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The Formal Definition of a Limit

Numeracy Workshop Adrian Dudek

Adrian Dudek The Formal Definition of a Limit 2 / 37

Introduction

These slides cover the formal definition of a limit, and aim to be helpful for students studying calculus to the level of MATH1001 or higher. Drop-in Study Sessions: Monday, Wednesday, Thursday, 10am-12pm, Meeting Room 2204, Second Floor, Social Sciences South Building, every week. Website: Slides, notes, worksheets. http://www.studysmarter.uwa.edu.au → Numeracy → Online Resources Email: geoff.coates@uwa.edu.au

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The Fibonacci Sequence

The study of limits is, in a sense, the study of closeness. Consider the famous Fibonacci sequence, defined by setting the first two terms of the sequence to 1, and furthermore defining each term to be the sum of the two before it. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . .

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The Fibonacci Sequence

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . Choose two terms in the Fibonacci sequence that sit next to each other, and divide the larger by the smaller. For example, we choose 89 and 55, and then 89 55 = 1.6181818 . . .

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The Fibonacci Sequence

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . Choose two terms in the Fibonacci sequence that sit next to each other, and divide the larger by the smaller. For example, we choose 89 and 55, and then 89 55 = 1.6181818 . . . Let’s make a whole new sequence by taking the Fibonacci sequence and dividing each term by the term before it: 1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, . . .

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The Golden Ratio

1 1 2 1 3 2 5 3 8 5 13 8 21 13 . . .

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The Golden Ratio

1 1 2 1 3 2 5 3 8 5 13 8 21 13 . . . If we write each of these in their decimal notation to 4 decimal places we get: 1.0000 2.0000 1.5000 1.6667 1.6000 1.6250 1.6154 . . .

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The Golden Ratio

1 1 2 1 3 2 5 3 8 5 13 8 21 13 . . . If we write each of these in their decimal notation to 4 decimal places we get: 1.0000 2.0000 1.5000 1.6667 1.6000 1.6250 1.6154 . . . These numbers appear to be converging to a value.

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The Golden Ratio

1 1 2 1 3 2 5 3 8 5 13 8 21 13 . . . If we write each of these in their decimal notation to 4 decimal places we get: 1.0000 2.0000 1.5000 1.6667 1.6000 1.6250 1.6154 . . . These numbers appear to be converging to a value. In fact, this sequence approaches the golden number φ, which is equal to 1 + √ 5 2 = 1.618033 . . .

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The Golden Ratio

1 1 2 1 3 2 5 3 8 5 13 8 21 13 . . . If we write each of these in their decimal notation to 4 decimal places we get: 1.0000 2.0000 1.5000 1.6667 1.6000 1.6250 1.6154 . . . These numbers appear to be converging to a value. In fact, this sequence approaches the golden number φ, which is equal to 1 + √ 5 2 = 1.618033 . . . We say the limit of the sequence is φ.

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Limits: Example

Recall that a function takes in x-values and outputs y-values.

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Limits: Example

Recall that a function takes in x-values and outputs y-values. Sometimes we are more concerned with the value of a function near a point, rather than at the point itself.

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Limits: Example

Recall that a function takes in x-values and outputs y-values. Sometimes we are more concerned with the value of a function near a point, rather than at the point itself. Usually this is because the function itself doesn’t exist at the point!

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Limits: Example

Recall that a function takes in x-values and outputs y-values. Sometimes we are more concerned with the value of a function near a point, rather than at the point itself. Usually this is because the function itself doesn’t exist at the point! From this, the notion of a limit arises.

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Limits: Example

Here is an example function f (x).

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Limits: Example

Here is an example function f (x). We can say that limx→−1 f (x) = −4.

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Limits: Example

The problem here is, we haven’t really justified this properly! We’ve just had a quick glance.

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Limits: Example

The problem here is, we haven’t really justified this properly! We’ve just had a quick glance. In fact, the function might be getting closer and closer to 4.000001, rather than 4.

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Limits: Example

The problem here is, we haven’t really justified this properly! We’ve just had a quick glance. In fact, the function might be getting closer and closer to 4.000001, rather than 4. So, we need a stronger argument for proving limits. Let’s introduce our argument with an example.

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Limits: Example

Example: Prove that lim

x→0(x + 4) = 4.

