# Epsilon-delta definition of limits

In summary, using the ε-δ definition of limit, the statement limx→2 x2 = 4 can be thought of as the result of some scratch work. However, there is the problem that x is in that expression. One way around this is to put a bound on |x + 2|. To explain that, think about this ε-δ game that's being played here. It's really only a challenge when ε is small. For instance, if δ =1, then you would need an ε ≥ 5.

## Homework Statement

Use the ε-δ definition of limits to prove that limx→2 x2 = 4.

## The Attempt at a Solution

|x2 - 4| < ε

0 < |x - 2| < δ

|x - 2| |x + 2| < ε

And that's where I get stuck, can I divide both sides by |x + 2| to yield

|x - 2| < ε/|x + 2| = δ

In which case, where do I go from there? Can I input 2 to get ε/4 = δ?

## Homework Statement

Use the ε-δ definition of limits to prove that limx→2 x2 = 4.

## The Attempt at a Solution

|x2 - 4| < ε

0 < |x - 2| < δ

|x - 2| |x + 2| < ε

And that's where I get stuck, can I divide both sides by |x + 2| to yield

|x - 2| < ε/|x + 2| = δ

In which case, where do I go from there? Can I input 2 to get ε/4 = δ?
First you need to be able to state, using the ε-δ definition of this limit, what is meant by the statement, limx→2 x2 = 4 .

Basically in ε-δ language we have:

Given any ε > 0, there exists (indeed we can find what its value is) a δ > 0 (this δ generally depends on ε) such that for any x that satisfies 0 < |x - 2| < δ it is true that |x2 - 4| < ε .

Often in figuring out what δ has to be, you will do some "scratch work" -- sort of working backwards. Then you turn things around and write up a nice proof, and in the process look like a genius.

Your statement ε/|x + 2| = δ can be thought of as the result of some of that scratch work. However, there is the problem that x is in that expression.

One way around this (for this particular limit) is put a bound on |x + 2| . To explain that, think about this ε-δ game that's being played here. It's really only a challenge when ε is small. For instance, if δ =1, then you would need an ε ≥ 5 . In other words, a δ of 1 will work just fine for any ε that's 5 or larger.

What is the largest value |x + 2| has for those values of x satisfying 0 < |x - 2| < 1 ?

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SammyS said:
First you need to be able to state, using the ε-δ definition of this limit, what is meant by the statement, limx→2 x2 = 4 .

Basically in ε-δ language we have:

Given any ε > 0, there exists (indeed we can find what its value is) a δ > 0 (this δ generally depends on ε) such that for any x that satisfies 0 < |x - 2| < δ it is true that |x2 - 4| < ε .

Often in figuring out what δ has to be, you will do some "scratch work" -- sort of working backwards. Then you turn things around and write up a nice proof, and in the process look like a genius.

Your statement ε/|x + 2| = δ can be thought of as the result of some of that scratch work. However, there is the problem that x is in that expression.

One way around this (for this particular limit) is put a bound on |x + 2| . To explain that, think about this ε-δ game that's being played here. It's really only a challenge when ε is small. For instance, if δ =1, then you would need an ε ≥ 5 . In other words, a δ of 1 will work just fine for any ε that's 5 or larger.

What is the largest value |x + 2| has for those values of x satisfying 0 < |x - 2| < 1 ?
Now I'm completely lost. Did you arbitrarily choose δ to equal 1? If δ = 1, why does it follow that ε ≥ 5? I get:

-δ < x - 5 < δ

5 - δ < x < δ + 5

δ = 1 ⇒ 4 < x < 6

ε = δ|x + 2|

= |x +2| (where x is between 4 and 6).

Edit: I don't know where I got 5 from, it should read:

-δ < x - 2 < δ

2 - δ < x < δ + 2

δ = 1 ⇒ 1 < x < 3

ε = δ|x + 2|

= |x +2| (where x is between 1 and 3).

Given some ##\epsilon >0 ## you need to show that there exists a ##\delta >0## such that if ## 2-\delta \leq x \leq 2+\delta ## then ## 4-\epsilon \leq x^2 \leq 4+ \epsilon ##.
Normally you are looking to define ##\delta ## as a function of ##\epsilon.##

RUber said:
Given some ##\epsilon >0 ## you need to show that there exists a ##\delta >0## such that if ## 2-\delta \leq x \leq 2+\delta ## then ## 4-\epsilon \leq x^2 \leq 4+ \epsilon ##.
Normally you are looking to define ##\delta ## as a function of ##\epsilon.##
How? I get no further than what I've written above.

Try squaring the limits on x, then setting the stuff that is added or subtracted from 4 equal to epsilon and then solving for delta in terms of epsilon.

