2 J}
(e) {x 2 R : x3 5x < 0}
18
ANDRZEJ ROSLANOWSKI
2. Real numbers
Ever wondered what are the real numbers? Many mathematicians did. It is
relatively easy to introduce/define/formalize first natural numbers, then integers
and after this the rational numbers. After this one gets stuck, at first one may even
hope that we got all numbers we need. But already the Ancient Greeks knew that
this is not the case.
Theorem 2.1 (Pythagoreans; perhaps Hippasus of Metapontum, 5th century BC).
There is no rational number whose square is 2.
Proof. You must have seen a proof of this theorem a couple of times. If you do not
remember the proof, just Google it up.
⇤
The modern answer to the question what real numbers are, is rather by giving
their properties than explaining their nature/character. It follows, in some sense,
the pattern given by Euclid in his Elements. So we just declare that the reals are
equipped with two algebraic operations, with a linear order and these have nice
properties: they form a complete ordered field.
Definition 2.2. A complete ordered field is a structure (R, +, ·, <) such that the
following demands are satisfied.
(0) +, · are binary operations on R, that is + : R⇥R ! R and · : R⇥R ! R,
and < is a binary relation on R.
(1) 8x, y, z 2 R (x + y) + z = x + (y + z) ^ (x · y) · z = x · (y · z) .
(2) 8x, y 2 R x + y = y + x ^ x · y = y · x .
(3) 8x, y, z 2 R x · (y + z) = (x · y) + (x · z) .
(4) There is a unique element 0 2 R such that (8x 2 R)(0 + x = x).
(5) For each x 2 R , there is a unique y 2 R such that x + y = 0, and we write
y = x.
(6) There is a unique element 1 2 R such that 0 6= 1 and (8x 2 R)(1 · x = x).
(7) For each x 2 R with x 6= 0, there is unique element y 2 R such that
x · y = 1, and we write y = x 1 or y = x1 .
(8) 8x, y, z 2 R x < y ) x + z < y + z .
(9) 8x, y, z 2 R (x < y ^ y < z) ) x < z .
(10) For any x, y 2 R, exactly one of the following is true:
x < y,
y < x, or x = y.
(11) 8x, y, z 2 R (x < y ^ z > 0) ) x · z < y · z .
(12) Every nonempty <–bounded from above subset of R has a least upper
bound.
Remark 2.3.
(1) Axioms 2.2(0)–(7) say that (R, +, ·) is a field.
(2) Axioms 2.2(9)-(10) are equivalent to saying that < is a strict linear order
on R.
(3) Axioms 2.2(0)–(11) say that (R, +, ·, <) is an ordered field.
(4) There are many examples of ordered fields, some of them are quite exotic.
(5) The last axiom, 2.2(12) is known as the Completeness Axiom
Theorem 2.4.
(1) There exists complete ordered fields.
NOTES ON INTRODUCTION TO ANALYSIS
19
(2) Moreover, any two complete ordered fields are isomorphic/identical:
if (R1 , +1 , ·1 , <1 ) and (R2 , +2 , ·2 , <2 ) are ordered fields then there is a bijection ' : R1 ! R2 such that for all x, y 2 R1 we have:
'(x +1 y) = '(x) +2 '(y) and
'(x ·1 y) = '(x) ·2 '(y) and
x <1 y , '(x) <2 '(y).
Definition 2.5.
(1) The number line R = (R, +, ·, <) is any of the (isomorphic) complete ordered fields. We refer to its elements as real numbers.
(2) The integers are all elements of R which can be obtained by repeated addition and/or subtraction of 1. The set Z of all integers is the subring of R
generated by {1}.
(3) The rationals are all real numbers of the form p · q 1 where p, q 2 Z and
q 6= 0. The set Q of all rational numbers is the subfield of R generated by
{1}.
All the properties of the real numbers that you are familiar with (or rather: you
believe in) can be shown from the axioms of complete ordered field. For instance
we have the following consequences of Definition 2.2.
Theorem 2.6. Let x, y and z be any real numbers. Then
(1) 0 · x = 0 = 0.
(2) ( x) · y = (x · y) and hence in particular ( 1) · ( 1) = 1,
(3) If x < y, then y < x.
(4) 0 < 1.
(5) If 0 < x < y, then 0 < y1 < x1 .
(6) If x < y and z < 0, then y · z < x · z.
(7) x2 0.
Proof. (1) First note that 0 + 0 = 0 by Axiom (4), and hence by the uniqueness
in Axiom (5) we have 0 = 0. Next, using Axioms (5,3) we get
x · 0 = x · (0 + 0) = (x · 0) + (x · 0)
and hence by Axiom (2)
0 · x = (0 · x) + (0 · x).
Consequently,
(0 · x) + [ (0 · x)] = [(0 · x) + (0 · x)] + [ (0 · x)]
and applying Axioms (1,5) we conclude 0 = 0 · x, as desired.
(2)
Note that
x + [( 1) · x] = [1 · x] + [( 1) · x] = (1 + ( 1)) · x = 0 · x = 0.
Hence
x = ( 1) · x (for all x 2 R). Now,
( x) · y = [( 1) · x] · y = ( 1) · (x · y) =
(x · y).
Concerning “in particular” part, note that now
( 1) · ( 1) =
(1 · ( 1)) =
( (1 · 1)) =
(Identify all Axioms used in the arguments above.)
( 1) = 1.
20
(3)
ANDRZEJ ROSLANOWSKI
Assume x < y. Then, by Axiom (8), we have
x + ( x) + ( y) < y + ( x) + ( y)
and using Axioms (1,2) we conclude
(4)
y<
x.
By Axiom (10) we know that exactly one of the following is true:
0 = 1,
0 < 1,
1 < 0.
Axiom (6) guarantees that 0 6= 1, so we just need to eliminate the third possibility
above. Suppose towards contradiction that 1 < 0. Then, by clause (3), 0 = 0 <
1 and by Axiom (11) and clause (2)
0 · ( 1) < ( 1) · ( 1) = 1,
a contradiction.
(5) Assume 0 < x < y.
We know that, by Axiom (10), exactly one of the following is true:
0=
1
,
x
0<
1
,
x
1
< 0.
x
Since x · x1 = 1 6= 0 we know that 0 6= x1 (remember clause (1)). We want to
eliminate the third possibility above, so suppose towards contradiction that x1 < 0.
Since x > 0 we may use Axiom (11) to get
1=x·
1
x
contradicting clause (4). Thus 0 <
0<x·
so
1
< x · 0 = 0,
x
and in a similar manner also 0 < y1 . Now:
⇣1⌘ ⇣1⌘
⇣1⌘ ⇣1⌘
·
<y·
·
x
y
x
y
0<
1
1
< .
y
x
(Identify all Axioms used in the arguments above.)
(6) Assume x < y and z < 0. Then 0 < z (by clauses (1), (3)) and (x · z) =
x · ( z) < y · ( z) = (y · z) (by clause (2), Axiom (11)). Using clause (3) again
we get y · z < x · z.
(7)
By Axiom (10), exactly one of the following is true:
0 = x,
0 < x,
x < 0.
If x = 0, then x2 = 0 · 0 = 0, so x2 0.
If x > 0 then x2 = x · x > x · 0 = 0, so x2 0.
If x < 0, then x > 0 and x2 = ( 1) · x · ( 1) · x = ( x) · ( x) > 0.
(Identify all Axioms/clauses used in the arguments above.)
⇤
Theorem 2.7. Every nonempty bounded from below set of reals has a greatest lower
bound.
NOTES ON INTRODUCTION TO ANALYSIS
21
Proof. Suppose ; 6= S ✓ R is bounded from below by m, i.e., m  s for all s 2 S.
Let
T = { s : s 2 S}.
Clearly, T 6= ; and if t 2 T then for some s 2 S we have t = s  m. Thus
the set T is bounded from above by M = m. By Axiom (12), the set T has least
upper bound. Let
B = sup(T ).
We are going to show that then
B = inf(S).
First, B is a lower bound for S: if s 2 S then s 2 T and thus s  B, so B  s.
Second, B is greatest lower bound: suppose B < C. Then C < B = sup(T )
and C is not an upper bound for T . Thus there is t 2 T such that C < t.
Necessarily, t = s for some s 2 S, and we have C < s or s < C (for this
s 2 S). But the latter inequality shows that C is not a lower bound for S.
⇤
Proposition 2.8. The set Z of all integers is not bounded from below.
Proof. Suppose towards contradiction that the set Z is bounded from below. Then,
by Theorem 2.7, the set Z has the greatest lower bound: let z0 = inf(Z). Since
z0 + 1 > z0 , the number z0 + 1 cannot be a lower bound of Z. Therefore there is
an integer n 2 Z such that n < z0 + 1. But then
n
1 < z0
and
n
1 2 Z,
⇤
contradicting the assumption that z0 is a lower bound of Z.
Theorem 2.9. Let x 2 R. Then there is an integer n such that
n  x < n + 1.
Proof. By Proposition 2.8 the number x cannot be a lower bound for Z, so the set
def
A = {n 2 Z : n  x} is not empty. It is bounded from above by x, so (by Axiom
(12)), it has least upper bound. Let z0 = sup(A). Since z0 1 < z0 we know that
z0 1 cannot be an upper bound for A. Hence there is n 2 A (so n 2 Z and n  x)
such that z0 1 < n. Then z0 < n + 1, so n + 1 2
/ A. Since n + 1 2 Z we conclude
n + 1 > x n.
⇤
Theorem 2.10. Between any two distinct real numbers, there is a rational number.
Proof. Let x, y 2 R, say x < y.
By Theorem 2.9 we may pick an integer N such that 0 < y 1 x < N . Then also
1 < (y x) · N . Using Theorem 2.9 again, we choose n 2 Z such that
n  N x < n + 1.
Then
n + 1  N x + 1 < N x + N (y
so N x < n + 1 < N y. Since N > 0, we have x <
proof is complete.
x) = N y,
n+1
N
< y. Since
n+1
N
2 Q, our
⇤
Corollary 2.11. Between any two distinct real numbers, there is infinitely many
rational numbers.
22
ANDRZEJ ROSLANOWSKI
Theorem 2.12. If p is a positive real number, there is a positive real number x
such that x2 = p.
Proof. Case 1: p 1
Let A = {z 2 R : z > 0 ^ z 2  p}. Clearly 1 2 A (so A 6= ;) and A is bounded
from above by p (as z > p implies z 2 > z > p). Consequently, the set A has a least
upper bound. Let x = sup(A) > 0. Our goal now is to show that x2 = p.
⇣
⌘
x2
Suppose x2 < p. Let = min 1, p2x+1
> 0 and note that
(x + )2 = x2 + 2 x +
2
 x2 + 2 x + = x2 + (2x + 1)  x2 + p
x2 = p.
Thus x < x + 2 A, contradicting x is an upper bound for A.
⇣
⌘
2
Suppose x2 > p. Let 0 < < min x, x 2x p and suppose that a 2 A. Then
a2  p = x2
(x2
p) < x2
2 x  x2
2 x+
2
= (x
)2 ,
and consequently a < x
(note that x
> 0). Thus we have shown that x
is an upper bound to A, contradicting x was the least upper bound.
Since neither x2 > p nor x2 < p is possible, we conclude that x2 = p.
Case 2: 0 < p < 1
Then 1/p > 1 so we may use the pervious arguments to find y > 0 such that
y 2 = 1/p. Now, p = 1/y 2 = (1/y)2 and 1/y > 0, so letting x = 1/y completes the
proof in this case.
⇤
Theorem 2.13. Between any two distinct real numbers there is an irrational number.
p
2
Proof. By Theorem 2.12 there is
p a real number a > 0 such that a = 2; call it 2.
By Theorem 2.1 we know that p2 2
/ Q.p
Now, suppose x < y. Since
x
2
< y p2, we may use Corollary 2.11 to may pick
p
r 2 Q such that r 6= 0 and x 2 < r < y 2. Then x < pr2 < y and pr2 2
/ Q.
⇤
Definition 2.14. For a real number a we define
⇢
a
if a 0,
|a| =
a if a < 0.
Theorem 2.15. Let a and b be any real numbers. Then
p
(1) 0  |a| = a2 .
(2) a  |a|.
(3) |ab| = |a| · |b|.
(4) If " > 0, then
(5) If " > 0, then
|a
|a|  "
if and only if
b| < "
if and only if
(6) |a + b|  |a| + |b|.
(7) |a| |b|  |a b|.
Proof. Left for Exercises.
"  a  ".
a
" < b < a + ".
⇤
NOTES ON INTRODUCTION TO ANALYSIS
23
Exercises.
Problem 2.1. Show that between any two distinct real numbers, there is infinitely
many rational numbers.
Problem 2.2. Let a and b be any real numbers. Prove the following assertions.
p
(1) 0  |a| = a2 .
(2) a  |a|.
(3) |ab| = |a| · |b|.
(4) If " > 0, then
(5) If " > 0, then
|a
|a|  "
if and only if
b| < "
if and only if
"  a  ".
a
" < b < a + ".
(6) |a + b|  |a| + |b|.
(7) |a| |b|  |a b|.
Problem 2.3. Prove that:
if x < y, then x < x+y
2 .
Problem 2.4. Prove that:
p
If x 0 and y 0, then xy 
p
p 2
[Hint: note that ( x
y)
0]
x+y
2 .
Problem 2.5. Prove that:
p
p
If 0 < a < b, then 0 < a2 < b2 and 0 < a < b.
Problem 2.6. Prove that:
If x, y, a, and b are greater than zero and
x
y
< ab , then
x
y
<
x+a
y+b
< ab .
Problem 2.7. Let A = {r : r is a rational number and r2 < 2 }. Prove that A has
no largest member.
r2
[Hint: If r2 < 2, and r > 0, pick a positive rational number < min 1, 22r+1
.
Prove that (r + )2 < 2. ]
Problem 2.8. Suppose that S ✓ R and x = sup S . Prove that:
for every " > 0, there is an a 2 S such that x " < a  x.
Problem 2.9. Suppose that S ✓ R and y = inf S Prove that:
for every " > 0, there is an a 2 S such that y  a < y + ".
Problem 2.10. Suppose that ; =
6 A ✓ B ✓ R and the set B is bounded from
above. Prove that then
sup(A)  sup(B)
Problem 2.11. Suppose that A, B ✓ R are non-empty and bounded from above.
Show that then A [ B is bounded from above (and non-empty) and
sup(A [ B) = max(sup(A), sup(B))
24
ANDRZEJ ROSLANOWSKI
3. Sequences and Convergence
Definition 3.1.
(1) A sequence is a function whose domain is the set J of
positive integers.
(2) A real sequence (or: a number sequence) is a sequence whose terms are real
numbers.
Sequences are denoted using notation of the form
1
n=1
an
Thus
⇣ 1 ⌘1
1 1 1 1
= , , , ,...
2n n=1
2 4 8 16
⇣
⌘1
( 1)n
=
1, 1, 1, 1, 1, 1, 1, . . . .
and
n=1
Definition 3.2. A real sequence an
1
n=1
8" > 0 9N 2 J
1
an n=1
converges to A 2 R if and only if
8n
N
|an
A| < " .
We say that the sequence
converges (or: is a convergent sequence) if it
converges to some A 2 R.
1
If the sequence an n=1 is not convergent, it is said to be divergent.
⇣q
⌘1
Example 3.3. The sequence
9 + n1
converges to 3.
n=1
Proof. We are going to show that
r
1
3 <" .
n
Suppose that "0 > 0 is arbitrary but fixed. Take a natural number N0 2 J such
that "10 < N0 . Then we will also have that N10 < "0 .
Now, suppose that n N0 is an arbitrary but fixed natural number. Then
q
q
r
9 + n1 3 ·
9 + n1 + 3
9+ 1 9
1
q
9+
3 =
= q n
=
n
9 + n1 + 3
9 + n1 + 3
8" > 0 9N 2 J
1
q
n 9+
1
n
+3

8n
N
9+
1
1

< "0 .
n
N0
Thus (as n
N0 was arbitrary) we have justified that
r
1
8n N0
9+
3 < "0
n
and consequently
r
1
9N 2 J 8n N
9+
3 < "0 .
n
Since "0 > 0 was arbitrary, we may now conclude that
r
1
9+
3 <" ,
8" > 0 9N 2 J 8n N
n
as desired.
⇤
NOTES ON INTRODUCTION TO ANALYSIS
Example 3.4. The sequence ( 1)n · n
1
n=1
does not converge.
Proof. Suppose towards contradiction that ( 1)n · n
A 2 R. Then
( ) 8" > 0 9N 2 J
⇤
8n
N
25
( 1)n · n
1
n=1
converges to a number
A <" .
Consider " = 1. Applying ( ) to this particular "⇤ we may choose an N ⇤ such
that
( ) 8n
N⇤
⇤
( 1)n · n
A <1 .
N ⇤ but
Let n0 = N + b|A|c + 2. Then n0
( 1)n0 · n0
A
2,
⇤
contradicting ( ).
Definition 3.5. A set U ✓ R is a neighbourhood of a real number x if and only
if U contains an interval of positive length centered at x — that is, if and only if
there is " > 0 such that (x ", x + ") ✓ U .
Example 3.6.
(1) If " > 0, then (x ", x + ") is a neighbourhood of each of
its members.
(2) If a < x < b, then (a, b) is a neighbourhood of x.
(3) If x 2 U ✓ V ✓ R and U is a neighbourhood of x, then V is a neighbourhood
of x.
(4) Q is not a neighbourhood of any point.
1
Lemma 3.7. A real sequence an n=1 converges to A 2 R if and only if each
neighbourhood of A contains all but a finite number of terms of the sequence.
Assume an
1
n=1
( ) 8" > 0 9N 2 J
8n
Proof. ())
converges to A 2 R, that is we assume that
N
|an
A| < " .
Suppose that U is a neighbourhood of A. Take "0 > 0 such that (A "0 , A+"0 ) ✓ U .
Applying ( ) to this particular "0 we may choose N0 2 J such that
N0 |an
(8n
A| < "0
Since the inequality |an A| < "0 is equivalent to A
Theorem 2.15(5)), we conclude that
(8n
N0 an 2 (A
"0 < an < A + "0 (remember
"0 , A + "0 )
and hence
(8n
1
an n=1
N0 an 2 U .
Thus all the terms of
that do not belong to U are among the first N0 1
terms of this sequence. In particular, U contains all but a finite number of terms
1
of the sequence an n=1 .
(() Suppose that each neighbourhood of A contains all but a finite number of
1
terms of the sequence an n=1 . We want to argue that then
( ) 8" > 0 9N 2 J
8n
N
|an
A| < " .
So suppose "0 > 0 is arbitrary but fixed. Consider the interval U = (A "0 , A + "0 ).
It is a neighbourhood of A, so (by our assumptions) it contains all but finitely many
26
ANDRZEJ ROSLANOWSKI
1
terms of the sequence an n=1 . Thus the set I = {n 2 J : an 2
/ U } [ {1} is finite.
Let N0 = max(I) + 1. Now, if n N0 then n 2
/ I, so an 2 U . Hence
8n
N0 an 2 (A
" 0 , A + "0 )
and (remembering Theorem 2.15(5))
8n
N0 |an
A| < "0 .
⇤
Now we easily see that ( ) holds indeed true.
Theorem 3.8. If
1
an n=1
converges to A and also to B, then A = B.
1
Proof. Assume towards contradiction that an n=1 converges to A and also to B,
but A 6= B. Without loss of generality, A < B. Let " = B 2 A and set
⇣ 3A B A + B ⌘
U0 = (A ", A + ") =
,
2
2
and
⇣ A + B 3B A ⌘
U1 = (B ", B + ") =
,
.
2
2
Then U0 is a neighbourhood of A, so it contains all but a finite number of terms of
1
the sequence an n=1 . Also, U1 is a neighbourhood of B, so it contains all but a
1
finite number of terms of the sequence an n=1 . Hence for all but a finite number
of terms we have an 2 U0 and an 2 U1 . But U0 \ U1 = ;, a contradiction.
⇤
Notation 3.9. In view of Theorem 3.8 we may use the notation
lim an = A
n!1
to signify that an
Definition 3.10.
(2) A sequence
1
n=1
converges to A.
(1) A sequence an
1
an n=1
9M
8n 2 J
1
n=1
is bounded from above if and only if
an  M .
is bounded from below if and only if
9m 8n 2 J
1
an n=1
m  an .
(3) A sequence
is bounded if and only if it is bounded from above and
bounded from below.
1
Proposition 3.11. Let an n=1 be a real sequence. The following conditions are
equivalent.