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Limits: Example

Example: Prove that lim

x→0(x + 4) = 4.

What we are trying to show that the sum of 4 and something very close to 0, is something very close to 4.

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Limits: Example

Example: Prove that lim

x→0(x + 4) = 4.

What we are trying to show that the sum of 4 and something very close to 0, is something very close to 4. The idea is as follows. You want to show that you can keep (x + 4) extremely close (as close as we like) to 4 by keeping x extremely close to 0.

Adrian Dudek The Formal Definition of a Limit 10 / 37

Limits: Example

Example: Prove that lim

x→0(x + 4) = 4.

What we are trying to show that the sum of 4 and something very close to 0, is something very close to 4. The idea is as follows. You want to show that you can keep (x + 4) extremely close (as close as we like) to 4 by keeping x extremely close to 0. It’s this idea of closeness that requires formalising.

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Limits: Example

You might be asked to keep 3.9 < x + 4 < 4.1. How would you do this?

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Limits: Example

You might be asked to keep 3.9 < x + 4 < 4.1. How would you do this? Adding −4 to each part of the inequality gives −0.1 < x < 0.1.

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Limits: Example

You might be asked to keep 3.9 < x + 4 < 4.1. How would you do this? Adding −4 to each part of the inequality gives −0.1 < x < 0.1. You might be asked to keep 3.9999 < x + 4 < 4.0001. How would you do this?

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Limits: Example

You might be asked to keep 3.9 < x + 4 < 4.1. How would you do this? Adding −4 to each part of the inequality gives −0.1 < x < 0.1. You might be asked to keep 3.9999 < x + 4 < 4.0001. How would you do this? Again, adding −4 to each part of the inequality gives −0.0001 < x < 0.0001.

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Limits: Example

Essentially, you are being asked how to keep 4 − ǫ < x + 4 < 4 + ǫ, and you are answering by saying that this is possible by −δ < x < δ.

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Limits: Example

Essentially, you are being asked how to keep 4 − ǫ < x + 4 < 4 + ǫ, and you are answering by saying that this is possible by −δ < x < δ. However, there are infinitely many ǫ’s that they can throw at you, and you don’t want to play this “choose δ” game forever! So you have to choose a δ that will beat any ǫ.

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A Silly Game

A particularly silly game which two people can play is “who can say the highest number”.

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A Silly Game

A particularly silly game which two people can play is “who can say the highest number”. This is a back-and-forth game just like the “choose delta” game which we just saw.

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A Silly Game

A particularly silly game which two people can play is “who can say the highest number”. This is a back-and-forth game just like the “choose delta” game which we just saw. Every time your oppenent says a number, you can just say the number that is one higher than their number to stay in the game!

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A Silly Game

A particularly silly game which two people can play is “who can say the highest number”. This is a back-and-forth game just like the “choose delta” game which we just saw. Every time your oppenent says a number, you can just say the number that is one higher than their number to stay in the game! However, rather than playing in this silly game, you could simply program a robot to respond for you with x + 1, where x is the number your opponent chooses.

Adrian Dudek The Formal Definition of a Limit 13 / 37

A Silly Game

A particularly silly game which two people can play is “who can say the highest number”. This is a back-and-forth game just like the “choose delta” game which we just saw. Every time your oppenent says a number, you can just say the number that is one higher than their number to stay in the game! However, rather than playing in this silly game, you could simply program a robot to respond for you with x + 1, where x is the number your opponent chooses. We want to do the same thing with epsilons and deltas!

Adrian Dudek The Formal Definition of a Limit 13 / 37

Limits: Example

Let’s get back to our example! So we wish to keep 4 − ǫ < x + 4 < 4 + ǫ by keeping −δ < x < δ.

Adrian Dudek The Formal Definition of a Limit 14 / 37

Limits: Example

Let’s get back to our example! So we wish to keep 4 − ǫ < x + 4 < 4 + ǫ by keeping −δ < x < δ. We simply do the same algebra we did to the specific examples: Add −4 to all sides of 4 − ǫ < x + 4 < 4 + ǫ to get: −ǫ < x < ǫ

Adrian Dudek The Formal Definition of a Limit 14 / 37

Limits: Example

Let’s get back to our example! So we wish to keep 4 − ǫ < x + 4 < 4 + ǫ by keeping −δ < x < δ. We simply do the same algebra we did to the specific examples: Add −4 to all sides of 4 − ǫ < x + 4 < 4 + ǫ to get: −ǫ < x < ǫ So in this case, ǫ = δ.

