Lecture 5

This lecture.• The definition of and topics related to the important concept

of “Continuity”.

The definition and first examples. Important classes of continuous functions:

Polynomials Rational functions Trigonometric functions Exponential functions

Using elementary operations to build new continuous functions from other continuous functions.

Continuous functions and limits. Compositions. The Intermediate Value Theorem.

An observationWe’ve been thinking a lot about limits recently.

Example. What is ?

Solution.

First consider the denominator:

lim𝑥→ 1

(𝑥2−2)=lim𝑥→ 1

𝑥2+ lim𝑥→1

(−2)

¿ ( lim𝑥→1 𝑥)2+ lim𝑥→1 (−2)¿ (1 )2+(−2)

¿−1

(By the limit law for sums.)

(By the limit law for products.)

(By the limit law for simple fns.)

Next consider the numerator. In the same way, we deduce:

lim𝑥→1

(𝑥+1)=lim𝑥→1

𝑥+ lim𝑥→1

(1)=2.

An observationExample. What is ?

Solution (cont.)

So, because:• exists,• exists and is not equal to zero,

we can apply the limit law for quotients to conclude that:• exists and equals

Here’s the observation:

Hang on… That was a lot of work just to derive the answer .Couldn’t we have gotten this answer just by substituting directly into the function ??? What’s the point of all this limit stuff?

ContinuityIt’s true that for many functions we can obtain the limit just by evaluating the function at the point we are taking the limit: .

But these are actually very special types of functions: the continuous functions.

Definition (‘Continuous at ’).Let be a function and let .The function is said to be continuous at the point if the following three things are true:

1. 2. The limit exists.3.

Continuity

Definition (‘Continuous at ’).

Let be a function and let .

The function is said to be continuous at the point if the following three things are true:

1. The limit exists.2.

3.

A few comments:

• Often we summarize these three conditions by a single equation: .

• This equation is sometimes called the “direct substitution property”.

• The intuitive idea of a continuous function is that “the graph can be drawn without lifting the pen off the paper”. Or – “a small change in away from only leads to a small change in ”.

Continuity: An example

Note that both of the following functions have limits at .

However:

Exercise:

Use the definition of continuity and standard properties of limits to show that the function is continuous at the point .

Solution: lim𝑥→ 4

𝑓 (𝑥)¿ lim𝑥→ 4

(𝑥2+√7−𝑥 )

+

¿16+√3¿ 𝑓 (4)

Summary:

Because:

1. The limit of at exists.2. The limit equals the value of at .

We conclude that is continuous at .

A function can fail to be continuous in several different ways...

This function is not continuous at because

the right 1-sided limit, , does not exist.

This function is not continuous at because

even though the one sided limits both exist, they

are not equal.

This function is not continuous at because

even though the function does actually have a limit

at , the actual value of the function there is

different to the limit.

“Essential discontinuity.”

“Jump discontinuity.”

“Removeable discontinuity.”

“Most” elementary functions are continuous.

Theorem.

Polynomial functions are continuous at every point .

Proof.

Now according to the definition of “continuous at ” we have three things to check:

1. The point lies in the domain of .2. The limit exists.3. And:

To begin, note that an arbitrary polynomial function is of the form:

for some constants .

The version of this expression we will use during the proof is: .

Theorem.

Polynomial functions are continuous at every point .

Proof (continued).

The domain of any polynomial function is all of , so it is obvious that will lie in the domain of the given .

Next consider the limit:

lim𝑥→𝑎

𝑓 (𝑥)=lim𝑥→𝑎 (∑𝑖=0

𝑘

𝑐𝑖 𝑥𝑖)

¿∑𝑖=0

𝑘

lim𝑥→𝑎

(𝑐 𝑖𝑥𝑖)

¿∑𝑖=0

𝑘

lim𝑥→𝑎

(𝑐𝑖 )( lim𝑥→𝑎(𝑥 ))𝑖

¿∑𝑖=0

𝑘

𝑐𝑖𝑎𝑖

¿ 𝑓 (𝑎 ) .

(By the limit law for sums.)

(By the limit law for products.)