1
(1) The sequence an n=1 is bounded.
(2) 9m 9M 8n 2 J m  an  M .
(3) 9M > 0 8n 2 J |an |  M .
⇤
Proof. Left for Exercises.
Theorem 3.12. Every convergent sequence is bounded.
Proof. Suppose that lim an = A, that is
n!1
(~)
8" > 0 9N 2 J
8n
N
an
A <" .
NOTES ON INTRODUCTION TO ANALYSIS
In order to show that the sequence an
8n 2 J
1
n=1
27
is bounded we will find m, M such that
m  an  M
(cf Proposition 3.11(2)). So, apply (~) for " = 1 to find N 2 J such that
8n
N
an
A <1 .
Set M = max{a1 , . . . , aN , A + 1} and m = min{a1 , . . . , aN , A
of N we know that for each n N we have
mA
1}. By the choice
1 < an < A + 1  M.
On the other hand, for n = 1, . . . , N we have m  an  M by the choice of m, M .
Together,
m  an  M
for all n 2 J.
⇤
Example 3.13. Let
1 1
1
+ + ... +
for n 2 J.
2 3
n
is unbounded, and hence it does not converge.
an = 1 +
Then the sequence an
Proof. Note that for n
1
n=1
3 we have
a 2n =
⇣
1 ⇣1 1⌘ ⇣1 1 1 1⌘
1
1⌘
+
+
+
+ + +
+ ... + n 1
+ ... + n >
2
3 4
5 6 7 8
2
+1
2
⇣1
1 ⇣1 1⌘ ⇣1 1 1 1⌘
1⌘
1+ +
+
+
+ + +
+ ... + n + ... + n =
2
4 4
8 8 8 8
2
2
1+
1
1
1
1
n
+ 2 · + 4 · + . . . + 2n 1 · n = 1 + .
2
4
8
2
2
2|M 1|
Consequently, if M 2 J, taking k > 2
will give us ak > a22|M
2|M 1|
=
1
+
|M
1|
M
.
2
1+
1|
> 1+
⇤
28
ANDRZEJ ROSLANOWSKI
Exercises.
1
Problem 3.1. Let an n=1 be a real sequence. Show that the following conditions
are equivalent.
1
(1) The sequence an n=1 is bounded.
(2) 9m 9M 8n 2 J m  an  M .
(3) 9M > 0 8n 2 J |an |  M .
Problem 3.2. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣
1 ⌘1
5+
n n=1
converges.
Problem 3.3. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣ 2 2n ⌘1
n
n=1
converges.
Problem 3.4. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣
⌘1
2 n
n=1
converges.
Problem 3.5. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣ 9n + 1 ⌘1
n + 3 n=1
converges to 9.
Problem 3.6. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣ n2 + n + 1 ⌘1
n2 + 3
n=1
converges to 1.
Problem 3.7. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣ p n 1 ⌘1
p
9n + 1 n=1
converges to 1/3.
NOTES ON INTRODUCTION TO ANALYSIS
29
Problem 3.8. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
s
⇣ 16pn + 25 ⌘1
p
n
n=1
converges to 4.
Problem 3.9. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣p
p ⌘1
n+1
n
n=1
converges to 0.
Problem 3.10. Directly from the definition of the limit, in a manner similar to
Example 3.3, show that the sequence
⇣
⌘1
ln(n + 1) ln(n)
n=1
converges to 0.
Problem 3.11. Let ↵ 2 R and N0 2 J. Suppose that a sequence (an )1
n=1 satisfy
an = ↵ for all n N0
Directly from the definition of the limit, show that the sequence (an )1
n=1 converges
to ↵
Problem 3.12. Directly from Definition 3.5 show that the set [0, 1] is a neighbourhood of 23 .
Problem 3.13. Show the statements given in Example 3.6.
Problem 3.14. Let x and y be distinct real numbers. Show that there are neighborhoods P of x and Q of y such that P \ Q = ;.
Problem 3.15. Show that the sequence
lower bounds for this sequence.
3n+7 1
n
n=1
is bounded and give upper and
Problem 3.16. Prove that a sequence (an )1
n=1 converges to A if and only if the
sequence (an A)1
n=1 converges to 0.
Problem 3.17. Suppose that a sequence (an )1
n=1 converges to A, and define a
an +an+1
new sequence (bn )1
by
b
=
for
all
n.
Prove that the sequence (bn )1
n
n=1
n=1
2
converges to A as well.
30
ANDRZEJ ROSLANOWSKI
1
1
Problem 3.18 (Sandwich Theorem). Suppose (an )1
n=1 , (bn )n=1 , and (cn )n=1 , are
sequences of real numbers such that
• (an )1
n=1 converges to A, and
• (bn )1
n=1 converges to A, and
• an  cn  bn for all n.
Prove that then the sequence (cn )1
n=1 converges to A as well.
Problem 3.19. Let (an )1
n=1 be a sequence of real numbers and A 2 R. Prove that
1
if (an )n=1 converges to A, then (|an |)1
n=1 colnverges to |A|.
Is the converse implication true? Justify.
NOTES ON INTRODUCTION TO ANALYSIS
31
4. Cauchy Sequences
Convergence is a property that involves both a sequence and its limit. Can we
check if a sequence converges without knowing its limit ? Surprisingly the answer
is: yes, in the real numbers, convergence is the property of the sequence alone, we
do not have to look at the space around it!
Definition 4.1. A sequence an
1
n=1
8" > 0 9N 2 J
is a Cauchy sequence if and only if
8m, n
|an
N
am | < " .
Theorem 4.2. Every convergent sequence is a Cauchy sequence.
Proof. Suppose that lim an = A, i.e.,
n!1
8" > 0 9N 2 J
(~)
1
an n=1
We want to argue that
8n
N
an
A <" .
is a Cauchy sequence, that is
(|)
8" > 0 9N 2 J 8m, n N |an am | < " .
To this end, suppose that "0 > 0 is arbitrary but fixed. Apply (~) to " = "0 /2 to
find N0 such that
8n N0 an A < "0 /2 .
Now, if m, n
|an
N0 , then |an
am | = |an
A| < "0 /2 and |A
am |  |an
A+A
am | < "0 /2. Hence
A| + |A
am | < "0 /2 + "0 /2 = "0 .
Now we easily conclude (|).
⇤
Theorem 4.3. Every Cauchy sequence is bounded.
Proof. Suppose that an
1
n=1
is a Cauchy sequence, that is
(|)
8" > 0 9N 2 J 8m, n N |an am | < " .
Applying (|) for " = 1 we may choose N0 2 J such that
|an
am | < 1
for all n, m
N0 .
Then, in particular,
(⇤) aN0 1 < an < aN0 + 1 for all n N0 .
Let M = max{a1 , . . . , aN0 } + 2 and m = min{a1 , . . . , aN0 }
(⇤⇤) m < an < M for all n  N0 .
It follows from (⇤) above that
(⇤ ⇤ ⇤) m < an < M for all n N0 .
Together, m < an < M for all n 2 J, so the sequence an
2. Clearly
1
n=1
is indeed bounded.
⇤
Definition 4.4. Let S ✓ R. A real number A is an accumulation point of S if and
only if every neighbourhood of A contains infinitely many points of S.
Let acc(S) denote the set of all accumulation points of S.
Lemma 4.5. Let S ✓ R, A 2 R. Then the following conditions are equivalent:
(a) every neighbourhood of A has a non-empty intersection with S \ {A},
(b) A is an accumulation point of S,
(c) there is a sequence a1 , a2 , a3 , . . . 2 S \ {A} such that lim an = A.
n!1
32
ANDRZEJ ROSLANOWSKI
Proof. (a) ) (b)
Assume (a). Let U be a neighbourhood of A. If the set U \ S
were finite, then we could set
s = min
|A
x| : x 2 U \ S \ {A}
> 0.
(Note that we know, by (a), that U \S\{A} =
6 ;, so also |A x| : x 2 U \S\{A} 6=
;.) Then (A s, A + s) \ S \ {A} = ;, contradicting the assumption (a).
(b) ) (c)
By the assumption (b) each of the sets
1
1
1
1
, A + ), (A
, A + ), (A
2
2
3
3
has infinite intersection with S. Therefore we may choose
(A
1, A + 1),
(A
a1 2 S \ (A
a2 2 S \ (A
a3 2 S \ (A
an 2 S \ (A
1, A + 1) \ {A}
1
1
, A + ) \ {A}
2
2
1
1
, A + ) \ {A}
3
3
..
.
1
1
, A + ) \ {A}
n
n
..
.
This way we choose a sequence an
A.
(c) ) (a)
1
n=1
1
1
, A + ), . . .
4
4
then |a1
then |a2
then |a3
then |an
A| < 1
1
A| <
2
1
A| <
3
A| <
1
n
of elements of S \{A} and plainly lim an =
n!1
Assume an 2 S \ {A} (for n 2 J) are such that lim an = A. Suppose
n!1
that U is a neighbourhood of A and let r > 0 be such that (A r, A + r) ✓ U .
Now, applying the definition of lim an = A (for " = r) we may find N0 2 J
n!1
such that for n N0 we have |an A| < r and thus an 2 (A r, A + r) ✓ U . Hence,
in particular, aN0 2 U \ S \ {A}, showing that the latter set is not empty.
⇤
Example 4.6.
(1) Let S = 1 + n1 : n 2 J . Then acc(S) = {1}.
(2) Let S = m + 2 n : m, n 2 J . Then acc(S) = J.
1
(3) Let S = n1 + m
: m, n 2 J . Then acc(S) = k1 : k 2 J [ {0}.
Proof. Left for Exercises.
⇤
Theorem 4.7. Every real number is an accumulation point of the set of rational numbers. Every real number is an accumulation point of the set of irrational
numbers.
Proof. Let x 2 R. Suppose that U is a neighbourhood of x and let " > 0 be such
that (x ", x + ") ✓ U . It follows from Theorems 2.10 and 2.13 that the interval
(x, x+") ✓ U \{x} contains both rational and irrational numbers. Now use Lemma
4.5.
⇤
Theorem 4.8 (1st Bolzano–Weierstrass Theorem). Every bounded infinite subset
of R has at least one accumulation point.
NOTES ON INTRODUCTION TO ANALYSIS
33
Proof. Suppose that S ✓ R is infinite and bounded. Let a, b 2 R be such that
a < x < b for all x 2 S. To show that S has an accumulation point we consider the
following procedure.
We define inductively ak , bk so that:
• a1 = a, b1 = b.
• Assume ak , bk has been chosen and the intersection [ak , bk ] \ S is infinite.
k
Let c = ak +b
. Consider the two intervals [ak , c] and [c, bk ]. One of them
2
contains infinitely many elements of the set [ak , bk ]\S so let ak+1 < bk+1 be
the endpoints of that one. Thus the intersection [ak+1 , bk+1 ] \ S is infinite.
Clearly this construction can be carried out. Note that for each k 2 J we have
• ak  ak+1 < bk+1  bk and
• bk ak = 2bk a1 and
• [ak , bk ] \ S is infinite.
Let T = {an : n 2 J}. The set T is nonempty and bounded from above, so it has
a least upper bound. Let t = sup(T ).
Claim 4.8.1. t 2 acc(S).
Proof of the Claim. Suppose that U is a neighbourhood of t and let " > 0 be such
that (t ", t + ") ✓ U . Since t " < t we know that t " is not an upper bound for
T . Therefore there is n 2 J such that t " < an . Take m > n such that 2bm a1 < ".
Then
b a
t " < an  am < bm  am + m 1 < am + "  t + ".
2
The interval [am , bm ] has infinite intersection with S, so also
(t
", t + ") \ S
must be infinite and hence U \ S is infinite.
⇤
⇤
1
Lemma 4.9. Assume an n=1 is a Cauchy sequence. If for some A the set {m 2
J : am = A} is infinite, then
lim an = A.
n!1
Proof. Let
(~)
1
an n=1
and A be as in the assumptions. We are going to show that
8" > 0 9N 2 J
8n
N
Let "0 > 0 be arbitrary but fixed. Since
such that
(|)0
8m, n
N0 |an
an
A <" .
1
an n=1
is Cauchy, we may choose N0 2 J
a m | < "0 .
Since the set {m 2 J : am = A} is infinite, we may find m0 > N0 such that
am0 = A. It follows from (|)0 that then for all n N0 we have
|an
A| = |an
a m 0 | < "0 .
Now we easily conclude (~).
Theorem 4.10. Every Cauchy sequence is convergent.
⇤
34
ANDRZEJ ROSLANOWSKI
1
Proof. Let an n=1 is a Cauchy sequence.
If for some A the set {m 2 J : am = A} is infinite, then by Lemma 4.9 we know
that
lim an = A.
n!1
So suppose that every value is repeated in an
def
1
n=1
finitely many times only. Then
the set S = {an : n 2 J} is infinite. It is also bounded (by Theorem 4.3). Hence,
by 1st Bolzano–Weierstrass Theorem 4.8, the set S has an accumulation point. Let
a 2 acc(S).
Claim 4.10.1. lim an = a.
n!1
Proof of the Claim. We will argue that
(~)
8" > 0 9N 2 J 8n N
an
a <" .
1
an n=1
Let "0 > 0 be arbitrary but fixed. Since
is a Cauchy sequence, we may
choose N0 2 J such that
(|)0
8m, n N0 |an am | < "20 .
Since a 2 acc(S), the set (a "20 , a + "20 ) \ S is infinite and therefore we may choose
m0 > N0 such that am0 2 (a "20 , a + "20 ). Now, for any n N0 we have
"0
"0
|a an | = |a am0 + am0 an |  |a am0 | + |am0 an | <
+
= "0 .
2
2
Now (~) follows.
⇤
⇤
NOTES ON INTRODUCTION TO ANALYSIS
35
Exercises.
Problem 4.1. Using Definition 4.1 only, show that the sequence
Cauchy.
Problem 4.2. Using Definition 4.1 only, show that the sequence
Cauchy.
1
2
1+n n=1
is
2n+1 1
n
n=1
is
Problem 4.3. Using Definition 4.1 only, show that the sequence
s
!1
1
9+ p
n
n=1
is Cauchy.
1
Problem 4.4. Suppose that (an )1
n=1 and (bn )n=1 are both Cauchy sequences.
1
Using Definition 4.1 only, show that the sequence an + bn n=1 is Cauchy.
1
Problem 4.5. Suppose that (an )1
n=1 and (bn )n=1 are both Cauchy sequences.
1
Using Definition 4.1 only, show that the sequence an · bn n=1 is Cauchy.
Problem 4.6. Prove the following claims.
(1) Let S = 1 + n1 : n 2 J . Then acc(S) = {1}.
(2) Let S = m + 2 n : m, n 2 J . Then acc(S) = J.
1
(3) Let S = n1 + m
: m, n 2 J . Then acc(S) = k1 : k 2 J [ {0}.
Problem 4.7. Give an example of a set S ✓ R such that acc(S) has exactly 2
elements. Justify your answer.
Problem 4.8. Determine the accumulation points of the set
1
{2n + : n, k 2 J}.
k
Problem 4.9. Does there exists a set S ✓ R such that acc(S) = Q ? Justify your
answer.
Problem 4.10. Let ; 6= S ✓ R. Suppose that the set S is bounded from above
(below, respectively) and let x = sup(S) (x = inf(S), respectively). Prove that
either x 2 S, or else x 2 acc(S).
Problem 4.11. Suppose that a sequence (an )1
n=1 converges to a number A and
the set of its term {an : n 2 J} is infinite. Show that A 2 acc({an : n 2 J}).
36
ANDRZEJ ROSLANOWSKI
5. Arithmetic Operations on Sequences
1
Theorem 5.1. If a sequence an n=1 converges to A, and a sequence bn
1
converges to B, then the sequence an + bn n=1 converges to A + B.
1
n=1
Proof. We assume that a sequence an
(⌦)A
8" > 0 9N 2 J
8n
N
8" > 0 9N 2 J
8n
N
8n
NB
and we assume that a sequence
(⌦)B
|an
1
bn n=1
1
n=1
converges to A, that is
A| < " ,
converges to B, that is
|bn
B| < " .
1
We want to show that the sequence an + bn n=1 converges to A + B and for this
we have to argue that
(⌦)A+B 8" > 0 9N 2 J 8n N |(an + bn ) (A + B)| < " .
So suppose that "0 > 0 is arbitrary but fixed. Apply (⌦)A for " = "0 /2 to choose
NA 2 J such that
8n NA |an A| < "0 /2 .
Then use (⌦)B to choose NB 2 J such that
|bn
B| < "0 /2 .
Let N0 = max{NA , NB }.
Suppose n N0 . By the choice of N0 , NA , NB we know that
|an
Now,
|(an + bn )
A| < "0 /2
(A + B)| = |(an
|bn
and
B| < "0 /2.
B)|  |an
A) + (bn
A| + |bn B| <
"0 /2 + "0 /2 = "0 .
Thus we have shown that
8n
N0 |(an + bn )
(A + B)| < "0
⇤
and hence (⌦)A+B readily follows.
1
an n=1
Theorem 5.2. If a sequence
converges to A, and a sequence
1
converges to B, then the sequence an · bn n=1 converges to A · B.
Proof. Let us assume that a sequence an
(⌦)A
8" > 0 9N 2 J
8n
N
8" > 0 9N 2 J
8n
N
converges to A, that is
|an
A| < " ,
|bn
B| < " .
and let us also assume that a sequence bn
(⌦)B
1
n=1
1
n=1
1
bn n=1
converges to B, that is
It follows from Theorem 3.12 that the sequence an
sition 3.11 we may choose M > 0 such that
( ) 8n 2 J |an |  M .
1
1
n=1
is bounded, so by Propo-
Our goal is to show that the sequence an · bn n=1 converges to A · B and for
this we have to argue that
(⌦)A·B 8" > 0 9N 2 J 8n N |(an · bn ) (A · B)| < " .
"0
To this end suppose that "0 > 0 is arbitrary but fixed. Clearly |B|+M
> 0, so using
"0
(⌦)A and (⌦)B for " = |B|+M we may choose NA , NB 2 J such that
( )A |an
A| <
"0
|B|+M
for all n
NA , and
NOTES ON INTRODUCTION TO ANALYSIS
( )B |bn
B| <
"0
|B|+M
for all n
37
NB .
Let N0 = max{NA , NB }. Suppose n N0 . Then both ( )A and ( )B apply to n
so we have
"0
"0
( ) |an A| < |B|+M
and |bn B| < |B|+M
.
Now,
|an | · |bn
|(an · bn ) (A · B)| = |(an bn an B) + (an B AB)| 
B| + |B| · |an A|  M · |bn B| + |B| · |an A| <
"0
"0
M · |B|+M
+ |B| · |B|+M
= "0 .
Thus we have shown that
8n
N0 |(an · bn )
(A · B)| < "0
and hence (⌦)A·B readily follows.
2 R. If lim an = A and lim bn = B, then the sequence
Corollary 5.3. Let ↵,
↵ · an +
·
1
bn n=1
⇤
n!1
n!1
converges to ↵A + B.
1
Lemma 5.4. If a sequence bn n=1 converges to B and B 6= 0, then there is a
positive real number M and a positive integer N such that, if n N , then |bn | M .
Proof. We know that B 6= 0 and
(⌦)B 8" > 0 9N 2 J 8n N
Let M =
|B|
2
|bn
B| < " .
|bn
B| < |B|/2 .
> 0. Applying (⌦)B to " = |B|/2 > 0 we may find N such that
Then for all n
8n
N
N we have
|B| = |bn + B
and hence
|bn |
|B|
|bn
bn |  |bn | + |B
|B|
B|
|B|/2 =
bn |
|B|
= M.
2
1
an n=1
Theorem 5.5. If a sequence
converges to A, and a sequence
converges to B with B 6= 0 and bn 6= 0 for all n, then the sequence
A
converges to B
.
Proof. We assume that a sequence an
(⌦)A
8" > 0 9N 2 J
8n
N
and we also assume that a sequence
1
n=1
|an
⇤
1
bn n=1
an 1
bn n=1
converges to A, that is
A| < " ,
1
bn n=1
converges to B, that is
(⌦)B 8" > 0 9N 2 J 8n N |bn B| < " .
It follows from Lemma 5.4 that there are M > 0 and N1 2 J such that |bn |
for all n N1 .
We are going to argue that
A
(⌦)A/B 8" > 0 9N 2 J 8n N | abnn
B| < " .
M
To this end suppose that "0 > 0 is arbitrary but fixed. Apply (⌦)A and (⌦)B to
M "0
" = 1+|A/B|
> 0 to choose NA and NB , respectively, so that
( )A |an
A| <
M "0
1+|A/B|
for all n
NA , and
38
ANDRZEJ ROSLANOWSKI
( )B |bn
B| <
M "0
1+|A/B|
for all n
NB .