Adrian Dudek The Formal Definition of a Limit 14 / 37

Limits: Example

Let’s get back to our example! So we wish to keep 4 − ǫ < x + 4 < 4 + ǫ by keeping −δ < x < δ. We simply do the same algebra we did to the specific examples: Add −4 to all sides of 4 − ǫ < x + 4 < 4 + ǫ to get: −ǫ < x < ǫ So in this case, ǫ = δ. This is all we have to do to prove a limit! Provide a response δ in terms of any ǫ. Of course, there is a little bit more to write out, but the hard work is done!

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Limits: Graphical Example

Suppose we have a function f (x), and we wish to show that limx→3 f (x) = 5.

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Limits: Graphical Example

Here ǫ = 1, and so we must choose a δ which works. We can see that δ = 1 is a fine choice.

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Limits: Graphical Example

Here ǫ = 1/2, and so we must choose a δ which works. We can see that δ = 1/2 is a fine choice.

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Limits: Graphical Example

Here ǫ = 0.25, and so we must choose a δ which works. We can see that δ = 0.25 is a fine choice.

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Limits: Example

Example: Prove lim

x→2 3x + 4 = 10

That is, show that as x gets really close to 2, then 3x + 4 gets really close to 10.

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Limits: Example

We want 3x + 4 to be really close to 10. We do this by specifying that the distance between them remain less than any positive number ǫ: |3x + 4 − 10| < ǫ

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Limits: Example

We want 3x + 4 to be really close to 10. We do this by specifying that the distance between them remain less than any positive number ǫ: |3x + 4 − 10| < ǫ We want to show that this can be accomplished by keeping the distance between x and 2 less than any amount δ: |x − 2| < δ

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Limits: Example

We want 3x + 4 to be really close to 10. We do this by specifying that the distance between them remain less than any positive number ǫ: |3x + 4 − 10| < ǫ We want to show that this can be accomplished by keeping the distance between x and 2 less than any amount δ: |x − 2| < δ The problem is solved by establishing an answer δ in terms of ǫ, so that you have an answer for any ǫ they throw at you!

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Limits: Example

We usually proceed by rearranging the demand |3x + 4 − 10| < ǫ

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Limits: Example

We usually proceed by rearranging the demand |3x + 4 − 10| < ǫ into an inequality of the form |x − 2| < δ. Then we can simply read off what δ must be.

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Limits: Example

We usually proceed by rearranging the demand |3x + 4 − 10| < ǫ into an inequality of the form |x − 2| < δ. Then we can simply read off what δ must be. We start by writing the above without absolute value brackets. −ǫ < 3x + 4 − 10 < ǫ

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Limits: Example

We usually proceed by rearranging the demand |3x + 4 − 10| < ǫ into an inequality of the form |x − 2| < δ. Then we can simply read off what δ must be. We start by writing the above without absolute value brackets. −ǫ < 3x + 4 − 10 < ǫ Simplifying slightly we get −ǫ < 3x − 6 < ǫ

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Limits: Example

−ǫ < 3x − 6 < ǫ Dividing by 3 we get −ǫ/3 < x − 2 < ǫ/3

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Limits: Example

−ǫ < 3x − 6 < ǫ Dividing by 3 we get −ǫ/3 < x − 2 < ǫ/3 which is the same as |x − 2| < ǫ/3

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Limits: Example

−ǫ < 3x − 6 < ǫ Dividing by 3 we get −ǫ/3 < x − 2 < ǫ/3 which is the same as |x − 2| < ǫ/3 Thus, for any ǫ > 0 we choose, we would set δ = ǫ/3.

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Limits: Example

−ǫ < 3x − 6 < ǫ Dividing by 3 we get −ǫ/3 < x − 2 < ǫ/3 which is the same as |x − 2| < ǫ/3 Thus, for any ǫ > 0 we choose, we would set δ = ǫ/3. That is, by keeping |x − 2| < δ = ǫ/3, we guarantee that |3x + 4 − 10| < ǫ.

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Limits: Example

That is, by keeping |x − 2| < δ = ǫ/3, we guarantee that |3x + 4 − 10| < ǫ.