RUber said:
Try squaring the limits on x, then setting the stuff that is added or subtracted from 4 equal to epsilon and then solving for delta in terms of epsilon.
If I understood you correctly:

4 - ε < x2 < ε + 4

2 - δ < x < 2 + δ ⇒ (2 - δ)2 < x2 < (2 + δ)2

(2 - δ)2 = 4 - ε ⇒ ε = 4δ - δ2

(2 + δ)2 = 4 + ε ⇒ ε = 4δ + δ2

4δ - δ2 = 4δ + δ2 ⇒ 2δ2 = 0 ⇒ δ = 0

But δ > 0 so this can't be right.

Now I'm completely lost. Did you arbitrarily choose δ to equal 1? If δ = 1, why does it follow that ε ≥ 5? I get:
-δ < x - 5 < δ
5 - δ < x < δ + 5
δ = 1 ⇒ 4 < x < 6
ε = δ|x + 2|
= |x +2| (where x is between 4 and 6).

Edit: I don't know where I got 5 from, it should read:

-δ < x - 2 < δ

2 - δ < x < δ + 2

δ = 1 ⇒ 1 < x < 3

ε = δ|x + 2|

= |x +2| (where x is between 1 and 3).
One thing you could do, since you edited that, use the strike-though or delete the unwanted part & leave a little note.
How? I get no further than what I've written above.
Now use some of that scratch work from the Original Post, together with these results.

To clean that up: You will never use δ > 1, Right? -- Note: δ will usually be much smaller than this.

So that x will be between 1 and 3 as you said. That means that | x+2 | is between 3 and 5, but the main thing is that it's less than 5.

So how does this help ?

Let's see what that does for your relation between between ε and δ .

You had ε = δ|x + 2| .

But what do you know about |x + 2| ?

(Well, we'll get to that scratch work in a minute.)

Do one at a time. The result should end up being the same when you take the absolute value.

SammyS said:
One thing you could do, since you edited that, use the strike-though or delete the unwanted part & leave a little note.
Now use some of that scratch work from the Original Post, together with these results.

To clean that up: You will never use δ > 1, Right? -- Note: δ will usually be much smaller than this.

So that x will be between 1 and 3 as you said. That means that | x+2 | is between 3 and 5, but the main thing is that it's less than 5.

So how does this help ?

Let's see what that does for your relation between between ε and δ .

You had ε = δ|x + 2| .

But what do you know about |x + 2| ?

(Well, we'll get to that scratch work in a minute.)
Sorry about that, I'll do it next time

3 < |x + 2| < 5

ε = δ|x + 2]

ε = 4δ ?

I think I'm just getting more confused, hahah.

RUber said:
Do one at a time. The result should end up being the same when you take the absolute value.
I'm not following; do what one at a time? :)

Sorry about that, I'll do it next time

3 < |x + 2| < 5

ε = δ|x + 2]

ε = 4δ ?

I think I'm just getting more confused, hahah.
All that's important here is the maximum possible value of |x + 2|.

It will all work out shortly.

So, if δ ≤ 1, what is the maximum value that |x + 2| can have?

SammyS said:
All that's important here is the maximum possible value of |x + 2|.

It will all work out shortly.

So, if δ ≤ 1, what is the maximum value that |x + 2| can have?
The maximum value is 4.999 or is it including 5? Making the maximum value 5 and making δ = ε/5?

The maximum value is 4.999 or is it including 5? Making the maximum value 5 and making δ = ε/5?
That's good.

Now, what's next ?

But you chose δ = ε/5 , right?

So | x - 2 | < ε/5 .

Now how does | x - 2 | relate to | x2 - 4 | ? (You have something anout this in your scratch-work.)

SammyS said:
That's good.

Now, what's next ?

But you chose δ = ε/5 , right?

So | x - 2 | < ε/5 .

Now how does | x - 2 | relate to | x2 - 4 | ? (You have something anout this in your scratch-work.)

|x - 2||x + 2| < e/5 |x + 2|

|x - 2||x + 2| < e/5 * 5

|x^2 - 4| < e

I don't want to interfere with a good discussion, but I have to go to bed, so if I don't post this now, it will have to wait until tomorrow. This is how I would describe what you should do:

Let ##\delta## be an arbitrary positive real number (for now), and let x be an arbitrary real number such that ##|x-2|<\delta##. Now try to derive an inequality of the form ##|x^2-4|<f(\delta)## where ##f## is some function. If this succeeds, then we know that it will be sufficient to find a positive ##\delta## such that ##f(\delta)=\varepsilon##. If you find this equation, solve it for ##\delta##, and one of the solutions turns out to be positive, then you will have found a sufficiently small ##\delta##.