Let N0 = max{N1 , NA , NB }. Suppose n N0 . Then both ( )A and ( )B apply
to n so we have
M "0
M "0
( ) |an A| < 1+|A/B|
and |bn B| < 1+|A/B|
, and also |bn |
M (by the
choice of N1 ).
Now,
an
A
a n B bn A
an B AB + AB bn A
|
|=|
|=|
|
bn
B
bn B
bn B
|
an
A
bn
|+
|A| · |bn B|
M "0
1
M "0
|A|
<
·
+
·
=
|bn | · |B|
1 + |A/B| |bn | 1 + |A/B| |bn | · |B|
⇣ 1
M "0
|A| ⌘
M "0
1 ⇣
|A| ⌘
·
+

·
· 1+
= "0 .
1 + |A/B|
|bn | |bn | · |B|
1 + |A/B| M
|B|
Thus we have shown that
an
A
8n N0 |
| < "0
bn
B
and hence (⌦)A/B readily follows.
⇤
Theorem 5.6. If lim an = A and lim bn = B with an  bn for all n 2 J, then
n!1
n!1
A  B.
Proof. Let us assume that an  bn for all n 2 J and that the sequence an
1
converges to A, and the sequence bn n=1 converges to B. Thus
(⌦)A
(⌦)B
8" > 0 9N 2 J
8" > 0 9N 2 J
8n
8n
N
N
|an
|bn
1
n=1
A| < " ,
B| < " .
Suppose towards contradiction that B < A. Then " = A 2 B > 0 and we may use
(⌦)A , (⌦)B to choose NA , NB so that
( )A |an A| < A 2 B for all n NA , and
( )B |bn B| < A 2 B for all n NB .
Let n = NA + NB . Then both ( )A and ( )B apply to n, so
( ) A A 2 B < an < A + A 2 B and B A 2 B < bn < B + A 2 B .
But
A B
A+B
A B
B+
=
=A
,
2
2
2
so ( ) implies that
A+B
bn <
< an ,
2
a contradiction.
⇤
Theorem 5.7. If a sequence an
bounded, then lim an bn = 0.
1
n=1
converges to 0 and a sequence bn
1
n=1
is
1
n=1
is
n!1
1
Proof. Let us assume that an n=1 converges to 0, and the sequence bn
bounded. Thus
(⌦)A 8" > 0 9N 2 J 8n N |an | < " ,
and we may find M > 0 such that
NOTES ON INTRODUCTION TO ANALYSIS
39
( ) |bn |  M for all n 2 J.
We want to argue that
(⌦) 8" > 0 9N 2 J 8n N |an bn | < " .
To this end suppose that "0 > 0 is arbitrary but fixed. Apply (⌦)A for " = "0 /M > 0
to find N 2 J such that
(⇤) |an | < "0 /M for all n N .
Then for every n N we will have
"0
|an bn | = |an | · |bn |  |an | · M <
· M = "0 .
M
Now (⌦) readily follows.
⇤
40
ANDRZEJ ROSLANOWSKI
Exercises.
1
Problem 5.1. Suppose (an )1
n=1 and (bn )n=1 are real sequences such that both
1
1
(an )n=1 and (an + bn )n=1 converge. Does (bn )1
n=1 converge? Justify your answer
1
Problem 5.2. Suppose (an )1
n=1 and (bn )n=1 are real sequences such that neither
1
1
(an )n=1 nor (bn )n=1 converge. Can (an + bn )1
n=1 converge? Justify your answer.
1
1
Problem 5.3. Suppose (an )1
n=1 and (bn )n=1 are real sequences such that (an )n=1
1
1
converges to A 6= 0 and (an · bn )n=1 converges. Does (bn )n=1 converge? Justify
your answer.
Problem 5.4. Assume that an
0 for all n 2 J and that the sequence (an )1
n=1
p
p
1
converges to a. Prove that
an n=1 converges to a.
p
p
a
[Hint: Consider two cases with a = 0 and a > 0. In the latter an
a = paann +p
.]
a
n
Problem 5.5. For natural numbers m  n, the binomial coefficient m
is defined
by
✓ ◆
n
n!
(n m + 1) · (n m + 2) · . . . · n
=
=
.
m
m! · (n m)!
1 · 2 · ... · m
Let k 2 J. Prove that the sequence
!1
n+k
k
(n + k)k
n=1
converges to 1/k!.
Problem 5.6. Let (an )1
n=1 be a real sequence and, for each n, define
def a1 + a2 + . . . + an
↵n =
.
n
(a) Prove that if lim an = A, then also lim ↵n = A.
n!1
n!1
(b) Does the converse hold true? Justify.
Problem 5.7. Find the limits of the sequences with the following general terms:
n2 + 4n
(a)
n2 5
cos(n)
(b)
n
sin n2
p
(c)
n
n
(d) 2
n
3
⇣r
⌘
1
(e) n ·
4
2
n
p
n
(f) ( 1)n ·
n+7
NOTES ON INTRODUCTION TO ANALYSIS
41
6. Subsequences and Monotone Sequences
1
n=1
Definition 6.1. Let an
integers such that
⇣
The sequence ank
⌘1
k=1
be a sequence and nk
1
k=1
be any sequence of positive
n1 < n2 < n3 < n4 < . . . .
is called as a subsequence of an
1
.
n=1
Example 6.2.
(1) Letting an = n1 and nk = 2k we see that the sequence
1
1
1 1
k 1
2
is a subsequence of 1/n n=1 . Similarly, both 2k
and k12 k=1
k=1
k=1
1
are subsequences of 1/n n=1 .
1
1
(2) The sequence 4 ` `=1 is a subsequence of 2 k k=1 (just consider k` = 2`).
1
So it is a subsequence of a subsequence of 1/n n=1 .
1
1
(3) In general, if an n=1 is a sequence, nk k=1 is a strictly increasing se1
quence of positive integers and k` `=1 is an increasing sequence of positive
⇣
⌘1
⇣
⌘1
integers, then ank`
is a subsequence of the subsequence ank
of
1
`=1
k=1
the sequence an n=1 .
(4) Let an = ( 1)n . Letting nk = 2k we see that the constant sequence
1
1
1 k=1 is a subsequence of ( 1)n n=1 . Similarly, the contant the constant
1
1
sequence
1 k=1 is a subsequence of ( 1)n n=1 . Both constant sub1
sequences converge, but the limits are distinct. The sequence ( 1)n n=1
does not converge.
Theorem 6.3. A sequence converges if and only if each of its subsequences converges. In fact, if every subsequence converges, they all converge to the same limit.
1
Proof. ()) Suppose that a sequence an n=1 converges to A and ank
1
subsequence. We are going to show that ank k=1 converges to A, i.e.,
(~) 8" > 0 9K 2 J
8k
K |ank
1
k=1
is its
A| < " .
So suppose that "0 > 0 is arbitrary but fixed. Since lim an = A, we may find
n!1
N0 2 J such that
(|) |an
A| < "0 for all n
N0 .
Let K0 = N0 . Since n1 < n2 < . . . < nk < . . . ..., we know that for k
nk
Consequently, for each k
K0 we have
K0 = N 0 .
K0 we have
|ank
(remember (|)). Thus (9K 2 J)(8k
arbitrary the demand (~) follows.
A| < "0
K)(|ank
1
A| < "0 ), and since "0 > 0 was
1
(() Suppose that each subsequences of an n=1 converges. In particular, an n=1
(which is its own subsequence) converges. Applying the argument from above we
also see that in this case all subsequences converge to the same limit.
⇤
Theorem 6.4 (2nd Bolzano–Weierstrass Theorem). Every bounded sequence has
a convergent subsequence.
42
ANDRZEJ ROSLANOWSKI
Proof. Suppose that a sequence an
plementary cases.
1
n=1
is bounded and let us consider two com-
Case 1: For some A the set {n 2 J : an = A} is infinite.
Then we may choose n1 < n2 < n3 < n4 < . . . such that ank = A (for each k 2 J).
1
1
Clearly ank k=1 is a convergent subsequence of an n=1 .
Case 2: Not Case 1, i.e., no value is repeated infinitely often in the sequence
1
an n=1 .
def
Then the set S = {an : n 2 J} is infinite and bounded. Hence, by 1st Bolzano–
Weierstrass Theorem 4.8, S has an accumulation point. Let a 2 acc(S). We will
1
construct a subsequence of an n=1 converging to a. For this we choose inductively
n1 < n2 < n3 < . . . so that |a ank | < 1/k for each k.
Suppose we have picked n1 < . . . < nk . Since a is an accumulation point of S
1
1
and the interval (a k+1
, a + k+1
) is a neighbourhood of a, we know that
1
1
,a +
) \ S is infinite.
k+1
k+1
Therefore we may find nk+1 > nk such that
1
1
a
< ank+1 < a +
.
k+1
k+1
(a
Since |a
to a.
ank | < 1/k one easily shows that the subsequence ank
1
k=1
converges
⇤
1
Theorem 6.5. Suppose xn n=1 is a bounded sequence. If all its convergent subsequences have the same limit, then the sequence is convergent.
1
Proof. Assume that xn n=1 is a bounded sequence and all its convergent subsequences have the same limit.
1
1
By Theorem 6.4 the sequence xn n=1 has a convergent subsequence xnm m=1 .
Let A = lim xnm . We will argue that A = lim xn , i.e.,
m!1
n!1
(~)
8" > 0 9N 2 J 8n N xn A < " .
Suppose towards contradiction that (~) is not true. Then its negation
¬(~)
9" > 0 8N 2 J 9n N xn A
"
holds true. So let "0 > 0 be such that
() for every N 2 J there is n N such that xn A
"0 .
Now we pick inductively n1 < n2 < n3 < . . . so that
x nk
A
"0 .
Suppose n1 , . . . , nk has been determined. Apply () to N = nk + 1 to find nk+1 >
nk such that xnk+1 A
"0 .
1
After the above procedure is completed we have a (bounded) sequence xnk k=1 .
1
1
By Theorem 6.4 the sequence xnk k=1 has a convergent subsequence xnk` `=1 .
1
It is also a subsequence of xn n=1 , so by the assumptions of the theorem we have
lim xnk` = A.
`!1
However, 0 < "0  xnk`
A for each `, a contradiction.
⇤
NOTES ON INTRODUCTION TO ANALYSIS
Definition 6.6. A sequence an
A sequence
1
an n=1
1
n=1
43
is increasing if
8n 2 J
an  an+1 .
is decreasing if
8n 2 J
an
an+1 .
A sequence is monotone if it is either increasing or decreasing.
Remark 6.7. If a sequence is both increasing and decreasing, then it is constant.
Theorem 6.8. A monotone sequence is convergent if and only if it is bounded.
Proof. ())
If a sequence converges, then it is bounded (by Theorem 3.12).
1
1
(() Suppose an n=1 is bounded and monotone; say an n=1 is increasing. The
set {an : n 2 J} is bounded (and nonempty), so it has a least upper bound. Let
s = sup({an : n 2 J}).
We are going to show that s = lim an , i.e.,
n!1
(~)
8" > 0 9N 2 J 8n N an s < " .
Suppose "0 > 0 is arbitrary but fixed.
Since s "0 < s, we know that s "0 is not an upper bound for {an : n 2 J}.
Therefore there must be an N 2 J such that s "0 < aN . Then for every n N
we have
s "0 < a N  a n  s < s + " 0 .
Now (~) follows.
⇤
1
Example 6.9. Consider the sequence sn n=1 defined as follows:
q
p
p
s1 = 2, sn = 2 + sn 1 for n 2.
Then sn
1
n=1
converges and its limit L satisfies the equation
L4
Proof. We will show that
1
sn n=1
4L2
L + 4 = 0.
is bounded and increasing.
Claim 6.9.1. 0  sn < 2 for all n 2 J.
Proof
of the Claim. We show this by induction on n 2 J. First, clearly 0  s1 =
p
2 < 2. Now, for the inductive step, suppose that 0  sn < 2. Then
q
p
p
0  sn+1 = 2 + sn < 2 + 2 = 2.
⇤
Claim 6.9.2. sn  sn+1 for all n 2 J.
Proof. We show this by induction on n 2 J. First,
q
p
p
s1 = 2 < 2 + s1 = s2 .
Now, for the inductive step, suppose that sn  sn+1 . Then
q
q
p
p
sn+2 = 2 + sn+1
2 + sn = sn+1 .
⇤
44
ANDRZEJ ROSLANOWSKI
Now, by Theorem 6.8 we know that the sequence
lim sn = L. Since for all n 2 J we have
n!1
q
p
sn+1 = 2 + sn
sn
1
n=1
converges.
we may use Theorem 6.3 and Problem 5.4 to claim that
q
p
L = 2 + L.
Hence L4
4L2
⇤
L + 4 = 0.
Example 6.10. Let 0 < b < 1. The sequence
1
bn n=1
Let
converges to 0.
Proof. First note that 0 < bn 1 (1 b) = bn 1 bn , and therefore the sequence
1
bn n=1 is decreasing. It should be also clear that
0 < bn < 1
for all n 2 J.
Therefore, by Theorem 6.8, the sequence bn
Theorems 6.3 and 5.2 we must have
1
n=1
converges. Let L = lim bn . By
n!1
L = lim b2n = lim bn · bn = L · L.
n!1
n!1
Consequently, either L = 0 or L = 1. The latter is clearly impossible, so lim bn =
n!1
0.
⇤
p
1
Example 6.11. Let 0 < c < 1. The sequence n c n=1 converges to 1.
Proof. Note that
p
p
1
n+1
n
c
c = c n+1
1
1
1
1
1
1
c n = c n+1 · 1 c n n+1 = c n+1 · 1 c n(n+1) > 0.
p
p
1
Therefore the sequence n c n=1 is increasing. It should be clear that 0 < n c < 1,
p
so our sequence is bounded and therefore it converges. Let L = lim n c. By
n!1
Theorem 6.3 and Problem 5.4 we must have
q
p
p
p
n
L = lim 2n c = lim
c = L.
n!1
n!1
p
Consequently, either L = 1 or L = 0. The latter is clearly impossible, so lim n c =
n!1
1.
⇤
NOTES ON INTRODUCTION TO ANALYSIS
45
Exercises.
Problem 6.1. Let an = ( 1)n 1
of the sequence (an )1
n=1 .
1
n
for n 2 J. Find a convergent subsequence
Problem 6.2. Let (an )1
n=1 be a real sequence and S = {an : n 2 J}. Suppose
x 2 acc(S). Prove that there is a subsequence of (an )1
n=1 that converges to x.
Problem 6.3. Assume that a sequence (an )1
n=1 is decreasing and bounded. Prove
that it converges. (This is the unproven part of Theorem 6.8.)
Problem 6.4. Assume c > 1. Show that the sequence
p
n
c
1
n=1
converges to 1.
1
Problem 6.5. Asssume that real sequences (xn )1
n=1 and (yn )n=1 both converge to
1
x0 . Define a sequence (zn )n=1 by
⇢
xk if n = 2k
zn =
yk if n = 2k 1
Prove that lim zn = x0 .
n!1
Problem 6.6. Define recursively a sequence (an )1
n=1 by
p
a1 = 6,
an = 6 + an 1 for n > 1.
Prove that (an )1
n=1 converges and find its limit.
Problem 6.7. Let (xn )1
n=1 be a bounded sequence and let
1
E = y 2 R : there is a subsequence (xnk )1
k=1 of (xn )n=1 converging to y .
We know from Theorem 6.4 that the set E is nonempty. Prove that E is bounded
and contains both sup(E) and inf(E).
Problem 6.8. Let T : J ! J be a one-to-one function. Assume that (xn )1
n=1
converges to A. Show that then the sequence (xT (n) )1
n=1 converges to A as well.
How does this relate to subsequences?
Problem 6.9. Assume 0  a  b. Prove that the sequence
⇣
⌘
1/n 1
a n + bn
n=1
converges and find the limit.
Problem 6.10. For k 2 J let ak =
k
P
n=1
p 1
.
k2 +n
or converge? If it converges, what is the limit?
Does the sequence (ak )1
k=1 diverge
46
ANDRZEJ ROSLANOWSKI
Problem 6.11. Show that every real number is the limit of a sequence of rational
numbers.
Problem 6.12. Show that every real number is the limit of a sequence of irrational
numbers.
Problem 6.13. Suppose that lim an = A and that B 2 acc {an : n 2 J} . Show
n!1
that A = B.
Problem 6.14. Prove that the sequence
⇣ p ⌘1
n
n
n=1
converges and find its limit.
NOTES ON INTRODUCTION TO ANALYSIS
47
7. Limits of Functions
Remember Definition 4.4: A number x0 is an accumulation point of a set S ✓ R
if and only if every neighbourhood of x0 contains infinitely many elements of the
set S. By Proposition 4.5, x0 is an accumulation point of S if and only if
8" > 0 (x0
", x0 + ") \ S \ {x0 } =
6 ; .
Definition 7.1. Suppose that f : D ! R and x0 is an accumulation point of D.
We say that f has a limit L at x0 if and only if
(8" > 0)(9 > 0)(8x 2 D)(0 < |x
x0 | <
) |f (x)
L| < ").
Remark 7.2. Definition 7.1 formalizes both the notion of two-sided and that of
one-sided limit: as long as x0 remains an accumulation point of D, we may restrict
the domain of the function under consideration.
Proposition 7.3. If f : D ! R, x0 2 R is an accumulation point of D and
(a) f has a limit A at x0 , and
(b) f has a limit B at x0 ,
then A = B.
Proof. So let us assume that x0 2 acc(D) and
(⇤)A (8" > 0)(9 > 0)(8x 2 D)(0 < |x x0 | < ) |f (x) A| < "), and
(⇤)B (8" > 0)(9 > 0)(8x 2 D)(0 < |x x0 | < ) |f (x) B| < ").
Suppose towards contradiction that A 6= B, say A < B. Let "0 = B 2 A > 0. Use
(⇤)A to choose A > 0 such that
|f (x) A| < "0 whenever x 2 D \ (x0
A , x0 + A ), x 6= x0 ,
and then use (⇤)B to choose B > 0 such that
|f (x) B| < "0 whenever x 2 D \ (x0
B , x0 + B ), x 6= x0 .
Let 0 = min( A , B ) > 0. Since x0 is an accumulation point of D we may find
x1 2 D \ (x0
0 , x0 + 0 ) \ {x0 }. Then
A+B
=B
2
"0 < f (x1 ) < A + "0 =
A+B
,
2
⇤
a contradiction.
Remark 7.4. Proposition 7.3 says that the limit of a function, if it exists at all, is
unique. This allows us to use the notation
lim f (x) = L
x!x0
to say
x0 is an accumulation point of the domain of f and f has a limit
L at x0 .
Example 7.5. Let f : (0, 1) ! R be defined by f (x) = x cos
limit 0 at 0.
1
x
. Then f has a
Proof. The domain of our function f is the interval (0, 1) and the number 0 is an
accumulation point of (0, 1). Let us argue that
8" > 0 9 > 0 8x 2 (0, 1) 0 < |x
So let "0 > 0 be arbitrary though fixed. Put
0| <
0
= "0 .
) |f (x)
0| < " .
48
ANDRZEJ ROSLANOWSKI
Suppose that x 2 (0, 1) is arbitrary and assume that it satisfies 0 < |x
Then, as 1  cos x1  1,
1
1
= |x| · cos
x
x
Thus, as x was arbitrary, we may conclude
|f (x)
 |x| = |x
0| = x cos
8x 2 (0, 1) 0 < |x
0| <
) |f (x)
0
0| <
0
0| <
0.
= "0 .
0| < "0 ,
and thus
9 > 0 8x 2 (0, 1) 0 < |x
) |f (x)
0| <
0| < "0 .
Since "0 > 0 was arbitrary we get
8" > 0 9 > 0 8x 2 (0, 1) 0 < |x
) |f (x)
0| <
0| < " ,
⇤
as required.
2
Example 7.6. Let f : ( 2, 0) ! R : x 7!
2.
2x +3x 2
.
x+2
Then f has a limit
5 at
Proof. It should be clear that 2 is an accumulation point of the declared domain
( 2, 0) of our function. We will argue that
8" > 0 9 > 0 8x 2 ( 2, 0) 0 < |x
( 2)| <
) |f (x)
( 5)| < " .
To this end, let "0 > 0 be arbitrary though fixed. Put 0 = "0 /2.