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Limits: Example

That is, by keeping |x − 2| < δ = ǫ/3, we guarantee that |3x + 4 − 10| < ǫ. We check as follows: |3x + 4 − 10| = |3x − 6|

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Limits: Example

That is, by keeping |x − 2| < δ = ǫ/3, we guarantee that |3x + 4 − 10| < ǫ. We check as follows: |3x + 4 − 10| = |3x − 6| = 3|x − 2|

Adrian Dudek The Formal Definition of a Limit 23 / 37

Limits: Example

That is, by keeping |x − 2| < δ = ǫ/3, we guarantee that |3x + 4 − 10| < ǫ. We check as follows: |3x + 4 − 10| = |3x − 6| = 3|x − 2|

Adrian Dudek The Formal Definition of a Limit 23 / 37

Limits: Example

That is, by keeping |x − 2| < δ = ǫ/3, we guarantee that |3x + 4 − 10| < ǫ. We check as follows: |3x + 4 − 10| = |3x − 6| = 3|x − 2| < 3ǫ/3

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Limits: Example

That is, by keeping |x − 2| < δ = ǫ/3, we guarantee that |3x + 4 − 10| < ǫ. We check as follows: |3x + 4 − 10| = |3x − 6| = 3|x − 2| < 3ǫ/3 < ǫ

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Limits: The Definition

In general, if we wish to show that lim

x→a f (x) = L

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Limits: The Definition

In general, if we wish to show that lim

x→a f (x) = L

then we need to show that the distance between f (x) and L can be made as small as we want, by making the distance between x and a sufficiently small.

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Limits: The Definition

In general, if we wish to show that lim

x→a f (x) = L

then we need to show that the distance between f (x) and L can be made as small as we want, by making the distance between x and a sufficiently small. That is, if somebody wants the distance between f (x) and L to be less than ǫ, then we need to show that there is some δ (in terms of ǫ) where keeping the distance between x and a less than δ guarantees this.

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Limits: The Definition

In general, if we wish to show that lim

x→a f (x) = L

then we need to show that the distance between f (x) and L can be made as small as we want, by making the distance between x and a sufficiently small. That is, if somebody wants the distance between f (x) and L to be less than ǫ, then we need to show that there is some δ (in terms of ǫ) where keeping the distance between x and a less than δ guarantees this. ∀ǫ > 0 ∃ δ > 0 s.t |x − a| < δ ⇒ |f (x) − L| < ǫ

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Limits: The Definition

Definition: We say that the limit of f (x) as x → a is L if ∀ǫ > 0 ∃ δ > 0 s.t |x − a| < δ ⇒ |f (x) − L| < ǫ

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Limits: The Definition

Definition: We say that the limit of f (x) as x → a is L if ∀ǫ > 0 ∃ δ > 0 s.t |x − a| < δ ⇒ |f (x) − L| < ǫ You need to remember this for tests and exams. Feel free to recite it at parties to test your memory!

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Infinite Limits

Sometimes we deal with limits as x → ±∞. One such example is: lim

x→∞

• 2 + 4

x

• = 2

This says, that as x gets really large, 2 + 4 x gets really close to 2.

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Infinite Limits

Sometimes we deal with limits as x → ±∞. One such example is: lim

x→∞

• 2 + 4

x

• = 2

This says, that as x gets really large, 2 + 4 x gets really close to 2. Once again, the way we prove this is the same! You want to show that the distance between 2 + 4 x and 2 can be made smaller than any positive number ǫ, by making x larger than a corresponding number N.

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Infinite Limits: Example

Example: Prove that lim

x→∞

• 2 + 4

x

• = 2

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Infinite Limits: Example

Example: Prove that lim

x→∞

• 2 + 4

x

• = 2

We want to keep

• 2 + 4

x − 2

• < ǫ. How large does x need to be to guarantee this? That

is, find a number N, where keeping x > N will guarantee this.

Adrian Dudek The Formal Definition of a Limit 27 / 37

Infinite Limits: Example

Example: Prove that lim

x→∞

• 2 + 4

x

• = 2

We want to keep

• 2 + 4

x − 2

• < ǫ. How large does x need to be to guarantee this? That

is, find a number N, where keeping x > N will guarantee this. Once again, we rearrange our original inequality for x.

