MAT01A1: The Precise Definition of a Limit

Dr Craig

17–20 March 2020

In the last section, we introduced the limit

laws and the direct substitution property. We

also saw the useful fact that if f (x) = g(x)

for x 6= a then limx→a f (x) = limx→a g(x)

(provided that both limits exist).

Now we explore what is called the precise

definition of a limit. This is what is used to

prove the various limit laws (see Appendix F

if you are interested in looking at the proofs).

At the end of these slides, there are video

links to some good videos from Khan

Academy explaining what this section is all

about. You might want to watch those

before or perhaps after reading these slides.

Now we recall the definition of a limit that

we saw in Chapter 2.2.

Definition of a limit of a function

Definition: Suppose f(x) is defined for x near*a. If we can get the values of f(x) as close to Las we like by taking x as close to a on either side(but not equal to a), then we write

limx→a

f(x) = L

and say

“the limit of f(x), as x approaches a, equals L”

*near: f(x) is defined on some open intervalincluding a, but possibly not at a itself.

Before, we give the precise definition, let us

think of the previous definition in a more

informal way.

Essentially, the definition of

limx→a

f (x) = L

is saying:

“If you give me any y value close to L, I can

give you an x value such that f (x) will be

closer than your y value to L.”

The importance of absolute values

In our first definition we said

“as close to L as we like”

and on the last slide we said

“a y value close to L”

Q: Do we want to the y value to be a little

bit below L or a little bit above L?

A: Both!

It shouldn’t matter if the y value is bigger or

smaller than L. What matters is the

distance from y to L.

The importance of absolute values

In our first definition we said

“as close to L as we like”

and on the last slide we said

“a y value close to L”

Q: Do we want to the y value to be a little

bit below L or a little bit above L?

A: Both!

It shouldn’t matter if the y value is bigger or

smaller than L. What matters is the

distance from y to L.

The importance of absolute values

In our first definition we said

“as close to L as we like”

and on the last slide we said

“a y value close to L”

Q: Do we want to the y value to be a little

bit below L or a little bit above L?

A: Both!

It shouldn’t matter if the y value is bigger or

smaller than L. What matters is the

distance from y to L.

The importance of absolute values

In our first definition we said

“as close to L as we like”

and on the last slide we said

“a y value close to L”

Q: Do we want to the y value to be a little

bit below L or a little bit above L?

A: Both!

It shouldn’t matter if the y value is bigger or

smaller than L. What matters is the

distance from y to L.

The importance of absolute values

In our first definition we said

“as close to L as we like”

and on the last slide we said

“a y value close to L”

Q: Do we want to the y value to be a little

bit below L or a little bit above L?

A: Both!

It shouldn’t matter if the y value is bigger or

smaller than L. What matters is the

distance from y to L.

Two important characters

ε “epsilon” and δ “delta”

These two Greek letters are often used in

mathematics to represent any small positive

real number.

By small, we usually mean “close to 0”.

Note: ε is slightly different to the ∈ symbol

that we use when writing p ∈ Z.

Two important characters

ε “epsilon” and δ “delta”

These two Greek letters are often used in

mathematics to represent any small positive

real number.

By small, we usually mean “close to 0”.

Note: ε is slightly different to the ∈ symbol

that we use when writing p ∈ Z.

Two important characters

ε “epsilon” and δ “delta”

These two Greek letters are often used in

mathematics to represent any small positive

real number.

By small, we usually mean “close to 0”.

Note: ε is slightly different to the ∈ symbol

that we use when writing p ∈ Z.

Here is the precise definition (sometimes called theformal definition) of a limit:

Definition:

We write limx→a

f(x) = L if the following holds:

for every ε > 0 there exists δ > 0 such that if0 < |x− a| < δ then |f(x)− L| < ε.

This is saying for any ε > 0, there exists a δ > 0(which depends on ε) such that whenever x iswithin δ from a, then f(x) will be within ε from L.

Here is the precise definition (sometimes called theformal definition) of a limit:

Definition:

We write limx→a

f(x) = L if the following holds:

for every ε > 0 there exists δ > 0 such that if0 < |x− a| < δ then |f(x)− L| < ε.

This is saying for any ε > 0, there exists a δ > 0(which depends on ε) such that whenever x iswithin δ from a, then f(x) will be within ε from L.

Here is the precise definition (sometimes called theformal definition) of a limit:

Definition:

We write limx→a

f(x) = L if the following holds:

for every ε > 0 there exists δ > 0 such that if0 < |x− a| < δ then |f(x)− L| < ε.

This is saying for any ε > 0, there exists a δ > 0(which depends on ε) such that whenever x iswithin δ from a, then f(x) will be within ε from L.

Definition:

We write limx→a

f(x) = L if the following holds:

for every ε > 0 there exists δ > 0 such that if0 < |x− a| < δ then |f(x)− L| < ε.

This is saying for any ε > 0, there exists a δ > 0(which depends on ε) such that whenever x iswithin δ from a, then f(x) will be within ε from L.

Note 1: x ∈ (a− δ, a+ δ) ⇐⇒ |x− a| < δ.Note 2: x 6= a since 0 < |x− a|.

Definition:

We write limx→a

f(x) = L if the following holds:

for every ε > 0 there exists δ > 0 such that if0 < |x− a| < δ then |f(x)− L| < ε.

This is saying for any ε > 0, there exists a δ > 0(which depends on ε) such that whenever x iswithin δ from a, then f(x) will be within ε from L.

Note 1: x ∈ (a− δ, a+ δ) ⇐⇒ |x− a| < δ.Note 2: x 6= a since 0 < |x− a|.

Definition:

We write limx→a

f(x) = L if the following holds:

for every ε > 0 there exists δ > 0 such that if0 < |x− a| < δ then |f(x)− L| < ε.