Suppose that x 2 ( 2, 0) is arbitrary and assume that it satisfies 0 < |x ( 2)| <
.
0 Then
|f (x)
|2x
( 5)| =
2x2 + 3x
x+2
2
+5 =
1 + 5| = |2x + 4| = 2 · |x
Thus, as x was arbitrary, we may conclude
8x 2 ( 2, 0) 0 < |x
(x + 2)(2x
x+2
( 2)| < 2 ·
( 2)| <
0
0
) |f (x)
1)
+5 =
= "0 .
( 5)| < "0 ,
and hence
9 > 0 8x 2 ( 2, 0) 0 < |x
( 2)| <
) |f (x)
( 5)| < "0 .
Since "0 > 0 was arbitrary we get
8" > 0 9 > 0 8x 2 ( 2, 0) 0 < |x
( 2)| <
) |f (x)
( 5)| < " ,
⇤
as required.
Example 7.7.
p Let f : (0, 1) ! R : x 7!
f has limit x at x.
p
x. Then for every x > 0 the function
Proof. The domain of the function f is declared to be the interval (0, 1) and each
number x > 0 is clearly an accumulation point of the domain (so we may consider
the limits of f at x > 0).
Let x0 > 0 be an arbitrary positive real number. We are going to show that
p
lim f (x) = x0 , that is
x!x0
(8" > 0)(9 > 0)(8x 2 (0, 1))(0 < |x
x0 | <
) |f (x)
p
x0 | < ").
NOTES ON INTRODUCTION TO ANALYSIS
49
To this end suppose that "0 > 0 is an arbitrary but fixed positive real number. Let
p
p
x0 — since x0 > 0 we know that 0 > 0.
0 = "0 ·
Suppose that x 2p(0, 1) is arbitrary
and assume that it satisfies 0 < |x x0 | < 0 .
p
p
p
Then, noting that x > 0 so x + x0 > x0 , we have
p
p
p
p
p
( x
x0 )( x + x0 )
p
p
p
|f (x)
x0 | = | x
x0 | =
=
p
x + x0
p
"0 · x 0
x x0
|x x0 |
|x x0 |
0
p
=

<
=
= "0 ,
p
p
p
p
p
x0
x0
x0
x + x0
| x + x0 |
p
x0 | < "0 . Hence, as x was arbitrary, we may conclude
p
8x 2 (0, 1) 0 < |x x0 | < 0 ) |f (x)
x0 | < "0 ,
p
so |f (x)
and thus
9 > 0 8x 2 (0, 1) 0 < |x
x0 | <
) |f (x)
p
x 0 | < "0 .
Since "0 > 0 was arbitrary we get
8" > 0 9 > 0 8x 2 (0, 1) 0 < |x
x0 | <
) |f (x)
p
x0 | < " ,
as required.
Example 7.8 (Riemann’s function). Let f : (0, 1) ! R be defined by
⇢
0 if x is irrational,
f (x) =
1
if x = pq , p, q 2 J and p, q are relatively prime.
q
Then for every x 2 (0, 1) the function f has limit 0 at x.
⇤
50
ANDRZEJ ROSLANOWSKI
Proof. First, let us note that every non-negative rational number x can be written
in a unique way as a fraction pq with p, q being a non-negative relatively prime
integers, q > 0. (The “relatively prime” means that the greatest common divisor
of p, q is 1, so the fraction pq cannot be simplified.) Consequently, the function f is
well defined.
The domain of the function f is declared to be the interval (0, 1) and each number
x 2 (0, 1) is clearly an accumulation point of the domain (so we may consider the
limits of f at x 2 (0, 1)).
Let x0 2 (0, 1) be arbitrary. We are going to show that lim f (x) = 0, that is
x!x0
(8" > 0)(9 > 0)(8x 2 (0, 1))(0 < |x
x0 | <
) |f (x)
0| < ").
To this end suppose that "0 > 0 is an arbitrary but fixed positive real number. Let
q0 be any natural number greater than 1/"0 + 5 (so then q10 < "0 ). Note that the
set
np
o
def
A =
: p, q 2 J, p < q < q0
q
is finite but it has at least 5 elements. Therefore the set A \ {x0 } is finite and
non-empty, so we may let
def
0
= min |x0
r| : r 2 A \ {x0 } .
Since each of the numbers |x0 r| (for r 2 A \ {x0 }) is positive, we see that 0 > 0.
Suppose that x 2 (0, 1) is arbitrary and assume that it satisfies 0 < |x x0 | < 0 .
Now we will consider two cases, depending on the rationality of x.
Case 1: x is irrational.
Then, by the definition of f , f (x) = 0 and thus
|f (x)
0| = |0
0| = |0| = 0 < "0 .
Case 2: x is rational.
Take two relatively prime numbers p, q 2 J such that 0 < p < q and x = pq . Since
|x x0 | < 0 we know that (by the choice of 0 ) x 2
/ A \ {x0 }, and since 0 < |x x0 |
we also know that x 6= x0 . Therefore x 2
/ A, which implies that q
q0 . Now,
looking back at the definition of f , we have
1
1
|f (x) 0| = |f (x)| =

< "0 .
q
q0
Thus in both possible cases we conclude |f (x) 0| < "0 . Hence, as x was
arbitrary, we get
and thus
8x 2 (0, 1) 0 < |x
x0 | <
9 > 0 8x 2 (0, 1) 0 < |x
Since "0 > 0 was arbitrary we get
) |f (x)
x0 | <
8" > 0 9 > 0 8x 2 (0, 1) 0 < |x
as required.
0
0| < "0 ,
) |f (x)
x0 | <
0| < "0 .
) |f (x)
0| < " ,
⇤
NOTES ON INTRODUCTION TO ANALYSIS
51
Exercises.
Problem 7.1. Directly from the definition of the limit of a function, in a manner
similar to Examples 7.5—7.8, show that the function
f : ( 2, 0) ! R : x 7!
has a limit at
x2 4
x+2
2.
Problem 7.2. Directly from the definition of the limit of a function, in a manner
similar to Examples 7.5—7.8, show that the function
f : ( 2, 0) ! R : x 7!
has a limit at
2x2 + 3x
x+2
2
2.
Problem 7.3. Directly from the definition of the limit of a function, in a manner
similar to Examples 7.5—7.8, show that the function
has a limit 5 at 2.
f : R ! R : x 7! x2 + 1
Problem 7.4. Directly from the definition of the limit of a function, in a manner
similar to Examples 7.5—7.8, show that the function
has a limit 2 at 1.
g : R ! R : x 7! x3 + x
Problem 7.5. Directly from the definition of the limit of a function, in a manner
similar to Examples 7.5—7.8, show that the function
has a limit 0 at 0.
sin : R ! R : x 7! sin(x)
Problem 7.6. Directly from the definition of the limit of a function, in a manner
similar to Examples 7.5—7.8, show that the function
has a limit 3 at 0.
h : R ! R : x 7! 3 + sin2 (x)
Problem 7.7. Give an example of a function f : (0, 1) ! R which has a limit at
every point of its domain except 1/2. Show the required properties of your example
directly from the definition of the limit of a function.
Problem 7.8. Give an example of a function f : R ! R which is bounded and
has a limit at every point of its domain except 2. Show the required properties
of your example directly from the definition of the limit of a function.
52
ANDRZEJ ROSLANOWSKI
Problem 7.9. Let
f : (0, 1) ! R : x 7! cos
1
.
x
Does f have a limit at 0? Justify.
Problem 7.10. Let
f : (0, 1) ! R : x 7! x cos
1
.
x
Does f have a limit at 0? Justify.
Problem 7.11. Let
x2 + x 1
.
x 1
Does f have a limit at 1? Justify directly from the definition of the limit of a
function.
f : (0, 1) ! R : x 7!
x3
Problem 7.12. Let
x+1
.
x2 1
Does f have a limit at 1? Justify directly from the definition of the limit of a
function.
f : (0, 1) ! R : x 7!
NOTES ON INTRODUCTION TO ANALYSIS
53
8. Limits of Functions vs Sequences
The limits of functions are closely related to limits of sequences. As a matter
of fact, Heinrich Heine proposed a definition of the limit of a function based on
the limits of sequences. Here we give it as the following theorem (sometimes called
Heine-Cauchy Theorem).
Theorem 8.1 (Theorem HC). Suppose that f : D ! R and x0 is an accumulation point of D. Then:
f has a limit L at x0
if and only if
for each sequence (xn )1
n=1 converging to x0 with xn 2 D \ {x0 } for all n 2 J, the
1
sequence f (xn ) n=1 converges to L.
Proof. Let f : D ! R and let x0 be an accumulation point of D.
We have to show two implications, and we will treat them separately.
())
Let us assume that f has a limit L at x0 , that is
( ) 8">0 9 >0 8x2D 0 < |x x0 | < ) |f (x) L| < " .
Suppose that (xn )1
n=1 is a sequence converging to x0 such that xn 2 D \ {x0 } for
all n 2 J. Then
(~)
(8" > 0)(9N 2 J)(8n N )(|xn x0 | < ").
1
Our goal is to show that the sequence f (xn ) n=1 converges to L, that is
( )
(8">0)(9N 2J)(8n N )(|f (xn ) L| < ").
To this end, let "0 > 0 be arbitrary but fixed. Applying ( ) (to this "0 ) we may
find 0 > 0 such that
( )
8x 2 D 0 < |x x0 | < 0 ) |f (x) L| < "0 .
Next, applying (~) to 0 (as " there) we may choose N 2 J such that
(⇥)
(8n N )(|xn x0 | < 0 ).
Suppose now that n
N is arbitrary. It follows from the choice of N (i.e., (⇥))
that |xn x0 | < 0 . Since also xn 6= x0 (by our assumption on (xn )1
n=1 ) we have
Now look at the choice of
right above we have
As n
0,
0 < |xn
x0 | <
0.
that is ( ): by it, as xn 2 D, and by the inequalities
|f (xn ) L| < "0 .
N was arbitrary, we have just shown that
(8n
N )(|f (xn )
L| < "0 ),
and thus
(9N 2 J)(8n N )(|f (xn ) L| < "0 ).
Finally, as "0 > 0 was arbitrary, we may conclude ( ), that is
as needed.
(8" > 0)(9N 2 J)(8n
N )(|f (xn )
L| < "),
(()
We will prove this implication by showing its contrapositive, that is proving
that
if f does not have limit L at x0
then there is a sequence (xn )1
n=1 such that
(a)
xn 2 D \ {x0 } for all n 2 J, and
54
ANDRZEJ ROSLANOWSKI
(b)
(c)
lim xn = x0 , but
n!1
the sequence f (xn )
1
n=1
does not converge to L.
So suppose that f does not have limit L at x0 . This means that
that is
¬ 8" > 0 9 > 0 8x 2 D 0 < |x
x0 | <
) |f (x)
9" > 0 8 > 0 9x 2 D 0 < |x
x0 | <
^ |f (x)
So pick "0 > 0 such that
(i) 8 > 0 9x 2 D 0 < |x
x0 | <
^ |f (x)
L|
L| < " ,
L|
" .
"0 .
Then, for each natural number n 2 J, we may apply (i) to = n1 to choose xn 2 D
such that
1
0 < |xn x0 | <
^ |f (xn ) L| "0 .
n
This way we have determined a sequence (xn )1
n=1 . We are going to show that it
satisfies the demands in (a)–(c).
Since 0 < |xn x0 | we know that xn 6= x0 and thus xn 2 D \ {x0 }. Thus (a) is
satisfied.
Note that by the choice of xn we have also |xn x0 | < n1 , that is
(for all n 2 J). Since lim n1
n!1
Problem 3.18 to conclude that
1
< (xn
n
= lim
n!1
lim (xn
n!1
1
n
= 0, we may use the result shown in
x0 ) <
1
n
x0 ) = 0.
Now we may use the result shown in Problem 3.16 to obtain
lim xn = x0 ,
n!1
so the demand in (b) is satisfied.
Finally, by the choice of the xn we know that |f (xn ) L| "0 (for all n 2 J),
that is
f (xn ) 2
/ (L "0 , L + "0 )
(for all n 2 J). In other words, the neighbourhood (L "0 , L + "0 ) of L contains no
1
1
terms of the sequence f (xn ) n=1 . Hence, by Lemma 3.7, the sequence f (xn ) n=1
does not converge to L. Thus the demand in (c) is satisfied as well and the proof
is complete.
⇤
Theorem 8.2. Suppose that f : D ! R and x0 is an accumulation point of D.
Then:
f has a limit at x0
if and only if
for each sequence (xn )1
n=1 converging to x0 with xn 2 D \ {x0 } for all n 2 J, the
1
sequence f (xn ) n=1 converges.
Proof. Should be clear by Theorem 8.1 (and Theorem 6.3).
⇤
Theorem 8.3. Suppose that f : D ! R and x0 is an accumulation point of D.
If for each sequence (xn )1
n=1 converging to x0 with xn 2 D \ {x0 } for all n 2 J, the
1
sequence f (xn ) n=1 is Cauchy, then f has a limit at x0 .
NOTES ON INTRODUCTION TO ANALYSIS
55
Proof. Should be clear by Theorem 8.2 (and Theorem 4.10).
⇤
Exercises.
Problem 8.1. Let a function f : (0, 1) ! R be defined by
f (x) = xx
for x > 0.
Assume that f has a limit at 0 and find that limit.
[Hint: Choose (an )1
n=1 so that lim an = 0 and so that you can determine lim f (an ).]
n!1
n!1
Problem 8.2. Suppose that x0 is an accumulation point of a set D and functions
f, g, h : D ! R are such that
(8x 2 D)(f (x)  g(x)  h(x))
and
lim f (x) = lim h(x) = L.
x!x0
x!x0
Show that the function g has limit L at x0 .
Problem 8.3. Suppose that x0 is an accumulation point of a set D and a function
f : D ! R has a limit at x0 . Show that then the function |f | : D ! R has a
limit at x0 and lim |f (x)| = | lim f (x)|.
x!x0
x!x0
Problem 8.4. Let f : R ! R be defined by
⇢
8x
if x is a rational number,
f (x) =
2x2 + 8 if x is an irrational number.
At which points x0 does the function f have a limit? Justify your answer.
Problem 8.5. Suppose that f : D ! R and x0 is an accumulation point of a set
D. Prove that then:
the function f has a limit at x0 if and only if
for each " > 0, there is a neighborhood Q of x0 such that, for any x, y 2 Q \ D,
x 6= x0 , y 6= x0 , we have |f (x) f (y)| < ".
Problem 8.6. Suppose that f : D ! [0, 1) and f has a limit at x0 . Show that
then the function
p
g : D ! [0, 1) : x 7! f (x)
q
p
has a limit at x0 and lim
f (x) =
lim f (x).
x!x0
x!x0
56
ANDRZEJ ROSLANOWSKI
9. Algebra of Limits
We may use Theorem 8.1 to “transfer” our results concerning limits of sequences
to the context of limits of functions. We will use this approach in Theorem 9.2(i,iii)
below. Of course, we always may consider a direct approach, avoiding sequences,
like in Theorem 9.2(ii).
Let us start with a function-version of Theorem 3.12.
Theorem 9.1. Assume that f : D ! R and x0 is an accumulation point of D.
Suppose also that f has a limit at x0 . Then there is a neighbourhood Q of x0 and
a real number M 2 R such that
(8x 2 Q \ D)(|f (x)|  M ).
Proof. Suppose that f : D ! R, x0 is an accumulation point of D, and f has a
limit L at x0 . The latter implies that
8" > 0 9 > 0 8x 2 D 0 < |x
so (considering " = 1) we may choose
8x 2 D 0 < |x
Now we define M as follows
⇢
max{|L
M=
max{|L
Let Q = (x0
Thus
0 , x0
+
0 ).
0
x0 | <
) |f (x)
L| < " ,
> 0 such that
x0 | <
0
) |f (x)
1|, |L + 1|}
1|, |L + 1|, |f (x0 )|}
L| < 1 .
if x0 2
/ D,
if x0 2 D.
It is a neighbourhood of x0 and:
• If x 2 Q \ D \ {x0 }, then 0 < |x x0 | < 0 and therefore (by the choice of
L| < 1. Hence L 1 < f (x) < L + 1 and |f (x)| < max{|L
0 ) |f (x)
1|, |L + 1|}  M .
• If x = x0 2 D then f (x) = f (x0 ) and hence |f (x)|  M .
(8x 2 Q \ D)(|f (x)|  M ),
as required.
⇤
Let us state and prove limit laws for functions.
Theorem 9.2. Suppose f, g : D ! R with x0 an accumulation point of D, and
further suppose that f and g have limits at x0 . Then:
(i) f + g has a limit at x0 , and
lim (f + g)(x) = lim f (x) + lim g(x).
x!x0
x!x0
x!x0
(ii) f g has a limit at x0 , and
lim (f g)(x) = [ lim f (x)] · [ lim g(x)].
x!x0
x!x0
x!x0
(iii) If g(x) 6= 0 for all x 2 D and lim g(x) 6= 0, then f/g has a limit at x0 , and
x!x0
lim
x!x0
⇣f ⌘
g
lim f (x)
(x) =
x!x0
lim g(x)
x!x0
.
NOTES ON INTRODUCTION TO ANALYSIS
57
Proof. (i) We are going to use Theorem 8.1. Assume f and g have limits at x0 ,
say lim f (x) = ↵ and lim g(x) = .
x!x0
x!x0
Suppose that a sequence (xn )1
n=1 converges to x0 with xn 2 D \ {x0 } for all
n 2 J. By Theorem 8.1 applied to f we know that lim f (xn ) = ↵, and by the
n!1
same theorem applied to the function g we know that lim g(xn ) = . If follows
n!1
from Theorem 5.1 now that
lim f (xn ) + g(xn ) = ↵ + .
n!1
Thus we have shown that
for each sequence (xn )1
n=1 converging to x0 with xn 2 D \ {x0 } for
1
all n 2 J, the sequence (f + g)(xn ) n=1 converges to ↵ + .
By Theorem 8.1 we conclude now that
lim (f + g)(x) = lim f (x) + lim g(x).
x!x0
(ii)
x!x0
x!x0
Assume f and g have limits at x0 , say lim f (x) = ↵ and lim g(x) =
x!x0
⇤
⇤
x!x0
.
Then by Theorem 9.1 we may pick
> 0 and M > 0 such that
⇤
(⌦)0 |f (x)|  M ⇤ for all x 2 (x0
, x0 + ⇤ ) \ D.
Also, our assumptions mean that
(⌦1 ) 8" > 0 9 > 0 8x 2 D 0 < |x x0 | < ) |f (x) ↵| < " , and
(⌦2 ) 8" > 0 9 > 0 8x 2 D 0 < |x x0 | < ) |g(x)
|<" .
We want to show that
(~) 8" > 0 9 > 0 8x 2 D 0 < |x x0 | < ) |f (x) · g(x) ↵ · | < " .
"0
To this end, suppose that "0 > 0 is arbitrary but fixed. Applying (⌦1 ) to | |+M
⇤ > 0
we may find 1 > 0 such that
"0
(⇥1 ) |f (x) ↵| < | |+M
x0 | < 1 , x 2 D.
⇤ whenever 0 < |x
Similarly using (⌦2 ) we may find 2 > 0 such that
"0
(⇥2 ) |g(x)
| < | |+M
x0 | <
⇤ whenever 0 < |x
⇤
2,
x 2 D.
Let 0 = min( , 1 , 2 ) > 0.
Suppose that x 2 D is such that 0 < |x x0 | < 0 . Then the properties
mentioned in (⌦0 ), (⇥1 ) and (⇥2 ) apply to our x and thus we have
• |f (x)|  M ⇤ , and
"0
• |f (x) ↵| < | |+M
⇤ , and
"0
• |g(x)
| < | |+M ⇤ .
Consequently,
f (x)g(x)
↵
= f (x)g(x)
f (x)g(x)
f (x)
f (x) · g(x)
M⇤ ·
"0
+
| | + M⇤
+ f (x)
· f (x)
+
·
f (x) + f (x)
↵
=
↵ <
"0
= "0
| | + M⇤
↵

58
ANDRZEJ ROSLANOWSKI
Thus we have shown that
8x 2 D 0 < |x
x0 | <
0
) |f (x)g(x)
↵ | < "0 ,
and (~) readily follows.