If I understood you correctly:

4 - ε < x2 < ε + 4

2 - δ < x < 2 + δ ⇒ (2 - δ)2 < x2 < (2 + δ)2

(2 - δ)2 = 4 - ε ⇒ ε = 4δ - δ2

OR

(2 + δ)2 = 4 + ε ⇒ ε = 4δ + δ2

If you solve either or both of these for delta, you will end up with very similar forms. Hint: Complete the square.

|x - 2||x + 2| < e/5 |x + 2|
At this point, I would insert something like:
We know that |x + 2| ≤ 5 so that (ε/5)⋅|x + 2| ≤ (ε/5)⋅5​
Therefore, ... (what you have next)
|x - 2||x + 2| < e/5 * 5

|x^2 - 4| < e
Now there are a few details to take care of. Then writing up a nice proof.

What we have so far is the following. You have demonstrated that if δ ≤ 1, then choosing δ = ε/5 gives what is needed to prove that limx→2 x2 = 4 using the ε-δ definition of limit, namely:
For any x satisfying 0 < |x - 2| < δ it follows that |x2 - 4| < ε​

How should we handle the fact that we restricted ourselves to δ ≤ 1 ? Instead of saying δ = ε/5, we simply say that δ = the minimum of { 1, ε/5 } .I take that this ε-δ proof stuff is rather new to you and that you're pretty unsure where some of the details here come from. We can go through some of that later, if you're up to it.

For now, write up a nice proof.

If I understood you correctly:

4 - ε < x2 < ε + 4

2 - δ < x < 2 + δ ⇒ (2 - δ)2 < x2 < (2 + δ)2

(2 - δ)2 = 4 - ε ⇒ ε = 4δ - δ2

(2 + δ)2 = 4 + ε ⇒ ε = 4δ + δ2

4δ - δ2 = 4δ + δ2 ⇒ 2δ2 = 0 ⇒ δ = 0

But δ > 0 so this can't be right.
This method is different from the method I was suggesting. With this approach, you find the maximum possible δ for any particular ε .

In my opinion, neither method is better than the other. For some particular case one of them may be easier to implement than the other.

The choice ##\delta=\frac{\varepsilon}{5}## doesn't work when ##\varepsilon## is large. But perhaps you can find a number ##r## such that the choice ##\delta=\min\big\{\frac\varepsilon 5, r\big\}## will work.

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If I understood you correctly:

4 - ε < x2 < ε + 4

2 - δ < x < 2 + δ ⇒ (2 - δ)2 < x2 < (2 + δ)2

This only follows if $2 - \delta \geq 0$, ie. $\delta \leq 2$. (The square function is strictly increasing only on the non-negative reals.)

(2 - δ)2 = 4 - ε ⇒ ε = 4δ - δ2

(2 + δ)2 = 4 + ε ⇒ ε = 4δ + δ2

You have two upper bounds: you need both $4\delta - \delta^2 < \epsilon$ and $4\delta + \delta^2 < \epsilon$. Since $$4\delta - \delta^2 < 4\delta + \delta^2$$ the one to concentrate on is $4\delta + \delta^2 < \epsilon$. Now you can either argue that you need $\delta \in (0,2]$ and $\delta$ to lie between the roots of $P(t) = t^2 + 4t - \epsilon$, or you can use the known upper bound of 2 and require $$4\delta + \delta^2 = \delta(4 + \delta) \leq 6\delta < \epsilon.$$

## 1. What is the epsilon-delta definition of limits?

The epsilon-delta definition of limits is a mathematical concept used to define the precise meaning of a limit in calculus. It involves using two variables, epsilon (ε) and delta (δ), to describe the behavior of a function as it approaches a particular value.

## 2. How is the epsilon-delta definition of limits used in calculus?

The epsilon-delta definition of limits is used to determine the behavior of a function as it approaches a particular value, specifically in the context of limits. It allows for a more rigorous and precise understanding of limits in calculus.

## 3. What is the purpose of using epsilon and delta in the definition of limits?

The use of epsilon and delta in the definition of limits provides a way to quantitatively describe how close the values of a function are to a specific limit. Epsilon represents the desired level of closeness, while delta represents the corresponding distance between the input values and the limit.

## 4. How is the epsilon-delta definition of limits different from other approaches to defining limits?

The epsilon-delta definition of limits is more rigorous than other approaches, such as the intuitive approach or the limit laws approach. It provides a precise and quantitative definition of limits, allowing for a more thorough understanding of their behavior.

## 5. Can you give an example of using the epsilon-delta definition of limits?

One example of using the epsilon-delta definition of limits is to prove that the limit of the function f(x) = x^2 as x approaches 2 is equal to 4. This involves choosing a value for epsilon, such as 0.01, and then finding a corresponding value for delta that satisfies the definition. In this case, delta would be 0.1, meaning that for all x values within 0.1 of 2, the corresponding y values will be within 0.01 of 4.

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