This is saying for any ε > 0, there exists a δ > 0(which depends on ε) such that whenever x iswithin δ from a, then f(x) will be within ε from L.

Note 1: x ∈ (a− δ, a+ δ) ⇐⇒ |x− a| < δ.

Note 2: x 6= a since 0 < |x− a|.

Definition:

We write limx→a

f(x) = L if the following holds:

for every ε > 0 there exists δ > 0 such that if0 < |x− a| < δ then |f(x)− L| < ε.

This is saying for any ε > 0, there exists a δ > 0(which depends on ε) such that whenever x iswithin δ from a, then f(x) will be within ε from L.

Note 1: x ∈ (a− δ, a+ δ) ⇐⇒ |x− a| < δ.Note 2: x 6= a since 0 < |x− a|.

limx→3

f (x) = 5

Example

Let f (x) = x3 + 6. Prove that lim

x→3f (x) = 7.

Solution: This type of proof has two steps.

First, we must come up with a ‘guess’ for δ.

This will be a value of δ stated in terms of ε.

We will then show that if we are given an

arbitrary ε > 0, then our guess for δ will

ensure that the implication

0 < |x− a| < δ → |f (x)− L| < ε

will be true.

Example

Let f (x) = x3 + 6. Prove that lim

x→3f (x) = 7.

Solution: This type of proof has two steps.

First, we must come up with a ‘guess’ for δ.

This will be a value of δ stated in terms of ε.

We will then show that if we are given an

arbitrary ε > 0, then our guess for δ will

ensure that the implication

0 < |x− a| < δ → |f (x)− L| < ε

will be true.

Example

Let f (x) = x3 + 6. Prove that lim

x→3f (x) = 7.

Solution: This type of proof has two steps.

First, we must come up with a ‘guess’ for δ.

This will be a value of δ stated in terms of ε.

We will then show that if we are given an

arbitrary ε > 0, then our guess for δ will

ensure that the implication

0 < |x− a| < δ → |f (x)− L| < ε

will be true.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε.

So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.

∣∣∣(x3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Example: Let f(x) = x3 + 6. Prove lim

x→3f(x) = 7.

Step 1: At the end of Step 2 we want to have|f(x)− L| < ε, i.e.

∣∣(x3 + 6

)− 7

∣∣ < ε. So we startwith that and work backwards with the aim to finishwith some statement about |x− a|, i.e. |x− 3|.∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ −ε < x

3+ 6− 7 < ε

∴ −ε < x

3− 1 < ε

∴ −3ε < x− 3 < 3ε

∴ |x− 3| < 3ε

Our ‘guess’ for δ will be 3ε.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work.

Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work. Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work. Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work. Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work. Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work. Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work. Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

Step 2: Let ε > 0. We must show that our ‘guess’for δ does indeed work. Suppose that |x− 3| < δ,i.e. that |x− 3| < 3ε.

∴ −3ε < x− 3 < 3ε

∴ −ε < x

3− 1 < ε

∴ −ε < x

3+ 6− 7 < ε

∴∣∣∣(x

3+ 6

)− 7

∣∣∣ < ε

∴ |f(x)− 7| < ε

We have shown that

limx→3

(x3+ 6

)= 7.

We emphasized already that there are two mainsteps in the proof of the limit of a linear function.First: come up with a ‘guess’ for δ and then second,show that this formula for δ will work for any ε > 0.

Besides these two main steps there are three smallerthings that you need to be able to do:

I Identify the f(x), a and L from the precisedefinition.

I When finding your δ, you must manipulate theinequalities to get one involving |x− a|.

I In Step 2, you must manipulate the inequalitiesin reverse so that you get one involving|f(x)− L|.

Look at Example 2 to see another example of

how to prove that limx→a

f (x) = L for f (x) a

linear function.

Note that we do not cover limit proofs for

functions like those in Example 3 and

Example 4. We only cover linear functions.

Understanding the formal definition

Tools to enrich calculus

Click on “Limits and Derivatives” and then

“Precise Definitions of Limits”. Move ε to a

value greater than zero and then see what δ

value gives you an interval (a− δ, a + δ) for

which the distance from f (x) to L will

always be less than ε (i.e. |f (x)− L| < ε).

Note that

x ∈ (a− δ, a + δ) ⇐⇒ |x− a| < δ.

In the previous lecture we stated eleven

different Limit Laws but we did not prove

any of them.

We couldn’t prove them because we didn’t

know how. With our new formal definition of

a Limit, we can prove the Limit Laws.

Proofs of the Limit Laws can be found in

Appendix F. These are not examinable in this

course, but please take a look at them if you

are keen to have a better understanding of

the ε-δ definition of a Limit.

Infinite Limits

Previously we said that f (x) has an infinite

limit as x→ a if we can make f (x) “as big

as we like” by taking x close to a.

Now we want to formalise this in the same

way that we formalised the definition of a

(non-infinite) limit.

Definition: Let f be a function defined on

some open interval that contains a, except

possibly at a itself. Then

limx→a

f (x) =∞

means that for every positive number M

there is a δ > 0 such that

if 0 < |x− a| < δ then f (x) > M .

Example: Show that limx→0

1

x2=∞.

Solution:

Example: Show that limx→0

1

x2=∞.

Solution:

Definition: Let f be a function defined on

some open interval that contains a, except

possibly at a itself. Then

limx→a

f (x) = −∞

means that for every negative number N

there is a δ > 0 such that

if 0 < |x− a| < δ then f (x) < N .

Khan Academy videos

The following sequence of four videos will be

very helpful in understanding the precise

definition of a limit:

Formal definition of limits review (on KA)

Or use the YouTube link:

https://youtu.be/5i8HLmVTcRQ