(iii) We are going to use Theorem 8.1. Assume f and g have limits at x0 , say
lim f (x) = ↵ and lim g(x) = 6= 0. Assume also that g(x) 6= 0 for x 2 D
x!x0
x!x0
Suppose that a sequence (xn )1
n=1 converges to x0 with xn 2 D \ {x0 } for all
n 2 J. By Theorem 8.1 applied to f we know that lim f (xn ) = ↵, and by the
n!1
same theorem applied to the function g we know that lim g(xn ) = . If follows
n!1
from Theorem 5.5 that
f (xn )
↵
lim
= .
n!1 g(xn )
Thus we have shown that
for any sequence (xn )1
n=1 converging to x0 with xn 2 D \ {x0 } for
1
all n 2 J, the sequence f/g(xn ) n=1 converges to ↵/ .
By Theorem 8.1 we conclude now that
lim f (x)
lim f/g(x) = ↵/ =
x!x0
x!x0
lim g(x)
.
x!x0
⇤
Corollary 9.3. Assume P (x), Q(x) are two real polynomials and zeroes of Q are
included in the set {r1 , . . . , rn }. Let D = R \ {r1 , . . . , rn } and let
f : D ! R : x 7!
P (x)
.
Q(x)
Then for every x0 2 D the function f has a limit at x0 and
lim f (x) =
x!x0
P (x0 )
= f (x0 ).
Q(x0 )
Theorem 9.4. Suppose f : D ! R and g : D ! R, x0 is an accumulation point
of D, and f and g have limits at x0 . If f (x)  g(x) for all x 2 D, then
lim f (x)  lim g(x).
x!x0
x!x0
Proof. Since x0 is an accumulation point of D, we may use Lemma 4.5 to choose
1
a sequence (xn )1
n=1 of members of D \ {x0 } such that (xn )n=1 converges to x0 . It
follows from Theorem 8.1 that
• the sequence f (xn )
• the sequence g(xn )
1
n=1
1
n=1
converges to lim f (x), and
x!x0
converges to lim g(x).
x!x0
By the assumptions of the functions f, g we know that for each n 2 J we have
f (xn )  g(xn ). But now we may apply Theorem 5.6 to conclude that lim f (xn ) 
lim g(xn ) and hence lim f (x)  lim g(x), as claimed.
n!1
x!x0
x!x0
The following theorem is a function analog of Theorem 5.7.
n!1
⇤
NOTES ON INTRODUCTION TO ANALYSIS
59
Theorem 9.5. Let f : D ! R and g : D ! R and x0 be an accumulation point
of D. If f is bounded in a neighbourhood of x0 and g has limit zero at x0 , then f g
has a limit at x0 and lim f (x)g(x) = 0.
x!x0
Proof. Since f is bounded in a neighbourhood of x0 , we may pick
M ⇤ > 0 such that
⇤
( 0 ) f (x) < M ⇤ whenever x 2 (x0
, x0 + ⇤ ) \ D.
The assumption that lim g(x) = 0 means that
⇤
> 0 and
x!x0
( )1 8" > 0 9 > 0 8x 2 D 0 < |x x0 | < ) |g(x)| < " .
In order to show that lim f (x)g(x) = 0 we have to argue that
x!x0
(~) 8">0 9 >0 8x 2 D 0 < |x x0 | < ) f (x) · g(x) < " .
So suppose that "0 > 0 is arbitrary but fixed. Apply ( 1 ) to " = "0/M ⇤ to pick
1 > 0 such that
( 1 ) |g(x)| < "0/M ⇤ whenever 0 < |x x0 | < 1 , x 2 D.
Let 0 = min( ⇤ , 1 ) > 0.
Suppose that x 2 D is such that 0 < |x x0 | < 0 . Then the properties
mentioned in ( 0 ) and ( 1 ) apply to our x and thus we have
• |f (x)| < M ⇤ , and
"0
• |g(x)| < M
⇤.
Consequently,
"0
f (x)g(x) = f (x) · g(x) < M ⇤ · ⇤ = "0
M
Thus we have shown that
8x 2 D 0 < |x
and (~) readily follows.
x0 | <
0
) |f (x)g(x)| < "0 ,
⇤
60
ANDRZEJ ROSLANOWSKI
Exercises.
Problem 9.1. Define
x3 + 6x2 + x
for x 2 (0, 1).
x2 6x
Prove that f has a limit at 0. Find the limit.
f (x) =
Problem 9.2. Define
p
1+x 1
for x 2 (0, 1).
x
Prove that g has a limit at 0. Find the limit.
g(x) =
Problem 9.3. Define
p
x 3
for x 2 (0, 1).
x
Prove that f has a limit at 0. Find the limit.
f (x) =
9
Problem 9.4. Suppose that g : D ! R is such that (8x 2 D)(g(x) 6= 0). Assume
x0 is an accumulation point of D and g has a limit L 6= 0 at x0 . Show that then
(9M > 0)(9 > 0)(8x 2 D)(|x
x0 | <
) |g(x)|
M ).
Problem 9.5. Show by example that, even though f and g fail to have limits at
x0 , it is possible for f + g to have a limit at x0 . Similarly for f · g and fg .
NOTES ON INTRODUCTION TO ANALYSIS
61
10. Limits of Monotone Functions
In this section we have a closer look at limits of monotone functions. Remember,
Definition 10.1. Let f : D ! R. The function f is said to be increasing
(decreasing, respectively) if and only if for all x, y 2 D with x  y,
f (x)  f (y)
(f (x)
f (y), respectively ).
We say that f is monotone if it is either increasing or decreasing.
If f, g : D ! R and f (x) = g(x) for all x 2 D, then f is increasing if and only
if g is decreasing. Therefore in our studies of limits of monotone functions we may
restrict ourselves to the case of increasing functions. The next definition introduces
the main technical tool for that case.
Definition 10.2. Assume ↵ <
x 2 (↵, ) we define
and let f : [↵, ]
Uf (x) = inf{f (z) : x < z < }
and
! R be increasing. For
Lf (x) = sup{f (y) : ↵ < y < x}.
Lemma 10.3. Let f : [↵, ] ! R be increasing. Then for every x0 2 (↵, ) we
have:
(1) the values Uf (x0 ), Lf (x0 ) are well defined and
(2) f (y)  Lf (x0 )  f (x0 )  Uf (x0 )  f (z) for all y 2 (↵, x0 ) and z 2 (x0 , ).
Assume
⇣ x0 ⌘2 (↵, ). Let⇣ A = {f
⌘ (z) : x0 < z < } and B = {f (y) : ↵ <
x0 +
↵+x0
y < x0 }. Then f
2 A and f
2 B, so both sets are nonempty. Also,
2
2
Proof. (1)
if x0 < z < , then f (x0 )  f (z) so the set A is bounded from below (e.g., by f (x0 ))
and consequently it has the greatest lower bound. Similarly, if ↵ < y < x0 , then
f (y)  f (x0 ) so the set B is bounded from above (e.g., by f (x0 )) and consequently
it has the least upper bound.
(2)
With the notation as above, we easily see that
Lf (x0 ) = sup(B)  f (x0 )  inf(A) = Uf (x0 ).
Moreover, if y 2 (↵, x0 ), then f (y) 2 B and if z 2 (x0 , ), then f (z) 2 A. Hence
also f (y)  sup(B) and inf(A)  f (z).
⇤
Lemma 10.4. Let f : [↵, ] ! R be increasing and x0 2 (↵, ). Then
f has a limit at x0 if and only if Uf (x0 ) = Lf (x0 ),
and in this case lim f (x) = f (x0 ) = Uf (x0 ) = Lf (x0 ).
x!x0
Proof. ())
Suppose that f has a limit at x0 , say A = lim f (x). Then
x!x0
(⌦) 8">0 9 >0 8x2[↵, ] 0 < |x x0 | < ) |f (x) A| < " .
Let us fix "0 > 0 for a moment. Applying (⌦) to " = "0 /2 we may find 0 > 0 such
that
8x 2 [↵, ] 0 < |x x0 | < 0 ) |f (x) A| < "0 /2 .
Now choose y, z 2 (↵, ) such that
x0
0
< y < x0 < z < x 0 +
0.
Then both
|f (y)
A| < "0 /2
and
|f (z)
A| < "0 /2.
62
ANDRZEJ ROSLANOWSKI
Hence, as f is increasing and by Lemma 10.3(2),
A
"0 /2 < f (y)  Lf (x0 )  Uf (x0 )  f (z) < A + "0 /2.
Hence in particular Uf (x0 ) Lf (x0 ) < "0 and |A Uf (x0 )| < "0 /2. Thus we have
shown that
8" > 0 0  Uf (x0 ) Lf (x0 ) < "
and
8" > 0 Uf (x0 ) " < A < Uf (x0 ) + " .
Consequently, Uf (x0 ) = Lf (x0 ) = A. By Lemma 10.3(2) we also have Uf (x0 ) =
Lf (x0 ) = f (x0 ).
(() Assume now that Uf (x0 ) = Lf (x0 ). By Lemma 10.3(2) we have Uf (x0 ) =
Lf (x0 ) = f (x0 ). We will argue that lim f (x) = f (x0 ), i.e.,
x!x0
(~) 8">0 9 >0 8x2[↵, ] 0<|x x0 |< ) f (x) f (x0 ) <" .
So let "0 > 0 be arbitrary but fixed. Then Lf (x0 ) "0 is not an upper bound to the
set {f (y) : ↵ < y < x0 }, so we may fix a y 2 (↵, x0 ) such that Lf (x0 ) "0 < f (y).
Similarly, Uf (x0 ) + "0 is not a lower bound to the set {f (z) : x0 < z < }, so we
may fix a z 2 (x0 , ) such that f (z) < Uf (x0 ) + "0 . Now let
0
Suppose that x0
f (x0 )
Thus |f (x)
0
= min(x0
< x < x0 +
"0 = Lf (x0 )
0.
y, z
x0 ) > 0.
Then y < x < z and, as f is increasing,
"0 < f (y)  f (x)  f (z) < Uf (x0 ) + "0 = f (x0 ) + "0 .
f (x0 )| < "0 whenever 0 < |x
x0 | <
0.
Now (~) plainly follows.
⇤
Theorem 10.5. Let f : [↵, ] ! R be monotone. Then
D = {x 2 (↵, ) : f does not have a limit at x}
is countable. If f has a limit at x0 , then lim f (x) = f (x0 ).
x!x0
Proof. We may assume that f is increasing. For each x 2 D we may fix a rational
number q(x) such that
Lf (x) < q(x) < Uf (x).
This determines a function q : D ! Q. It follows from Lemma 10.3(2) that
Uf (x)  Lf (x0 ) whenever x < x0 . Therefore, the function q is strictly increasing,
and thus injective. Since the set Q of rational numbers is countable, the set D must
be countable as well.
⇤
NOTES ON INTRODUCTION TO ANALYSIS
63
Exercises.
Problem 10.1. Formulate a definition and a lemma similar to Definition 10.2 and
Lemma 10.3 for decreasing functions. Prove your lemma.
Problem 10.2. Suppose that a < b and f : [a, b] ! R is a monotone function.
Show that then f has a limit at a and f has a limit at b.
Problem 10.3. Suppose that f : [a, b] ! R is a bounded function (i.e., there is
M > 0 such that (8x 2 [a, b])(|f (x)| < M )). Assume also that x0 2 (a, b) and the
function f has a limit f (x0 ) at x0 . Let a function g : [a, b] ! R be defined by
g(x) = sup{f (t) : a  t  x}
Prove that then g has a limit at x0 .
(for x 2 [a, b]).
Problem 10.4. Suppose that a function f : R ! R satisfies
(a) (8x, y 2 R)(f (x + y) = f (x) · f (y)), and
(b) f has a limit at 0.
Prove that then:
(i) either f (x) = 0 for all x 2 R or lim f (x) = 1, and
x!0
(ii) f has a limit at every point x 2 R.
64
ANDRZEJ ROSLANOWSKI
11. Continuity of a Function at a Point
Remember Definition 7.1: If f : E ! R and x0 is an accumulation point of E,
then f has a limit L at x0 if and only if
(8" > 0)(9 > 0)(8x 2 E)(0 < |x
x0 | <
) |f (x)
Thus, if additionally x0 2 E, then f has a limit f (x0 ) at x0 i↵
(8" > 0)(9 > 0)(8x 2 E)(0 < |x
But note that if |x
x0 | <
x0 | = 0, then |f (x)
) |f (x)
L| < ").
f (x0 )| < ").
f (x0 )| = 0 < " . . . . . ..
Definition 11.1. Suppose E ✓ R and f : E ! R. If x0 2 E, then f is continuous
at x0 if and only if
8" > 0 9 > 0 8x 2 E |x
86 Chapter 3 Continuity
x0 | <
x0- 6
) |f (x)
xo
f (x0 )| < " .
xo+ 6
Figure 3.1
EASE
definition
in the previous
chapter
Example Compare
11.2. Thethis
sine
function with
sin : the
R definition
! R is continuous
at every
pointconcerning the
limit of a function at a point xo. First, for continuity at xo, the number a must belong
x0 2 R.
E, xbut2itR.need
not betoanshow
accumulation
of E. Indeed,
iff have
:E --,
Proof.toLet
In order
that sin(x)point
is continuous
at x we
to R with xo E E
0
argue and
thatxo not an accumulation point
and
x E E,9 then
= a;hence
(~) 8">0
>0 x8x2R
|x x0 | <
of E, then there is 6
) | sin(x)
0
> 0 such that if
lx - xol < 6
sin(x0 )| < " .
for every E > 0. In other words, if xo is not an accumulation point of E and xo E E,
then f is continuous at xo by default. Thus, the only interesting case is when a is an
accumulation point of E. In this case, the definition requires that f has a limit at xo
and that that limit be f (xo). This fact is a part of the next theorem, but before stating
and proving that theorem, compare Figure 3.1 with Figure 2.1 of Chapter 2. Thus, f
NOTES ON INTRODUCTION TO ANALYSIS
65
To this end assume "0 > 0 is arbitrary but fixed. Let 0 = "0 .
Suppose x 2 R satisfies |x x0 | < 0 . Then
⇣x + x ⌘
⇣x x ⌘
⇣x x ⌘
0
0
0
| sin(x) sin(x0 )| = 2 cos
sin
 2 sin
.
2
2
2
↵
(Above we used the formula sin(↵) sin( ) = 2 cos( ↵+
2 ) sin( 2 ).) Since 0 
sin(↵)  ↵ and sin( ↵) = sin(↵) for all ↵ 0, we have
⇣x x ⌘
0
2 sin
 |x x0 | < 0 = "0 .
2
Now (~) readily follows.
⇤
Example 11.3. The cosine function cos : R
x0 2 R.
! R is continuous at every point
Proof. Let x0 2 R. In order to show that cos(x) is continuous at x0 we have to
argue that
(|) 8">0 9 >0 8x2R |x x0 | < ) | cos(x) cos(x0 )| < " .
To this end assume "0 > 0 is arbitrary but fixed. Let 0 = "0 .
Suppose x 2 R satisfies |x x0 | < 0 . Then by the argument as in the proof of
Example 11.2 we have
⇣ ⇡
⌘
(
x) ( ⇡
x0 )
2
sin( ⇡2 x) sin( ⇡2 x0 )  2 sin 2
=
⇣
⌘2
x0 x
2 sin
 x x0 .
2
Since cos(↵) = sin( ⇡2
cos(x)
↵) for any ↵ 2 R, we get
cos(x0 ) = sin(⇡/2
x)
sin(⇡/2
x0 )  x
x0 <
0
= "0 .
⇤
Now (|) readily follows.
Observation 11.4. If f : E ! R and x0 2 E is not an accumulation point of E,
then f is continuous at x0 .
Theorem 11.5. Let f : E ! R with x0 2 E and x0 an accumulation point of E.
Then (i) through (iii) are equivalent:
(i) f is continuous at x0 .
(ii) f has a limit at x0 and lim f (x) = f (x0 ).
x!x0
1
(iii) For every sequence xn n=1 converging to x0 with xn 2 E for each n 2 J,
1
the sequence f (xn ) n=1 converges to f (x0 ).
Proof. Straightforward. Read carefully Definitions 7.1 and 11.1, remember Theorem 8.1.
⇤
Example 11.6.
(1) Let f : [0, 1] ! R be defined by
⇢
x cos x1
if x 2 (0, 1]
f (x) =
0
if x = 0.
Then f is continuous at 0. p
(2) Let g : (0, 1) ! R : x 7! x. Then for every x > 0 the function g is
continuous at x.
66
ANDRZEJ ROSLANOWSKI
(3) Let P (x), Q(x) are two real polynomials with r1 , . . . , rn being zeroes of Q.
P (x)
Let h : R\{r1 , . . . , rn } ! R : x 7! Q(x)
. Then the function h is continuous
as every x0 2 R \ {r1 , . . . , rn }.
(4) Let f : (0, 1) ! R be defined by
⇢
0 if x is irrational,
f (x) =
1
if x = pq , p, q 2 J and p, q are relatively prime.
q
If x 2 (0, 1) is irrational, then the function f is continuous at x, if x 2 (0, 1)
is rational, then the function f is discontinuous at x.
(5) Let F : R ! R be defined by
⇢
0 if x is irrational,
F (x) =
1 if x is rational.
Then the function F is discontinuous at any x 2 R.
Proof. (1)
By Example 7.5 and Theorem 11.5.
(2)
By Example 7.7 and Theorem 11.5. ,
(3)
By Corollary 9.3 and Theorem 11.5.
(4)
By Example 7.8 and Theorem 11.5.
(5) Let x0 2 R. By Theorem 4.7 the number x0 is an accumulation point of both
the set Q of the rational numbers and of its complement R \ Q. Therefore, using
1
1
Lemma 4.5 we may find sequences yn n=1 and zn n=1 such that
lim yn = lim zn = x0
n!1
n!1
and
yn 2 Q, and zn 2
/ Q for all n 2 J.
Then lim F (yn ) = 1 and lim F (zn ) = 0, so by Theorem 11.5 the function F
n!1
n!1
cannot be continuous at x0 .
⇤
NOTES ON INTRODUCTION TO ANALYSIS
67
Exercises.
Problem 11.1. Let f : R ! R : x 7! sin(x) + cos(x). Directly from the definition
of continuity, like in Examples 11.2,11.3, show that f is continuous at ⇡/2.
Problem 11.2. Let f : R ! R be defined by
f (x) = 3x2
2x + 1.
Show that the function f is continuous at 2
Problem 11.3. Let f : [ 4, 0] ! R be defined by
⇢ 2x2 18
if x 6= 3,
x+3
f (x) =
12
if x = 3.
Show that f is continuous at
3.
Problem 11.4. Assume that x0 2 E0 \ acc(E0 ) and f : E ! R. Prove that, for
every sequence (xn )1
n=1 converging to x0 with xn 2 E for all n 2 J, the sequence
1
f (xn ) n=1 converges to f (x0 ).
Problem 11.5. Define f : (0, 1) ! R by
1
f (x) = p
x
r
x+1
.
x
Can one define f (0) to make f continuous at 0? Justify.
p
Problem 11.6. Prove that f (x) = x is continuous for all x
0.
Problem 11.7. A function f : R !pR is continuous and satisfies f (r) = r2 for
each rational number r. Determine f ( 2). Justify your answer.
Problem 11.8. A function f : (a, b) ! R is continuous and satisfies f (r) = 0 for
each rational number r 2 (a, b). Prove that f (x) = 0 for all x 2 (a, b).
Problem 11.9. Define f : (0, 1) ! R by f (x) = x sin
to make f continuous at 0? Justify.
1
x
. Can one define f (0)
Problem 11.10. Let f : E ! R be continuous at x0 , and x0 2 F ✓ E. Let
g : F ! R be defined by g(x) = f (x) for all x 2 F . Prove that g is continuous at
x0 .
68
ANDRZEJ ROSLANOWSKI
12. Algebra of Continuous Functions
Lemma 12.1. Let g : D ! R be continuous at x0 2 D with g(x0 ) 6= 0. Then
there are > 0 and ↵ > 0 such that
if |x x0 | < and x 2 D, then |g(x)| ↵.
Proof. Let ↵ = |g(x2 0 )| > 0. Applying the continuity of g at x0 to " = ↵ we may
find > 0 such that
8x 2 D |x
Then for every x 2 (x0
x0 | <
) |g(x)
g(x0 )| < ↵ .
, x0 + ) \ D we will have
|g(x0 )|
= |g(x0 )| ↵ < |g(x0 )| |g(x) g(x0 )|.
2
Since |g(x0 )| = |g(x)+(g(x0 ) g(x))|  |g(x)|+|g(x0 ) g(x)| = |g(x)|+|g(x) g(x0 )|,
we have |g(x0 )| |g(x) g(x0 )|  |g(x)|. Consequently,
↵=
↵ < |g(x0 )|
|g(x)
g(x0 )|  |g(x)|.