(By the limit laws for simple fns.)

Thus the limit exists, and equals , as required by the definition.

Some important classes of continuous functions.

Properties of continuous functions.

We’re going to look at a few topics here:

1. Ways to combine continuous functions to construct new continuous functions.

2. How continuous functions interact with limits. Compositions of continuous functions.

3. One-sided continuity. Continuity on an interval.4. The intermediate value theorem.

Combining continuous functions.

The general principle:

Any function constructed from continuous functions by doing additions, multiplications, and divisions, is itself going to be continuous at every point of its domain.

Example:

If ,, and are functions which are continuous at each point of their domains, then the following function is also continuous at each point of its domain:

𝐹 (𝑥 )=( 𝑓 (𝑥 ) )5−12h (𝑥 ) 𝑔(𝑥)

𝑓 (𝑥)

1−1

1000+ 𝑓 (𝑥 )+𝑔 (𝑥 )+h(𝑥 )

Why? This is essentially because of the limit laws.

Combining continuous functions.

Exercise

Consider two functions: and . Assume that both of and are continuous at some point .Explain why the function is also continuous at the point .

Solution

To begin, we’ll write down explicitly the information we know already about and :

About :

1. .

About :

1. .

To explain why is also continuous at we have to answer the same 3 questions about , using this information.

Solution (continued)

About :

1) .

About :

a) .

Question 1. Why is ?

According to (1) and (a) above, lies in both and .So , which is exactly .

Question 2. Why does the limit exist?

We can apply the “limit law for sums” to this limit, because (2) and (b) above tell us that the assumptions of this law are satisfied.

Question 3. Why is ?

lim𝑥→𝑎

( 𝑓 (𝑥 )+𝑔 (𝑥 ))= lim𝑥→𝑎

𝑓 (𝑥 )+ lim𝑥→𝑎

𝑔 (𝑥 )

¿ 𝑓 (𝑎 )+𝑔 (𝑎 )=( 𝑓 +𝑔)(𝑎)

(By the limit law for sums.)

(By (3) and (c).)

Continuous functions, limits, and compositions

Question

Consider two functions: and . Is the following equation always true?

Maybe we just need to assume that exists?Illustrate your answer with an example.

Answer

No, it’s definitely not true.For example, consider: and .

. ?

?

But it turns out that this equation is actually true, with one extra assumption about :

Continuous functions, limits, and compositions (cont.)

Theorem (‘You can pass limits into continuous functions’).Let and be functions and let .Assume that:

• exists.• is continuous at .

Then:

Continuous functions, limits, and compositions (cont.)

Exercise

What is ?

Solution

Earlier in this lecture we stated that trigonometric functions are continuous functions at every point of their domains. (We’ll prove this later on.)

This means that we can exploit the theorem we just introduced to pass the limit operation into the function :

(Because is continuous at .)

Continuous functions, limits, and compositions (cont.)

Another way to formulate the theorem that we just met is as follows:

Theorem (‘Composition of continuous functions is continuous’).Let and be functions and let .Assume that:

• is continuous at .• is continuous at .

Then is continuous at .

Exercise.

On this week’s problem set you will be asked to show that this version is exactly equivalent to the theorem we met a few slides ago.

Continuous functions, limits, and compositions (cont.)

Problem.

At what points is the function continuous?

Solution.

We’ll begin by understanding the domain of .Note that it is a composition of two functions:

• with domain .• with domain .

We deduce in the standard way that the domain of is .

Now observe that:• is continuous at every point.• is continuous at every point except and .

It follows from the previous theorem that is continuous at every point of , which is the whole domain of .

Functions can be continuous “from one side”.

Question. Is the function continuous at the point ?

Here is the graph:

Answer. According to our intuition for the concept “continuous”, this function should be continuous at , but it actually isn’t according to the definition we have so far. Why?

Definition (‘Continuous function’).Let be a function and let .The function is said to be continuous at the point if the following three things are true:

1. The limit exists.2. 3.

Question.

Why is the function not continuous at according to the following definition?

Answer: According to our definition of limit, for a limit to exist at the function must be defined on both sides of . But at the function is only defined to the right.