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Infinite Limits: Example

We start with

• 2 + 4

x − 2

• < ǫ

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Infinite Limits: Example

We start with

• 2 + 4

x − 2

• < ǫ

which gives −ǫ < 4 x < ǫ

Adrian Dudek The Formal Definition of a Limit 28 / 37

Infinite Limits: Example

We start with

• 2 + 4

x − 2

• < ǫ

which gives −ǫ < 4 x < ǫ Multiplying through by x (keeping the inequality signs as they are, because x is positive) gives −ǫx < 4 < ǫx

Adrian Dudek The Formal Definition of a Limit 28 / 37

Infinite Limits: Example

We start with

• 2 + 4

x − 2

• < ǫ

which gives −ǫ < 4 x < ǫ Multiplying through by x (keeping the inequality signs as they are, because x is positive) gives −ǫx < 4 < ǫx The right hand side of this says that 4 < ǫx. We rearrange this to get x > 4 ǫ .

Adrian Dudek The Formal Definition of a Limit 28 / 37

Infinite Limits: Example

We start with

• 2 + 4

x − 2

• < ǫ

which gives −ǫ < 4 x < ǫ Multiplying through by x (keeping the inequality signs as they are, because x is positive) gives −ǫx < 4 < ǫx The right hand side of this says that 4 < ǫx. We rearrange this to get x > 4 ǫ . So N = 4 ǫ .

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Infinite Limits: Example

So, we claim that

• 2 + 4

x − 2

• < ǫ whenever x > N = 4

ǫ .

Adrian Dudek The Formal Definition of a Limit 29 / 37

Infinite Limits: Example

So, we claim that

• 2 + 4

x − 2

• < ǫ whenever x > N = 4

ǫ . To finish off the proof nicely we will show the first statement is true under the assumption of the second:

Adrian Dudek The Formal Definition of a Limit 29 / 37

Infinite Limits: Example

So, we claim that

• 2 + 4

x − 2

• < ǫ whenever x > N = 4

ǫ . To finish off the proof nicely we will show the first statement is true under the assumption of the second:

• 2 + 4

x − 2

• =
• 4

x

• Adrian Dudek

The Formal Definition of a Limit 29 / 37

Infinite Limits: Example

So, we claim that

• 2 + 4

x − 2

• < ǫ whenever x > N = 4

ǫ . To finish off the proof nicely we will show the first statement is true under the assumption of the second:

• 2 + 4

x − 2

• =
• 4

x

• The second statement re-arranges to give

4 x < ǫ.

Adrian Dudek The Formal Definition of a Limit 29 / 37

Infinite Limits: Example

So, we claim that

• 2 + 4

x − 2

• < ǫ whenever x > N = 4

ǫ . To finish off the proof nicely we will show the first statement is true under the assumption of the second:

• 2 + 4

x − 2

• =
• 4

x

• The second statement re-arranges to give

4 x < ǫ. Hence

• 2 + 4

x − 2

• =
• 4

x

• Adrian Dudek

The Formal Definition of a Limit 29 / 37

Infinite Limits: Example

So, we claim that

• 2 + 4

x − 2

• < ǫ whenever x > N = 4

ǫ . To finish off the proof nicely we will show the first statement is true under the assumption of the second:

• 2 + 4

x − 2

• =
• 4

x

• The second statement re-arranges to give

4 x < ǫ. Hence

• 2 + 4

x − 2

• =
• 4

x

• < |ǫ| = ǫ (since ǫ > 0).

Adrian Dudek The Formal Definition of a Limit 29 / 37

Infinite Limits: Example

So, we claim that

• 2 + 4

x − 2

• < ǫ whenever x > N = 4

ǫ . To finish off the proof nicely we will show the first statement is true under the assumption of the second:

• 2 + 4

x − 2

• =
• 4

x

• The second statement re-arranges to give

4 x < ǫ. Hence

• 2 + 4

x − 2

• =
• 4

x

• < |ǫ| = ǫ (since ǫ > 0).

We have proven the limit.

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Infinite Limits: The Definition

Definition: We say that the limit of f (x) as x → ∞ is L if ∀ǫ > 0 ∃ N > 0 s.t x > N ⇒ |f (x) − L| < ǫ

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Infinite Limits: The Definition

Definition: We say that the limit of f (x) as x → ∞ is L if ∀ǫ > 0 ∃ N > 0 s.t x > N ⇒ |f (x) − L| < ǫ Exercise: Try to write out the definition of a limit as x → −∞.