⇤
Theorem 12.2. Suppose f : D ! R and g : D ! R are continuous at x0 2 D.
Then
(i) f + g is continuous at x0 .
(ii) f g is continuous at x0 .
(iii) If g(x0 ) 6= 0, then f /g is continuous at x0 .
Proof. If x0 is not an accumulation point of D, then the conclusions follow from
Observation 11.4.
So suppose that x0 is an accumulation point of D.
1
Let xn n=1 converge to x0 , xn 2 D for all n 2 J. Then, by Theorem 11.5 we
know that
lim f (xn ) = f (x0 ) and
lim g(xn ) = g(x0 ).
n!1
n!1
By Theorems 5.1, 5.2 we conclude now that
lim f + g (xn ) = f + g (x0 )
n!1
and
lim (f · g)(xn ) = (f · g)(x0 ).
n!1
Hence, using Theorem 11.5 we may conclude that both f +g and f ·g are continuous
at x0 .
Almost the same argument works for the quotient f/g as well. One may worry
about the domain of f/g being possibly smaller than D and x0 not being an accumulation point of that domain. However, if that could happened, then f/g would
be continuous at x0 by Observation 11.4. But this actually cannot happen anyway: by Lemma 12.1, for some > 0 the function g does not take value zero for
x 2 D \ (x0
, x0 + ).
⇤
Corollary 12.3. All trigonometric functions are continuous on their domains.
⇤
Proof. By Examples 11.2, 11.3 and Theorem 12.2.
Theorem 12.4. If f : D ! R and g : E ! R with rng(f ) ✓ E, where f is
continuous at x0 2 D and g is continuous at f (x0 ), then g f is continuous at x0 .
Proof. Our assumptions on f and g mean that
(⇤)f 8">0 9 >0 8x2D |x x0 | < ) |f (x)
f (x0 )| < " ,
NOTES ON INTRODUCTION TO ANALYSIS
69
(⇤)g 8">0 9 >0 8y2E |y f (x0 )|< ) |g(y) g(f (x0 ))| < " .
Noting that the domain of the composition g f is D, we have to argue that
(~) 8">0 9 >0 8x2D |x x0 |< ) |g f (x)
g f (x0 ) |<" .
To this end assume "0 > 0 is arbitrary but fixed. Apply (⇤)g to " = "0 to find
1 > 0 such that
(⇤⇤)g 8y 2 E |y f (x0 )| < 1 ) |g(y) g(f (x0 ))| < "0 .
Then apply (⇤)f to " = 1 to choose 0 > 0 such that
(⇤⇤)f 8x 2 D |x x0 | < 0 ) |f (x) f (x0 )| < 1 .
Now suppose that x 2 D is such that |x x0 | < 0 . Then by (⇤⇤)f we have
|f (x) f (x0 )| < 1 and next by (⇤⇤)g we conclude
Now (~) readily follows.
|g(f (x))
g(f (x0 ))| < "0 .
⇤
Corollary 12.5. The following functions are continuous at every point of their
domains:
p
(1) f (x) = sin x2 + 1),
3
2
p +x2 +x+1)) ,
(2) g(x) = sin(cos(x
(3) h(x) =
1+cos (x)
sin(x)+sin2 (x)+sin3 (x)
cos(x)+cos(x2 )+cos(x3 ) .
70
ANDRZEJ ROSLANOWSKI
Exercises.
Problem 12.1. Suppose that f : D ! R is continuous at x0 2 D. Show that
there is M > 0 and a neighborhood Q of x0 such that
8x 2 Q \ D |f (x)|  M .
Problem 12.2. Assume that functions f, g : D ! R are both continuous on D.
Let h : D ! R be defined by h(x) = max{f (x), g(x)}. Prove that h is continuous
on D.
Problem
p 12.3. Suppose f : D ! [0, 1). Prove that, if f is continuous at x0 2 D,
then f is continuous at x0 .
Problem 12.4. Suppose f : D ! R. Prove that, if f is continuous at x0 2 D,
then |f | is continuous at x0 .
NOTES ON INTRODUCTION TO ANALYSIS
71
13. Uniform Continuity
! R is continuous at every point of its domain. This means
Suppose f : D
that
8x 2 D 8" > 0 9 > 0 8y 2 D) |x
) f (x)
y| <
f (y) < " .
Thus, for an x 2 D and an " > 0 we may find > 0 such that . . .. So we FIRST
are given x and " and only then we find ; the depends on BOTH x and ". Look
at this example.
Example 13.1. The function
h : (0, 1) ! (1, 1) : x 7!
1
x
is continuous at every point x of its domain (0, 1).
Proof. We want to show that
8x 2 (0, 1) 8" > 0 9 > 0 8y 2 (0, 1)) |x
y| <
)
1
x
1
<" .
y
So suppose x0 2 (0, 1). Assume that "0 > 0 is arbitrary but fixed. Pick
⇣ x " · (x )2 ⌘
0
0
0
=
min
,
.
0
2
2
(Note: the 0 above depends on both "0 and x0 !)
Now let y 2 (0, 1) be such that |x0 y| < 0 . Then
• x20 < y < 3x2 0 , and
• |x0 y| <
Consequently,
"0 ·(x0 )2
2
1
x0
< "0 · y · x 0 .
1
y x0
|y x0 |
"0 · y · x 0
=
=
<
= "0
y
yx0
yx0
yx0
⇤
and we easily finish our proof.
On the other hand, look at this example.
Example 13.2. The function
h : (0, 1) ! (0, 1) : x 7! x2
is continuous at every point x of its domain (0, 1).
Proof. We want to show that
8x 2 (0, 1) 8" > 0 9 > 0 8y 2 (0, 1)) |x
y| <
) x2
y2 < " .
So suppose x0 2 (0, 1). Assume that "0 > 0 is arbitrary but fixed. Let
"0
.
0 =
2
(Note: the 0 above depends on "0 only, and x0 is not relevant for it!)
Now let y 2 (0, 1) be such that |x0 y| < 0 . Then
• 0 < x0 + y < 2, and
• |x0 y| < "20 .
72
ANDRZEJ ROSLANOWSKI
Consequently,
(x0 )2
y 2 = (x0
y)(x0 + y) = |y
x0 | · |x0 + y| <
"0
· 2 = "0
2
⇤
and we easily finish our proof.
What is the di↵erence between Example 13.1 and Example 13.2? In the latter
we actually proved
8" > 0 9 > 0 8x, y 2 (0, 1)) |x
) x2
y| <
y2 < " ,
which is a stronger statement then continuity at every point of (0, 1). In the first
one we showed just that
8x 2 (0, 1) 8" > 0 9 > 0 8y 2 (0, 1)) |x
1
x
)
y| <
1
<"
y
and we will see later that we cannot get much more.
Definition 13.3. A function f : D ! R is uniformly continuous if and only if
8" > 0 9 > 0 8x, y 2 D |x
y| <
) |f (x)
f (y)| < " .
Observation 13.4. If f is uniformly continuous (on its domain) then it is continuous at every point of its domain.
Example 13.5.
(1) The function h : (0, 1) ! (0, 1) : x 7! x2 is uniformly
continuous.
(2) Consider the function f : [2.5, 3] ! R defined by f (x) = x 3 2 . Then f is
uniformly continuous.
Proof. (1)
(2)
This is what we actually showed in Example 13.2.
We have to show that
8" > 0 9 > 0 8x, y 2 [2.5, 3] |x
y| <
3
)
x
To this end suppose "0 > 0 is arbitrary but fixed. Put
x, y 2 [2.5, 3] are such that |x y| < 0 . Note that
0.5  |x
2| and 0.5  |y
3
2
0
=
y
2
1
12 "0
<" .
> 0. Suppose
2|,
and therefore
3
x
3
2
y
2
=
3(y x)
3(y x)

< 12
(x 2)(y 2)
0.25
0
= "0 .
Now we easily finish our proof.
⇤
Theorem 13.6. Let f : D ! R be uniformly continuous. Then, if x0 is an
accumulation point of D, f has a limit at x0 .
Proof. Let x0 be an accumulation point of D. We will argue that
1
(~) if xn n=1 is a sequence of members of D \ {x0 } converging to x0 ,
1
then f (xn ) n=1 is a Cauchy sequence.
NOTES ON INTRODUCTION TO ANALYSIS
73
So let xn 2 D \ {x0 } and lim xn = x0 .
n!1
Suppose "0 > 0 is arbitrary but fixed. Since f is uniformly continuous on D we
may pick 0 > 0 such that
Since xn
that
1
n=1
8x, y 2 D |x
y| <
) |f (x)
0
f (y)| < "0 .
converges, it is a Cauchy sequence. Therefore there is N0 2 J such
By the choice of
8n, m
0
N0
xn
xm <
0
.
we may conclude that
8n, m
1
N0
f (xn )
f (xm ) < "0 .
Thus, f (xn ) n=1 is Cauchy.
Finally note that by Theorem 8.3 the statement (~) implies that f has a limit
at x0 .
⇤
Corollary 13.7. The function
1
x
is continuous at every point x of its domain (0, 1) but it is not uniformly continuous.
h : (0, 1) ! (1, 1) : x 7!
Example 13.8. The function
g : R ! R : x 7! x2
is continuous at every point x of its domain R but it is not uniformly continuous.
(Thus Theorem 13.6 gives a necessary condition for uniform continuity, but one
that is not sufficient.)
Proof. Consider " = 1. Suppose
y0
x0 =
0
> 0. Take 0 < x0 < y0 such that
0 /2
and x0 + y0 = 3/ 0 .
Then
g(y0 )| = (y0 )2
|g(x0 )
(x0 )2 = (y0
x0 )(x0 + y0 ) =
0
2
·
3
=
0
3
> 1.
2
Consequently,
¬ 9 > 0 8x, y 2 R |x
) |g(x)
y| <
g(y)| < 1 ,
showing that " = 1 witnesses g is not uniformly continuous.
⇤
Lemma 13.9 (Eduard Heine and Émile Borel).
S Suppose {G : 2 } is an indexed
family of open intervals such that [↵, ] ✓
G . Then there are finitely many
indices
1, . . . , n
2
2 , n 2 J, such that [↵, ] ✓ G
Proof. Suppose
{G :
S
[↵, ] ✓
G . Let
2
1
[ . . . [ G n.
} is an indexed family of open intervals such that
2
def
Z =
r 2 [↵, ] : for some 1 , . . . , n 2 , n 2 J,
we have [↵, r] ✓ G 1 [ . . . [ G n .
74
ANDRZEJ ROSLANOWSKI
For some 1 2 we must have ↵ 2 G 1 , so then {↵} = [↵, ↵] ✓ G 1 . Hence ↵ 2 Z
and Z 6= ;. By its definition, Z ✓ [↵, ], so Z is a bounded set. Therefore Z has a
least upper bound, say
s = sup(Z).
Note that s 2 [↵, ].
Claim 13.9.1. s 2 Z
Proof of the Claim. Since s 2 [↵, ] we may find ⇤ 2
such that s 2 G ⇤ =
(a ⇤ , b ⇤ ). Let = min s a ⇤ , b ⇤ s > 0. Then (s , s+ ) ✓ (a ⇤ , b ⇤ ) = G ⇤ .
Now, s
< s so s
cannot be an upper bound for Z. Consequently, there is
r 2 Z such that s
< r.
Since r 2 Z, we may choose 1 , . . . , n 2 , n 2 J, such that
Since s
< r  s and (s
Hence
and therefore s 2 Z.
[↵, r] ✓ G
1
[ . . . [ G n.
, s + ) ✓ G ⇤ , we get
[r, s] ✓ (s
[↵, s] ✓ G
, s + ) ✓ G ⇤.
1
[ ... [ G
n
[G
⇤
⇤
Claim 13.9.2. s =
Proof of the Claim. We know that ↵  s  , so suppose towards contradiction
that s < .
It follows from Claim 13.9.1 that for some 1 , . . . , n 2
we have [↵, s] ✓
G 1 [ . . . [ G n . Let i  n be such that s 2 G i = (a i , b i ). Let = min s
a i , d i s > 0. Then (s
, s + ) ✓ (a i , b i ) = G i . Let ⇤ = min 2 , 2 s and
⇤
⇤
put s = s + . Then
• s < s⇤ < and
• s⇤ < s + , so [s, s⇤ ] ✓ G i , and [↵, s⇤ ] ✓ G 1 [ . . . [ G n .
Therefore s⇤ 2 Z, contradicting s⇤ > s = sup(Z).
⇤
Claims 13.9.1, 13.9.2 clearly complete our proof.
⇤
Theorem 13.10. If f : [↵, ] ! R is continuous, then it is uniformly continuous.
Proof. We want to show that
(~) 8" > 0 9 > 0 8x, y 2 [↵, ] |x y| < ) |f (x) f (y)| < " .
To this end assume "0 > 0 is arbitrary but fixed.
Let us fix an x 2 [↵, ] for a moment. Since f is continuous at x we may choose
x > 0 such that
( )1 8y 2 [↵, ] |x y| < x ) |f (x) f (y)| < "0 /2 .
x
x
Let Ux = x
2 , x + 2 ; it is an open interval and x 2 Ux .
Unfixing
S x, let us consider the family {Ux : x 2 [↵, ]} of open intervals. Clearly
[↵, ] ✓
Ux : x 2 [↵, ] , so by Lemma 13.9 we may find x1 , x2 , . . . , xn 2 [↵, ],
n 2 J, such that
[↵, ] ✓ Ux1 [ . . . [ Uxn .
⇣
⌘
x1
xn
Let 0 = min 2 , . . . , 2 > 0.
We claim that
NOTES ON INTRODUCTION TO ANALYSIS
75
( )2 8x, y 2 [↵, ] |x y| < 0 ) |f (x) f (y)| < "0 .
So suppose x, y 2 [↵, ] are such that |x y| < 0 . Choose i among 1, . . . , n such
xi
xi
that x 2 Uxi = xi
2 , xi + 2 . Then
|x
Since |x
y| <
0

xi
2
|y
Thus, by the choice of
xi | <
xi
2
.
we have also have
xi |  |y
xi
x| + |x
xi | <
xi
2
+
xi
2
=
xi .
we obtain
f (x)
f (xi ) < "0 /2
and
f (y)
f (x)
f (y)  f (x)
f (xi ) + f (y)
f (xi ) < "0 /2.
Consequently,
and ( )2 follows.
f (xi ) < "0 ,
⇤
76
ANDRZEJ ROSLANOWSKI
Exercises.
Problem 13.1. Suppose that f, g : D ! R are uniformly continuous. Prove that
then f + g : D ! R is uniformly continuous too. What can be said about the
product function f g? Justify.
Problem 13.2. Suppose that f : A ! B and g : B ! C are both uniformly
continuous. What can be said about the composition g f : A ! C? Justify.
Problem 13.3. Let f : [3.4, 5] ! R : x 7! x 2 3 . Directly from the definition of the
uniform continuity, like in Example 13.5(2), show that f is uniformly continuous.
Problem 13.4. Let f : (2, 7) ! R : x 7! x3 x + 1. Directly from the definition of the uniform continuity, like in Example 13.5(2), show that f is uniformly
continuous.
Problem 13.5. Suppose that a function f : R ! R satisfies:
• f is continuous, and
• f is periodic, i.e., for some T > 0 we have 8x 2 R f (x + T ) = f (x) .
Prove that then f is uniformly continuous.
Problem 13.6. Give an example of sets A and B and a continuous, but not
uniformly continuous, function f : A [ B ! R such that its restrictions f A and
f B are uniformly continuous (on A and on B, respectively).
Problem 13.7. Give an example of a continuous and bounded function f : R ! R
which is not uniformly continuous.
NOTES ON INTRODUCTION TO ANALYSIS
77
14. Properties of Continuous Functions
Theorem 14.1. If f : E ! R is uniformly continuous and E is a bounded set,
then f [E] is a bounded set.
Proof. Applying the uniform continuity of f to " = 1 we may find 0 > 0 such that
( )1 8x, y 2 E x y < 0 ) f (x) f (y) < 1 .
Since the set E is bounded, we may fix ↵ < such that E ✓ [↵, ]. Let N 2 J be
such that N > 2( 0 ↵) and for i = 0, 1, 2, . . . , N put xi = ↵ + i · 0 /2. Then
E ✓ [↵, ] ⇢
N
[
(xi
0 /2, xi
+
0 /2).
i=0
For i = 0, 1, . . . , N let yi 2 E be chosen so that
Set
if (xi
0 /2, xi + 0 /2) \ E 6= ;,
then yi 2 (xi
0 /2, xi + 0 /2) \ E 6= ;.
m = min{f (yi ) : i = 0, 1, . . . , N }
By the choice of
1
and
M = max{f (yi ) : i = 0, 1, . . . , N } + 1.
0
we know that if x 2 (xi
m  f (yi )
0 /2, xi
+
0 /2)
1 < f (x) < f (yi ) + 1  M.
Consequently, for every x 2 E we have m < f (x) < M .
\ E, then
⇤
Lemma 14.2. Let (xn )1
n=1 be a sequence of points in a closed interval [↵, ]. Then
1
there is x0 2 [↵, ] and a subsequence xnk k=1 that converges to x0 .
1
Proof. Let (xn )1
n=1 be a sequence of points in [↵, ]. Then the sequence (xn )n=1
is bounded, so by the 2nd Bolzano–Weierstrass Theorem 6.4, this sequence must
1
1
have a convergent subsequence xnk k=1 ; say xnk k=1 converges to x0 . It follows
from Lemma 4.5 that either x0 = xnk for some k (and then x0 2 [↵, ]) or x0 2
acc [↵, ] = [↵, ]. In any possible case x0 2 [↵, ], and the Lemma follows.
⇤
Theorem 14.3 ⇥(3rd Bolzano–Weierstrass
Theorem). Let f : [↵, ] ! R be con⇤
tinuous. Then f [↵, ] = rng(f ) is bounded and acc rng(f ) ✓ rng(f ).
⇥
⇤
Proof. It follows from Theorems 13.10+14.1 that f [↵, ] is bounded, so we have
to show that acc rng(f ) ✓ rng(f ) only.
Suppose that y0 2 acc rng(f ) . By Lemma 4.5 there is a sequence (yn )1
n=1 of
elements of rng(f ) convergent to y0 . Each yn is the value of the function f at some
xn 2 [↵, ], yn = f (xn ). Applying Lemma 14.2 we may find x0 2 [↵, ] and a
1
subsequence xnk k=1 of the sequence (xn )1
n=1 such that
lim xnk = x0 .
k!1
It follows from Theorem 11.5 that then
y0 = lim ynk = lim f (xnk ) = f (x0 ).
k!1
Consequently, y0 2 rng(f ).
k!1
⇤
78
ANDRZEJ ROSLANOWSKI
Corollary 14.4. If f : [↵, ] ! R is continuous, then there are x1 , x2 2 [↵, ]
such that, for all x 2 [↵, ],
f (x1 )  f (x)  f (x2 ).
Proof. By Theorem 14.3 the set rng(f ) is bounded and acc rng(f ) ✓ rng(f ).
Consequently,
inf rng(f ) = min rng(f ) 2 rng(f ) and sup rng(f ) = max rng(f ) 2 rng(f ).
Let x1 , x2 2 [↵, ] be such that f (x1 ) = min rng(f ) and f (x2 ) = max rng(f ) .
Then for every x 2 [↵, ] we will have
f (x1 )  f (x)  f (x2 ).
⇤
Example 14.5.
(1) Consider f : (0, 1) ! R defined by f (x) = x1 . The
function is continuous but not bounded from above. It is bounded from
below, but inf f [(0, 1)] = 1 2
/ f [(0, 1)].
(2) The function g : (0, 1) ! R defined by g(x) = x is certainly bounded, but
inf g[(0, 1)] = 0 2
/ g[(0, 1)] and sup g[(0, 1)] = 1 2
/ g[(0, 1)].
Example 14.6. Let E = [0, 1] [ [2, 3) and define f : E ! R by
⇢
x
if 0  x  1,
f (x) =
4 x if 2  x < 3.
One easily checks that
• f is continuous on E, and
• f is one-to-one on E.