We invent some new concepts to include situations like this in our discussion.

Definition (‘Continuous from the right’).Let be a function and let .The function is said to be continuous from the right at the point if

1. exists.2. .

Definition (‘Continuous from the left’).Let be a function and let .The function is said to be continuous from the left at the point if

1. exists.2. .

And:

Example.

The function is continuous from the right at the point .

𝑓 (𝑥 )=√𝑥

Question. Consider a function whose graph is:

Answer:At what points is 1. Continuous.2. Continuous from the right.3. Continuous from the left.

1.

Continuity on an interval

Thus far we have only been discussing continuity properties at a single point.

Functions which are continuous at every point of a whole interval have some really nice properties. To discuss these properties precisely it is useful to introduce another definition, which is a little bit unnatural, unfortunately.

Definition (‘Continuous on an interval’).Let be a function.

Consider an interval . (Note that by “interval” we are including all of the following possibilities: , , , , , , , and so on.)

We say that is continuous on if it is continuous at every point , according to the interpretation of “continuous” at the endpoints clarified below:

• At the boundary points of the form “” if is only defined on the right of , then “continuous” means “continuous from the right”.

• At the boundary points of the form “” if is only defined on the left of , then “continuous” means “continuous from the left”.

Exercise.

Name some intervals on which the function graphed here is continuous:

Answers.

There are countless answers actually.

?

The intermediate value theorem

Example.

Consider a function with domain whose graph is shown below. I have concealed part of the graph under a box.

Question.

If I tell you that at no point does the graph of intersect the line then what can you conclude?

Answer.

The function cannot be continuous on the whole interval .

(𝑏 , 𝑓 (𝑏))

(𝑎 , 𝑓 (𝑎 ))

𝑦=𝑁

The intermediate value theorem

This “observation” we just made is actually an important theorem:

Theorem (‘The intermediate value theorem’).Let be a function, and let , , and be real numbers with .Assume that:

• is continuous on the closed interval .• .• lies between and .

Then: there exists a number such that:

•

In particular, must be continuous from the right at and continuous from the left at .

Theorem (‘The intermediate value theorem’).

Let be a function, and let , and be real numbers.

Assume that:• is continuous on the closed interval .• .• lies between and .

Then: there exists a number such that:

•

Question.

• The I.V.T. is a statement that has “enough” numbers to solve certain types of equations.

• The I.V.T. actually reflects a deep property of , known as the completeness axiom. This axiom is the only difference in the lists of axioms of and .

• The I.V.T. seems “obvious” from an intuitive point of view, but when we are building up a mathematical theory from logical foundations, even obvious things need to be proved before we completely understand the theory. For example, if we couldn’t deduce this, then probably the axioms we were using were wrong, and our theory would have problems.

Comments

This seems obvious when you think about it. What’s the point?

The I.V.T: examples.

Prove that there exists at least one solution to the equation:

Example.

Solution.

Define a function .

We’ll try and apply the Intermediate Value Theorem to on the closed interval to show that such a must exist.

The first thing we need to do, then, is check whether the assumptions of the theorem are satisfied.

Any point where will be a solution of the given equation.

Solution (continued).

Question 1: Is continuous on the interval ?

Yes.

We have two cases to check.

At the function is only defined on the right, so at that point to check continuity we have to test a limit from the right:

At every other point , the function is defined on both sides, so the calculation we have to test in that case is:

Solution (continued).

Question 2: Does lie between and ?

This will be trivial to check:

•

As required: .

Conclusion:

The assumptions of the Intermediate Value Theorem are satisfied, so we conclude that there exists a such that . This will be a solution of the original equation.

Problem

Solution

Prove that for every positive integer , the number exists.

Well, what on earth does this mean? It means: Prove there exists a real number whose square is .

In other words: Prove that there exists a solution to the equation .

In other words: Prove that there exists a number where the function

has the value zero: .

We’ll prove that such a exists by applying the I.V.T. to this function on the interval . Note that:

• is a polynomial, so it is continuous at every point.

So the I.V.T. implies that such a solution exists. E.g. exists!