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Limit of a Sequence

What about limits of sequences? We might want to show that the sequence 1, 1/2, 1/3, 1/4 . . . converges to 0.

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Limit of a Sequence

The limit of a sequence requires the same sort of approach as for infinite limits.

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Limit of a Sequence

The limit of a sequence requires the same sort of approach as for infinite limits. If we believe a sequence an → L, then we must show that we can keep an arbitrarily close to L, by starting our sequence far enough to the right.

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Limit of a Sequence

The limit of a sequence requires the same sort of approach as for infinite limits. If we believe a sequence an → L, then we must show that we can keep an arbitrarily close to L, by starting our sequence far enough to the right. Starting our sequence far enough to the right means that the index n of our sequence an commences after some positive number N.

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Limit of a Sequence: Example

Let (an)n≥1 be the sequence defined by an = 1/(1 + n). Show that lim

n→∞ an = 0

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Limit of a Sequence: Example

Let (an)n≥1 be the sequence defined by an = 1/(1 + n). Show that lim

n→∞ an = 0

We want to show that |1/(1 + n) − 0| < ǫ whenever we keep n > N for some positive number N.

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Limit of a Sequence: Example

Let (an)n≥1 be the sequence defined by an = 1/(1 + n). Show that lim

n→∞ an = 0

We want to show that |1/(1 + n) − 0| < ǫ whenever we keep n > N for some positive number N. This will work just as before! We want to find N in terms of ǫ.

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Limit of a Sequence: Example

We start with |1/(1 + n) − 0| < ǫ

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Limit of a Sequence: Example

We start with |1/(1 + n) − 0| < ǫ which gives −ǫ < 1/(1 + n) < ǫ.

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Limit of a Sequence: Example

We start with |1/(1 + n) − 0| < ǫ which gives −ǫ < 1/(1 + n) < ǫ. Multiplying through by (1 + n) gives: −ǫ(1 + n) < 1 < ǫ(1 + n)

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Limit of a Sequence: Example

We start with |1/(1 + n) − 0| < ǫ which gives −ǫ < 1/(1 + n) < ǫ. Multiplying through by (1 + n) gives: −ǫ(1 + n) < 1 < ǫ(1 + n) Dividing through by ǫ gives: −(1 + n) < 1/ǫ < 1 + n

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Limit of a Sequence: Example

We start with |1/(1 + n) − 0| < ǫ which gives −ǫ < 1/(1 + n) < ǫ. Multiplying through by (1 + n) gives: −ǫ(1 + n) < 1 < ǫ(1 + n) Dividing through by ǫ gives: −(1 + n) < 1/ǫ < 1 + n

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Limit of a Sequence: Example

−(1 + n) < 1/ǫ < 1 + n Now rearranging the rightmost part of this inequality gives us n > 1/ǫ − 1.

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Limit of a Sequence: Example

−(1 + n) < 1/ǫ < 1 + n Now rearranging the rightmost part of this inequality gives us n > 1/ǫ − 1. So, to keep |1/(1 + n) − 0| < ǫ, we need to start the sequence off after N = 1/ǫ − 1. We show that this works as follows:

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Limit of a Sequence: Example

−(1 + n) < 1/ǫ < 1 + n Now rearranging the rightmost part of this inequality gives us n > 1/ǫ − 1. So, to keep |1/(1 + n) − 0| < ǫ, we need to start the sequence off after N = 1/ǫ − 1. We show that this works as follows: | 1 1 + n − 0| = 1 1 + n < 1 1 + 1/ǫ − 1 = 1 1/ǫ = ǫ and we are done.

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Limit of a Sequence: The Definition

Definition: We say that the limit of an as n → ∞ is L if ∀ǫ > 0 ∃ N > 0 s.t n > N ⇒ |an − L| < ǫ

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Limit of a Sequence: The Definition

Definition: We say that the limit of an as n → ∞ is L if ∀ǫ > 0 ∃ N > 0 s.t n > N ⇒ |an − L| < ǫ We often say an → L, rather than limn→∞ an = L.

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Using STUDYSmarter Resources This resource was developed for UWA students by the STUDYSmarter team for the numeracy program. When using our resources, please retain them in their original form with both the STUDYSmarter heading and the UWA crest.

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