3.4 Properties of Continuous Functions
103
The graph of f looks like this:
graph off
Figure 3.2
is not continuous. Of course, E must be noncompact. Let
anddefinef(x)=xforO 5 x 5 1 a n d f ( x ) = 4 - x f o r 2 < x < 3. It islefttothe
Figure 3.2
is not continuous. Of course, E must be noncompact. Let
NOTES ON INTRODUCTION TO ANALYSIS
79
anddefinef(x)=xforO
1 aEnsatisfies
d f ( x ) = 4 - x f o r 2 < x < 3. It islefttothe
Now, f [E] = [0, 2] and f 1 :5[0,x2] 5 !
readers to satisfy themselves ⇢
that f is indeed 1-1 and continuous. See Figure 3.2 for
x
if 0  x  1,
1
(x)==[0,2] and for 0 5 x
the graph off. Nowf f(E)
1,f-'(x) = x while for 1 < x 5 2,
4 x if 1 < x  2.
f-'(x) = 4 - x. See Figure 3.3 for the graph of f-'.
<
The graph of f
1
looks like this:
graph off
Figure 3 3
Plainly, f
1
is not continuous at x0 = 1.
Theorem 14.7. Suppose f : [↵, ] ! R is continuous and one-to-one. Then the
inverse function f 1 : rng(f ) ! [↵, ] is continuous.
Proof. Suppose y0 2 rng(f ).
Assume that a sequence (yn )1
n=1 converges to y0 , where each yn 2 rng(f ). We
1
want to argue that the sequence f 1 (yn ) n=1 converges to f 1 (y0 ). Let xn =
f 1 (yn ) 2 [↵, ] (for n = 0, 1, 2, . . .). Since the sequence (xn )1
n=1 is bounded, we
may use Theorem 6.5, and thus it is enough to show that
all convergent subsequences of (xn )1
n=1 converge to x0 .
1
⇤
Assume that a subsequence xnk k=1 of the sequence (xn )1
n=1 converges to x .
⇤
Since ↵  xnk  (for all k) we know that x 2 [↵, ]. By the continuity of f we
have
lim f (xnk ) = f (x⇤ ).
k!1
⇤
Since f (xnk ) = ynk and the sequence (yn )1
n=1 converges to y0 , we conclude f (x ) =
⇤
y0 = f (x0 ). But f is one-to-one, so x = x0 .
⇤
Theorem 14.8 (The Bolzano Intermediate–Value Theorem). Let
f : [a, b] ! R be a continuous function. Suppose f (a) < y < f (b) (or f (b) < y <
f (a)). Then there is c 2 (a, b) with f (c) = y.
80
ANDRZEJ ROSLANOWSKI
Proof. Assume f (a) < f (b) and let
def
A =
x 2 [a, b] : f (x)  y .
Then a 2 A, so A 6= ;. Also A is (by its definition) bounded, and therefore it has
a least upper bound. Let c = sup(A)  b.
Claim 14.8.1. f (c)  y.
Proof of the Claim. If c 2 A then clearly f (c)  y. Otherwise we may choose a
sequence (an )1
n=1 such that an 2 A and lim an = c. Then f (an )  y (for each
n!1
n 2 J) and lim f (an ) = f (c) (as f is continuous at c). But now it follows from
n!1
⇤
Theorem 5.6 that f (c)  y.
Claim 14.8.2. f (c)
y.
Proof of the Claim. Suppose towards contradiction that f (c) < y. Put "0 =
0. By the continuity of f at c we may find 0 > 0 such that
8x 2 [a, b] |x
c| <
0
) |f (x)
y f (c)
2
>
f (c)| < "0 .
Since y < f (b) we have f (c) 6= f (b) and hence c < b. Consequently we may find
x 2 [a, b] such that c < x < c + 0 . Then |f (x) f (c)| < "0 so
f (c)
y + f (c)
=
<y
2
2
and hence x 2 A. However, this contradicts c = sup(A) < x.
f (c)
"0 < f (x) < f (c) + "0 = f (c) +
y
⇤
⇤
Corollary 14.9. If p is a polynomial of odd degree with real coefficients, then the
equation p(x) = 0 has at least one real root.
Theorem 14.10. If f : [a, b] ! R is continuous, there are c and d such that
⇥
⇤
f [a, b] = [c, d].
Proof. By Corollary 14.4 there are x1 , x2 2 [a, b] such that
⇥
⇤
8x 2 [a, b] f (x1 )  f (x)  f (x2 ) .
Consequently, f [a, b] ✓ [f (x1 ), f (x2 )]. By the Bolzano
Theo⇥ Intermediate–Value
⇤
rem 14.8 we may conclude that also [f (x1 ), f (x2 )] ✓ f [a, b] . Together,
⇥
⇤
f [a, b] = [f (x1 ), f (x2 )].
⇤
Theorem 14.11. Let f : [a, b]
monotone.
! R be continuous and one-to-one. Then f is
Proof. Suppose towards contradiction that f is continuous, one-to-one but not
monotone. Then there are x, y, z 2 [a, b] such that x < y < z, and
• either f (x) < f (y) and f (z) < f (y),
• or f (x) > f (y) and f (z) > f (y).
Let us suppose the latter; in fact, let us suppose f (x) > f (z) > f (y). The other
cases may be handled in a similar fashion. By the Intermediate-Value Theorem
14.8, there is w 2 (x, y) such that f (w) = f (z), contrary to f being one-to-one. ⇤
NOTES ON INTRODUCTION TO ANALYSIS
81
Exercises.
Problem 14.1. Find an interval of length 1 that contains a root of the equation
xex = 1.
Problem 14.2. Find an interval of length 1 that contains a root of the equation
x3 6x2 + 2.826 = 0.
Problem 14.3. Let f : [↵, ] ! [↵, ] be a continuous function. Show that there
is at least one x 2 [↵, ] such that f (x) = x (such an x is called a fixed point).
Problem 14.4. Let f : [↵, ] ! R. Assume that
(1) f is one-to-one, and
(2) f has the intermediate-value property (i.e., if y is between f (u) and f (v),
then there is x between u and v such that f (x) = y).
Prove that f is continuous. [Hint: First show thatf is monotone.]
Problem 14.5. Prove that there is no continuous function f : R ! R such that,
for each c 2 R, the equation f (x) = c has exactly two solutions.
Problem 14.6. Suppose that a function f : [a, b] ! R is continuous. Define a
function g : [a, b] ! R by
g(t) = sup f (x) : a  x  t
Prove that g is continuous. (Cf Problem 10.3.)
for t 2 [a, b].
82
ANDRZEJ ROSLANOWSKI
15. The Derivative of a Function
Definition 15.1. Let f : (↵, ) ! R with x0 2 (↵, ). For each x 2 (↵, ) with
x 6= x0 define
f (x) f (x0 )
Tf (x) =
.
x x0
(Thus Tf : (↵, ) \ {x0 } ! R.)
The function f is said to be di↵erentiable at x0 (or has a derivative at x0 ) if and
only if the function Tf has a limit at x0 , and we write
lim Tf (x) = f 0 (x0 ).
x!x0
The number f 0 (x0 ) is called the derivative of f at x0 .
If
di↵erentiable
for each x 2 (↵, ), we say f is di↵erentiable on (↵, ).
114f isChapter
4 Differentiation
slope=f’(x_0)
slope=T(x)
Figure 4.1
This means15.2.
that the
of xthe2function
f atQ(xo,
Observation
Let slope
f : (↵,of )the!graph
R with
(↵, ). Let
: (↵yo) is
x0 ,the limit
0
quotient
as
x
approaches
a.
Does
this
sound
familiar?
It
should.
Read
on.
x0 ) \ {0} ! R be defined by
of this
f (x0 + t) f (x0 )
,
for t 2 (↵ x0 ,
x0 ), t 6= 0.
t
Then f is di↵erentiable at x0 if and only if Q has a limit at zero and in this case
f 0 (x0 ) = lim Q(t).
Q(t) =
t!0
DEFINITION Let f : D + R with xo an accumulation point of D and Q E D.
Proof. Straightforward – left for Exercises.
⇤
For each x E D with x # xo, define
Theorem 15.3. Suppose f : (↵, ) ! R and x0 2 (↵, ). Then
The function f is said to be differentiable at xo (or has a derivative at xo) iff T
has a limit at xo, and we write lim,,
T(x) =f '(no). The number f '(a)
is called
the derivative off at xo. Iff is differentiable for each x E E C D , we say f is
differentiable on E.
-
NOTES ON INTRODUCTION TO ANALYSIS
83
• f is di↵erentiable at x0
if and only if
1
• for every sequence xn n=1 of points of (↵, ) \ {x0 } converging to x0 , the
⇣
⌘1
sequence f (xxnn) fx0(x0 )
converges.
n=1
Proof. It follows from Theorem 8.2 (applied to the function Tf as in Definition
15.1).
⇤
Example 15.4.
(1) Let f : R ! R : x !
7 |x|. Then f is NOT di↵erentiable
at 0.
(2) Let g : R ! R by defined by
(
1 ⇣ ⌘ if x = 0
g(x) =
cos x1
if x 6= 0.
Then g is NOT di↵erentiable at 0.
(3) Let h : R ! R by defined by
(
0
⇣ ⌘
h(x) =
x sin x1
if x = 0
if x 6= 0.
Then h is NOT di↵erentiable at 0.
Proof. (1)
For x 2 R \ {0} let
Tf (x) =
Then
Tf (x) =
and therefore Tf has no limit at 0.
(2)
⇢
For x 2 R \ {0} let
Tg (x) =
For n 2 J let xn =
Tg (xn ) =
cos
1
(2n+1)⇡ .
1
xn
1
xn
1
|0|
.
0
if x > 0,
if x < 0,
1
cos x1
g(0)
=
0
x
g(x)
x
1
.
Then
cos (2n + 1)⇡
=
1
n=1
Clearly the sequence xn
f (0)
|x|
=
0
x
f (x)
x
1
1
(2n+1)⇡
=
1
1
1
(2n+1)⇡
=
converges to 0 while the sequence
Tg (xn )
1
n=1
=
2(2n + 1)⇡
1
n=1
does not converge. Consequently, Tg has no limit at 0.
(3)
For x 2 R \ {0} let
Th (x) =
For n 2 J let xn =
1
n⇡
h(x)
x
and yn =
x sin
h(0)
=
0
x
1
2n⇡+⇡/2
Th (xn ) = sin
1
x
0
0
= sin
. Then
1
= sin n⇡ = 0
xn
1
.
x
2(2n + 1)⇡.
84
ANDRZEJ ROSLANOWSKI
and
Th (yn ) = sin
1
= sin 2n⇡ + ⇡/2 = 1.
yn
1
1
Now, both sequences xn n=1 and yn n=1 converge to 0, but the sequences Th (xn )
1
and Th (yn ) n=1 have di↵erent limits. Consequently, Th has no limit at 0 (remember Theorem 8.1).
⇤
Example 15.5.
(1) Let f : R ! R : x 7! x2 and x0 2 R. Then f is
di↵erentiable at x0 and f 0 (x0 ) = 2x0 .
(2) Let g : R ! R : x 7! sin(x) and x0 2 R. Then g is di↵erentiable at x0
and g 0 (x0 ) = cos(x0 ).
(3) Let k : R ! R by defined by
(
0
⇣ ⌘ if x = 0
k(x) =
2
x sin x1
if x 6= 0.
Then k is di↵erentiable at 0 and k 0 (0) = 0.
Proof. (1)
For x 2 R \ {x0 } let
Tf (x) =
f (x0 )
x2
=
x0
x
f (x)
x
x20
(x
=
x0
x0 )(x + x0 )
= x + x0 .
x x0
Then lim Tf (x) = 2x0 and thus f 0 (x0 ) = 2x0 .
x!x0
(2)
For x 2 R \ {x0 } let
g(x)
x
Tg (x) =
g(x0 )
sin(x)
=
x0
x
sin(x0 )
.
x0
↵
Using the formula sin(↵) sin( ) = 2 cos( ↵+
2 ) sin( 2 ), we may write
⇣
⌘
⇣
⌘
⇣
⌘
x x0
x x0
0
⇣
⌘
2 cos x+x
sin
sin
2
2
2
x + x0
Tg (x) =
= cos
·
.
x x0
x x0
2
2
Accepting the well known fact that lim
z!0
sin
lim
x!x0
⇣
x x0
2
x x0
2
⌘
=1
and
sin(z)
z
= 1 we have
lim cos
x!x0
⇣x + x ⌘
0
2
= cos(x0 )
(for the latter remember Example 11.3). Hence lim Tg (x) = cos(x0 ) = g 0 (x0 ).
x!x0
(3)
For x 2 R \ {0} let
Tk (x) =
k(x)
x
x2 sin
k(0)
=
0
x
1
x
0
0
= x sin
1
.
x
It follows from Theorem 9.5 that lim Tk (x) = 0, and hence k 0 (0) = 0.
x!0
⇤
Theorem 15.6. Suppose f : (↵, ) ! R and x0 2 (↵, ). If f is di↵erentiable at
x0 , then f is continuous at x0 .
1
n=1
NOTES ON INTRODUCTION TO ANALYSIS
Proof. For x 2 (↵, ) \ {x0 } let
f (x)
x
Note that for x 2 (↵, ) \ {x0 } we have
Tf (x) =
85
f (x0 )
.
x0
f (x) f (x0 )
(x x0 ) + f (x0 ) = Tf (x)(x x0 ) + f (x0 ).
x x0
We know that lim Tf (x) = f 0 (x0 ), so easily f has limit f (x0 ) at x0 . By Theorem
f (x) =
x!x0
11.5 we conclude f is continuous at x0 .
⇤
86
ANDRZEJ ROSLANOWSKI
Exercises.
Problem 15.1. Let f : (↵, ) ! R with x0 2 (↵, ). Let Q : (↵
{0} ! R be defined by
x0 ,
x0 ) \
f (x0 + t) f (x0 )
,
for t 2 (↵ x0 ,
x0 ), t 6= 0.
t
Prove that f is di↵erentiable at x0 if and only if Q has a limit at zero and in this
case f 0 (x0 ) = lim Q(t).
Q(t) =
t!0
Problem 15.2. Directly from Definition 15.1 show that the function
f : R ! R : x 7! x3
x
is di↵erentiable at every x0 2 R (and find the derivative).
Problem 15.3. Directly from Definition 15.1 show that the function
p
f : (0, 1) ! (0, 1) : x 7! x
is di↵erentiable at every x0 > 0 (and find the derivative).
Problem 15.4. Directly from Definition 15.1 show that the function
p
g : R ! R : x 7! x2 + 1
is di↵erentiable at every x0 2 R (and find the derivative).
Problem 15.5. Directly from Definition 15.1 show that the function
f : R ! R : x 7! cos(x)
is di↵erentiable at every x0 2 R (and find the derivative).
Problem 15.6. Suppose that a function f : (↵, ) ! R is di↵erentiable at point
x 2 (↵, ). Prove that
f (x + h)
f (x
h)
= f 0 (x).
2h
Give an example of a function where this limit exists, but the function is not
di↵erentiable.
lim
h!0
Problem 15.7. Suppose that a function f : (↵, ) ! R is continuous on the
interval (↵, ) and di↵erentiable at a point x0 2 (↵, ). For x 2 (↵, ) define
⇢ f (x) f (x0 )
if x 2 (↵, ) \ {x0 }
x x0
g(x) =
0
f (x0 )
if x = x0 .
Show that g is continuous on (↵, ).
Problem 15.8. Assume that
• f : (↵, ) ! R satisfies 0  f (x) for all x 2 (↵, ), and
• x0 2 (↵, ) and f (x0 ) = 0, and
• the function f di↵erentiable at x0 .
NOTES ON INTRODUCTION TO ANALYSIS
Show that the f 0 (x0 ) = 0.
Problem 15.9. Let f, g, h : (↵, ) ! R and x0 2 (↵, ). Assume that
• f (x)  g(x)  h(x) for all x 2 (↵, ), and
• f (x0 ) = h(x0 ), and
• the functions f and h are di↵erentiable at x0 .
Show that then g is di↵erentiable at x0 . What is g 0 (x0 )?
87
88
ANDRZEJ ROSLANOWSKI
16. The Algebra of Derivatives
Theorem 16.1. Suppose that f, g : (↵, ) ! R are di↵erentiable at x0 2 (↵, ).
Then
(1) f + g is di↵erentiable at x0 and
(f + g)0 (x0 ) = f 0 (x0 ) + g 0 (x0 ),
(2) f g is di↵erentiable at x0 and
(f g)0 (x0 ) = f (x0 )g 0 (x0 ) + f 0 (x0 )g(x0 ),
(3) if g(x) 6= 0 for x 2 (↵, ), then f/g is di↵erentiable at x0 and
f /g
Proof. (1)
0
(x0 ) =
f 0 (x0 )g(x0 )
g(x0 )
Let h = f + g and Th (x) =
Th (x) =
f (x) + g(x) f (x0 )
x x0
g(x0 )
f (x)
x
g(x0 )
.
x0
f (x0 ) g(x)
+
x0
x
Hence
f (x0 )g 0 (x0 )
=
2
h(x) h(x0 )
x x0
f (x)
.
for x 2 (↵, ) \ {x0 }. Note that
f (x0 ) + g(x)
x x0
g(x0 )
=
(f + g)0 (x0 ) = lim Th (x) =
x!x0
lim
x!x0
✓
f (x)
x
f (x0 )
x0
◆
+ lim
x!x0
✓
g(x)
x
g(x0 )
x0
◆
=
f 0 (x0 ) + g 0 (x0 ).
(2)
Let h = f · g and Th (x) =
h(x) h(x0 )
x x0
Th (x) =
f (x)g(x) f (x0 )g(x0 )
f (x)g(x)
=
x x0
for x 2 (↵, ) \ {x0 }. Note that
f (x)g(x0 ) + f (x)g(x0 )
x x0
f (x0 )g(x0 )
f (x) g(x) g(x0 )
g(x0 ) f (x) f (x0 )
+
.
x x0
x x0
Remembering that by Theorem 15.6 we have lim f (x) = f (x0 ), we get
x!x0
0
(f · g) (x0 ) = lim Th (x) =
lim
x!x0
✓
x!x0
g(x)
f (x) ·
x
g(x0 )
x0
◆
f (x0 )g 0 (x0 ) + g(x0 )f 0 (x0 ).
+ lim
x!x0
✓
f (x)
g(x0 ) ·
x
f (x0 )
x0
◆
=
=
NOTES ON INTRODUCTION TO ANALYSIS
Let h = f/g and Th (x) =
(3)
f (x)/g(x)
Th (x) =
x
Also
h(x) h(x0 )
x x0
f (x0 )/g(x0 )
x0
=
89
for x 2 (↵, ) \ {x0 }. Then
f (x)g(x0 ) f (x0 )g(x)
g(x)g(x0 )
x
=
x0
f (x)g(x0 ) f (x0 )g(x)
x x0
g(x)g(x0 )
.
f (x)g(x0 ) f (x0 )g(x)
=
x x0
f (x)g(x0 )
g(x0 )
f (x)
x
f (x0 )g(x0 ) + f (x0 )g(x0 )
x x0
f (x0 )
x0
f (x0 )
g(x)
x
f (x0 )g(x)
=
g(x0 )
.
x0
Consequently,
f (x)g(x0 ) f (x0 )g(x)
= g(x0 )f 0 (x0 ) f (x0 )g 0 (x0 ),
x!x0
x x0
and remembering that by Theorem 15.6 we have lim g(x) = g(x0 ), we may conlim
x!x0
clude
✓ ◆0
f
f 0 (x0 )g(x0 ) f (x0 )g 0 (x0 )
(x0 ) =
.
2
g
g(x0 )
⇤
Theorem 16.2 (Chain Rule). Suppose f : (↵, ) ! R and g : (↵ , ) ! R
with rng(f ) ✓ (↵0 , 0 ). If f is di↵erentiable at x0 2 (↵, ) and g is di↵erentiable at
f (x0 ) 2 (↵0 , 0 ), then the composition g f is di↵erentiable at x0 and
0
0
(g f )0 (x0 ) = g 0 f (x0 ) · f 0 (x0 ).
Proof. Let y0 = f (x0 ) and for y 2 (↵0 ,
(
h(y) =
0
) let
g(y) g(y0 )
y y0
g 0 (y0 )
if y 6= y0
if y = y0 .
Since g is di↵erentiable at y0 we know that lim h(y) = g 0 (y0 ) = h(y0 ), so h is
y!y0
continuous at y0 . By Theorem 12.4 we have that h f is continuous at x0 .
Let Tf : D \ {x0 } ! R be defined by
f (x) f (x0 )
.
x x0
By our assumptions, Tf has a limit at x0 , namely, f 0 (x0 ). Note that for x 2
(↵, ) \ {x0 } we have
Tf (x) =
g(f (x)) g(f (x0 ))
= (h f )(x) · Tf (x).
x x0
Since both h f and Tf has limits at x0 , we conclude that g f is di↵erentiable at
x0 and
g(f (x)) g(f (x0 ))
(g f )0 (x0 ) = lim
=
x!x0
x x0
lim (h f )(x) · lim Tf (x) = g 0 (f (x0 )) · f 0 (x0 ).
x!x0
x!x0
90
ANDRZEJ ROSLANOWSKI
⇤
Remark 16.3. Why isn’t the proof of Theorem 16.2 reduced to writing just
g(f (x)) g(f (x0 ))
g(f (x)) g(f (x0 )) f (x) f (x0 )
=
·
?
x x0
f (x) f (x0 )
x x0
Because it could happen that f (x) = f (x0 ) for x in any neighbourhood of x0 !
Theorem 16.4.
(1) If f : R ! R is a constant function then f is di↵erentiable for all x 2 R and f 0 (x) = 0.
(2) If n 2 J and f (x) = xn for all x 2 R, then f is di↵erentiable for all x 2 R
and f 0 (x) = nxn 1 .
(3) If n 2 Z, n < 0 and f (x) = xn for all x 2 R \ {0}, then f is di↵erentiable
for all x 6= 0 and f 0 (x) = nxn 1 .
Proof. (1)
If f (x) = c for all x 2 R and x0 2 R, then
f (x) f (x0 )
c c
=
=0
x x0
x x0
If f (x) = x1 for all x 2 R and x0 2 R, then
for all x 2 R.
Tf (x) =
(2)
f (x) f (x0 )
x x0
=
= 1 · x00 for all x 2 R.
x x0
x x0
Thus f (x) = x1 is di↵erentiable and f 0 (x) = 1 · x1 1 .
Now we proceed inductively. Suppose that we know that the function x 7! xn
is di↵erentiable and (xn )0 = nxn 1 . Then x 7! xn+1 = x · xn is di↵erentiable by
Theorem 16.1(2) and
Tf (x) =
(xn+1 )0 = (x)0 · xn + x · (xn )0 = xn + x · n · xn
1
= xn + nxn = (n + 1)xn .
(3) Assume n 2 Z, n < 0 and let m = n 2 J. Then f (x) = xn =
di↵erentiable at every x 6= 0 by Theorem 16.1(3), and
⇣ ⌘0
0 m
m 0
m 1
(xn )0 = x1m = (1) ·x x2m1·(x ) = mx
=
x2m
m 1 2m
m 1
n 1
mx
= mx
= nx
.
1
xm ,
so f is
⇤
Exercises.
Problem 16.1. Let f : R ! R : x 7!
on R and find the derivative.
p
x2 + x + 1. Show that f is di↵erentiable
rq
p
Problem 16.2. Let f : (0, 1) ! R : x 7!
f is di↵erentiable on R and find the derivative.
p
x + 1 + 1 + 1 + 1. Show that
Problem 16.3. Let f : R ! R be a di↵erentiable function. For x 2 R put
g(x) = sin(x) · f (x2 ). Show that g is di↵erentiable on R and find its derivative.
hence, f '(xo) 5 0. Therefore, f '(xo) = 0. The case in which f has a relative
minimum at xo is left to the reader.
The reader should now realize that an important fact in this proof is that we are
free to choose points
in the domain on either side of xo as close as we91please. Since
NOTES ON INTRODUCTION TO ANALYSIS
both the maximum and the minimum occurred at end points in the preceding example,
the theorem did not
17. apply.
The Mean–Value Theorem
The following theorem, known as Rolle's Theorem, is an application of TheoDefinition 17.1. Let f : D ! R. A point x0 2 D is a relative maximum
rem 4.6.of We
postpone
a discussion
of its geometric
interpretation
(minimum)
f if shall
and only
if there
is a neighbourhood
U of x0 such
that if x 2until we have
completed
the proof.
U \ D,
then
f (x)  f (x0 )
(f (x)
f (x0 )).
4.7 ROLLE'S THEOREM Suppose f : [a, b] -+R is continuous on [a, b] and
A relative maximum is also called a local maximum; similarly for minimum.
f is differentiable on (a, b). Then iff (a) = f (b) = 0, there is c E (a, b) such that
f'(c)
Theorem
17.2= 0.
(Fermat’s Theorem). Suppose that f : [a, b] ! R and suppose
that f has either a relative maximum or a relative minimum at x0 2 (a, b). If f is
di↵erentiable at x0 , then f 0 (x0 ) = 0.
Proof Iff (x) = 0 for all x E [a, b], then f '(x) = 0 for all x E [a,b], and the
theorem
Proof. Assume
thatisfproved.
has a local maximum at x0 2 (a, b) and let > 0 be such
that
Suppose f(x) # 0 for some x E [a, b]. By Theorem 3.7, [a, b] is compact
bounded; hence, by Corollary 3.11, f assumes its maximum
(⇤)1 (xsince
, it
x0is+closed
) ✓ (a,and
b), and
0
(⇤)2 f (x)
f (x0 ) for all
(x0 say ,at
x0xl
+ and
). x2, respectively. Since f is not identically
andminimum
in x[a,2b],
and f(a)
at least one of xl and x2 must
For n 2 Jzero
let yon
x0 b] n+1
and x=
= x0 =
+0,
n =[a,
n f(b)
n+1 . Then
say xl E1(a, b). Now,1 by Theorem 4.6, f'(xl) = 0.
(⇤)3 both xn n=1 and yn n=1 converge to x0 , so
⇣
⇣
⌘1
⌘1
(⇤)4 both f (xxnn) fx0(x0 )
and f (yynn) fx0(x0 )
converge to f 0 (x0 ), but
belong to (a, b),
In essence, the theorem
states that if the
n=1
n=1graph of a differentiable function touches
and b there is a horizontal tangent.
0 4.2.
See
Figure
Consequently, f (x0 ) = 0.
⇤
Note that the theorem states that there is at least one c such that f'(c) = 0. There
Theorem
17.3
(Rolle’s
Suppose that
f : [a, b]4.2.! R is continuous on
may be
more
such Theorem).
points as indicated
in Figure
[a, b] and If
f is
di↵erentiable
on
(a,
b).
Then
if
f
(a)
f (b)a=geometric
0, there ispoint
c 2 (a,
we examine the graph in Figure 4.2 =
from
ofb)view-that is,
such that f 0 (c) = 0.
ignoring the choice of coordinate system-it would appear that the tangent line at c
f (xn ) f (x0 )
f (yn ) f (x0 )
(⇤)the
0 and
0 for all
n 2 J. a
5 x-axis
and
at b, then
between
xn x0at a
yn xsomewhere
0
c
h
Figure 4.2
Proof. If f (x) = 0 for all x 2 [a, b], then f 0 (x) = 0 for all x 2 (a, b) and we are
done.
So suppose f (x) 6= 0 for some x 2 [a, b]. By Corollary 14.4 there are x1 , x2 2 [a, b]
such that
f (x1 )  f (x)  f (x2 )
for all x 2 [a, b].
At least one of x1 , x2 must belong to (a, b), and by Fermat’s Theorem 17.2 the
derivative at this point must be zero.
⇤
92
ANDRZEJ ROSLANOWSKI
Theorem 17.4 (Lagrange Mean Value Theorem). If f : [a, b] ! R is continuous
on [a, b] and di↵erentiable on (a, b), then there is a c 2 (a, b) such that
0
f (c) =
124 Chapter 4 Differentiation
f (b)
b
f (a)
.
a
Figure 4 3
Proof.is Let
g : [a,to
b] the
!R
be connecting
defined by the
parallel
line
endpoints of the curve. This is essentially the
content of the Mean-Value Theorem
to follow.
f (b) f (a)
g(x) = f (x)
(x a) f (a).
Suppose we have a smoothb curve,
the graph off, connecting the points (a, f (a))
a
f (b)). We
and translate
coordinate axes so that the points lie on
Thenand
g is (b,
continuous
on can
[a, b],rotate
di↵erentiable
on (a,the
b) and
the new x-axis.0 Then 0Rolle'sf (b)
Theorem
f (a) will guarantee a point on the curve where the
= f (x)
x 2 (a,
b),
tangent line isg (x)
parallel
to the new
x-axis, for
which
contains
the segment connecting the
b a
points
(a,
f
(a))
and
(b,
f(b)).
(See
Figure
4.3.)
and g(a) = 0 = g(b). Therefore, by Rolle’s Theorem 17.3 there is c 2 (a, b) such
The slope of the segment joining (a, f(a)) and (b, (b)) is
that
0 = g 0 (c) = f 0 (c)
Thus f 0 (c) = bf -(b)a
b
f (b)
b
f (a)
.
a
f (a) 9
.
a
⇤
Theorem
(Cauchy
Theorem).
Letthat
f, gis,
: [a, b] ! R be continuso this17.5
is the
slope Mean
of theValue
tangent
line at c;
ous on [a, b] and di↵erentiable on (a, b). Then there is c 2 (a, b) such that
f (b)
f (a) · g 0 (c) = g(b)
g(a) · f 0 (c).
Proof. Define
= fstatement
(b) f (a) and
· g(t)
g(a) · f (t)
The h(t)
precise
proofg(b)
follow.
for t 2 [a, b].
Then h is continuous on [a, b] and di↵erentiable on (a, b), and
4.8 MEAN-VALUE
THEOREM
Iff=:h(b).
[a, b] + R is continuous on [a, b] and
h(a) = g(a)f
(b) g(b)f (a)
differentiable
on
(a,
b),
then
there
is
a
c
E (a, b) such that
By Lagrange Mean Value Theorem 17.4, there is c 2 (a, b) such that
0=
h(b)
b
h(a)
= h0 (c) = f (b)
a
f (a) · g 0 (c)
g(b)
g(a) · f 0 (c).
⇤
. Proof
To prove this theorem, we shall find a linear function L such that f - L
satisfies Rolle's Theorem and shall then apply Rolle's Theorem to obtain the
desired result. This corresponds to the change of coordinate system mentioned
..before this theorem. The function L must be linear and satisfy
L(a) =f(a)
and L(b) =f (b).
NOTES ON INTRODUCTION TO ANALYSIS
93
Theorem 17.6. Suppose f : [a, b] ! R is continuous on [a, b] and di↵erentiable
on (a, b). Then:
(1)
(2)
(3)
(4)
(5)
(6)
If
If
If
If
If
If
f 0 (x) 6= 0
f 0 (x) = 0
f 0 (x) > 0
f 0 (x) < 0
f 0 (x) 0
f 0 (x)  0
for
for
for
for
for
for
all
all
all
all
all
all
x 2 (a, b),
x 2 (a, b),
x 2 (a, b),
x 2 (a, b),
x 2 (a, b),
x 2 (a, b),
then
then
then
then
then
then
f
f
f
f
f
f
is
is
is
is
is
is
one-to-one.
constant.
strictly increasing.
strictly decreasing.
increasing.
decreasing.
Proof. (1) If f is not one-to-one then we may find x1 < x2 in [a, b] with f (x1 ) =
f (x2 ). By the Mean Value Theorem 17.4 there is c 2 (x1 , x2 ) such that
f 0 (c) =
f (x1 )
x1
f (x2 )
= 0.
x2
(2) If f is not constant then we may find x1 < x2 in [a, b] with f (x1 ) 6= f (x2 ). By
the Mean Value Theorem 17.4 there is c 2 (x1 , x2 ) such that
f 0 (c) =
f (x1 )
x1
f (x2 )
6= 0.
x2
(3) Suppose that a  x1 < x2  b. By the Mean Value Theorem 17.4 there is
c 2 (x1 , x2 ) such that
f (x1 ) f (x2 )
0 < f 0 (c) =
.
x1 x2
Hence f (x1 ) < f (x2 ).
(4) Suppose that a  x1 < x2  b. By the Mean Value Theorem 17.4 there is
c 2 (x1 , x2 ) such that
f (x1 ) f (x2 )
0 > f 0 (c) =
.
x1 x2
Hence f (x1 ) > f (x2 ).
(5,6)
⇤
Similarly.
Theorem 17.7. Suppose that f and g are continuous on [a, b] and di↵erentiable
on (a, b) and that f 0 (x) = g 0 (x) for all x 2 (a, b). Then there is a real number k
such that f (x) = g(x) + k for all x 2 [a, b].
Proof. Let h : [a, b] ! R : x 7! f (x)
di↵erentiable on (a, b) and
h0 (x) = f 0 (x)
g(x). Then h is continuous on [a, b] and
g 0 (x) = 0
for all x 2 (a, b).
By Theorem 17.6(2), h is a constant function, say h(x) = k for all x 2 [a, b]. Then
f (x) = g(x) + k for all x 2 [a, b].
⇤
Corollary 17.8 (Bernoulli’s inequality). Suppose p > 1 and h > 0. Then
(1 + h)p > 1 + ph.
94
ANDRZEJ ROSLANOWSKI
Proof. Let f (x) = (1 + x)p for x 2 [0, h]. Then f continuous on [0, h] and differentiable on (0, h). By the Mean Value Theorem 17.4 there is t 2 (0, h) such
that
⌘
f (h) f (0)
1 ⇣
f 0 (t) =
= · (1 + h)p 1 .
h 0
h
Since f 0 (t) = p · (1 + t)p 1 we conclude
and since (1 + t)
p 1
(1 + h)p = h · p · (1 + t)p
1
+ 1,
> 1 we get
(1 + h)p > h · p + 1.
⇤
0
Lemma 17.9. If f is di↵erentiable on (a, b) and f (x) 6= 0 for all x 2 (a, b), then
either f 0 (x) > 0 for all x 2 (a, b) or f 0 (x) < 0 for all x 2 (a, b).
Proof. If f 0 (x) 6= 0 for all x 2 (a, b), then f is one-to-one on (a, b). By Theorem
14.11, f is monotone.
If f is increasing, then for all distinct x, y we have f (x)x fy (y)
0 and hence
0
f (x) 0 for all x.
If f is decreasing, then for all distinct x, y we have f (x)x fy (y)  0 and hence
f 0 (x)  0 for all x.
⇤
Theorem 17.10. Suppose f is di↵erentiable on (a, b) and ↵ <
Assume also that 2 R satisfies f 0 (↵) < < f 0 ( ) or f 0 ( ) <
there is c 2 (a, b) such that f 0 (c) = .
are from (a, b).
< f 0 (↵). Then
Proof. Define g(x) = f (x)
x for x 2 (a, b). Then g is di↵erentiable on (a, b) and
g 0 (x) = f 0 (x)
for all x 2 [a, b].
If f 0 (↵) < < f 0 ( ) then g 0 (↵) < 0 < g 0 ( ). By Lemma 17.9 there is c 2 (a, b)
such that g 0 (c) = 0, i.e., f 0 (c) = .
Similarly when f 0 ( ) < < f 0 (↵).
⇤
NOTES ON INTRODUCTION TO ANALYSIS
95
Exercises.
Problem 17.1. Let f : R ! R : x 7!
and find it.
1
1+x2 .
Problem 17.2. Show that the equation x3
the interval [ 1, 1].
Show that f attains maximum value
3x = b has at most one solution in
Problem 17.3. Show that the equation x3 + x2 + 4x = cos(x) has exactly one
solution in the interval [0, ⇡/2].
Problem 17.4. Suppose that f, g : [↵, ] ! R are both di↵erentiable on (↵, )
and f (↵) = g(↵) and f 0 (x) > g 0 (x) for all x 2 (↵, ). Show that
8x 2 (↵, ] g(x) < f (x) .
Problem 17.5. Let f : (↵, ) ! R be di↵erentiable and |f 0 (x)|  M for all
x 2 (↵, ). Show that f is uniformly continuous on (↵, ).
Problem 17.6. Find an example of a di↵erentiable one-to-one function f : R ! R
such that f 0 (x) = 0 for some x 2 R.
Problem 17.7. Suppose that f : [↵, ] ! R is di↵erentiable at x0 2 (↵, ) and
f 0 (x0 ) > 0. Show that there is x 2 (x0 , ) such that f (x) > f (x0 ).
96
ANDRZEJ ROSLANOWSKI
18. L’Hospital’s Rule
Theorem 18.1 (L’Hospital’s Rule). Suppose f and g are continuous on [a, b] and
di↵erentiable on (a, b). If x0 2 [a, b], and
(i) g 0 (x) 6= 0 for all x 2 (a, b),
(ii) f (x0 ) = g(x0 ) = 0, and
(iii) f 0 /g 0 has a limit at x0 ,
then f /g has a limit as x0 and
lim
x!x0
f (x)
f 0 (x)
= lim 0
.
g(x) x!x0 g (x)
1
Proof. We will use the HC Theorem 8.1. So suppose that xn n=1 is a sequence
converging to x0 with xn 2 (a, b) \ {x0 } for all n 2 J.
By Cauchy Mean Value Theorem 17.5, for each n 2 J we may pick a number cn
between x0 and xn such that
[f (xn )
f (x0 )] · g 0 (cn ) = [g(xn )
g(x0 )] · f 0 (cn )
It follows from Theorem 17.6(1) that the function g is one–to–one, and therefore
g(xn ) 6= g(x0 ) = 0. Since f (x0 ) = g(x0 ) = 0 we get
f 0 (cn )
f (xn )
=
0
g (cn )
g(xn )
f (x0 )
f (xn )
=
.
g(x0 )
g(xn )
Now, the sequence (cn )1
n=1 converges to x0 and by the assumption (iii) we
⇣ 0know
⌘
(cn )
0
0
that f /g has a limit at x0 . Hence, by Theorem 17.5, also the sequence fg0 (c
n)
0
(x)
(xn )
converges to limx!x0 fg0 (x)
and consequently the sequence fg(x
converges to the
n)
same limit.
Now, using Theorem 8.1, we easily conclude that f /g has a limit at x0 and
lim
x!x0
f (x)
f 0 (x)
= lim 0
.
g(x) x!x0 g (x)
⇤
Example 18.2. Let f, g : [2, 3] ! R be defined by
p
p
p
p
f (x) = x
2 + x 2 and g(x) = x2
4.
Then
• f, g are continuous on [2, 3] and di↵erentiable on (2, 3),
• f (2) = g(2) = 0,
1
• f 0 (x) = 2p
+ 2px1 2 , and g 0 (x) = pxx2 4 ,
x
•
f 0 (x)
g 0 (x)
=
p
p
p
x+2( x 2+ x)
p
,
2x x
f 0 (x)
0
x!2 g (x)
so lim
= 1/2.
Using Theorem 18.1 we may conclude that
p
p
p
x
2+ x
p
lim
x!2
x2 4
2
=
1
.
2
Theorem 18.3. Suppose that f : [a, b] ! R is continuous and di↵erentiable
with f 0 (x) 6= 0 for all x 2 [a, b]. Then f is one–to–one, f 1 is continuous and
NOTES ON INTRODUCTION TO ANALYSIS
97
di↵erentiable on rng(f ), and
f
1 0
f (x) =
1
f 0 (x)
for all x 2 [a, b].
Proof. It follows from Theorem 17.6(1) that f is one–to–one and by Theorem 14.7
we know that the inverse function f 1 is continuous on its domain. Moreover,
Theorem 14.10 implies that
⇥
⇤
dom(f 1 ) = rng(f ) = f [a, b] = [c, d]
for some c < d.
Let y0 2 [c, d]. In order to show that f 1 is di↵erentiable at y0 suppose a sequence
1
(yn )1
(yn )
n=1 converges to y0 with yn 2 [c, d] \ {y0 } for all n 2 J. Let xn = f
1
for n = 0, 1, 2, 3, . . .. By the continuity of f
(and Theorem 11.5) we know that
(xn )1
n=1 converges to x0 . Hence, as f is di↵erentiable at x0 ,
f (xn ) f (x0 )
= f 0 (x0 ).
xn x0
By our assumptions, f 0 (x0 ) 6= 0 and since f is one–to–one we also know that
f (xn ) f (x0 )
6= 0. Hence, by Theorem 5.5, the sequence
xn x0
⇣ x
⌘1
⇣ f 1 (y ) f 1 (y ) ⌘1
x0
n
n
0
=
f (xn ) f (x0 ) n=1
yn y0
n=1
lim
n!1
converges to
1
f 0 (x0 ) .
By Theorem 15.3 we may now conclude that f 1 is di↵erentiable at y0 = f (x0 )
and
1
0
f 1 f (x0 ) = 0
.
f (x0 )
⇤
Exercises.
Problem 18.1. Find the following limits:
ln(x)
x!1 x 1
lim xx 1
x!0 e
lim sin(x)
x
x!0
(1) lim
(2)
(3)
x2 sin(x)
x!0 sin(x) x cos(x)
lim ln(1+x)
x
x!1
(4) lim
(5)