# Epsilon Delta and The Triangle Inequality

• DieCommie
In summary: Now we have \epsilon \ge (\delta + 6) \delta Since \delta \le 1 we have 7 \delta \le 7 Therefore we can choose \delta = min \left (\frac{\epsilon}{7}, 1 \right ) And that's it.
DieCommie
I must prove $$\lim_{x\rightarrow 3\\} x^2 = 9$$
I get this...
$$\mid x+3\mid\mid x-3\mid < \epsilon$$ if $$0< \mid x-3 \mid < \delta$$

then it says with the triangle inequality we see that
$$\mid x+3\mid = \mid (x-3)+6\mid \le \mid x-3\mid +6$$

therefore if $$0< \mid x-3 \mid < \delta$$ , then
$$\mid x+3\mid\mid x-3\mid \le (\mid x-3\mid+6)\mid x-3\mid < (\delta+6)\delta$$

What I don't understand is how to get this term.. $$(\delta+6)\delta$$ ?

Thanks

Let's go a little back...

$$\mid x+3\mid\mid x-3\mid < \epsilon$$ if $$0< \mid x-3 \mid < \delta$$

we could try to define a positive constant A such as

$$|x+3| < A$$

therefore

$$|x+3||x-3| < A|x-3|$$

Now you need to make $A|x-3| < \epsilon$ so $|x-3| < \frac{\epsilon}{A} = \delta$

can you take it from there?

Im afraid I am not very good at this...

I don't understand

$$|x+3| < A$$ how this
therefore
$$|x+3||x-3| < A|x-3|$$ leads to this

Does $$A|x-3| = \epsilon$$ ?

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Ok, well say for example |x+3| < 6. Then would you agree that 3|x+3| < 3(6) 18? Its the same idea.

You'll find that in a lot of e-d problems you come across, you'll need to use the fact that |x-a| < d in some way. Oh and just one more thing(which you probably shouldn't think about too much until you're able to understand this problem). Consider limits where x -> (a negative number). You'll find that a slightly different strategy is required for a lot of problems of this type.

Thx for the help guys, I understand a little more, but I don't get it.

This is what the book says

"Using the triangle inequality, we see that
$$|x+3| = |(x-3)+6| \le |x-3|+6$$ "
I understand how this works, but i don't quite understand why they are doing it. I see the (x-3) is the goal...

The text continues...
"Therfore, if $$0< |x-3| < \delta$$, then
$$|x+3||x-3| \le (|x-3| + 6) |x-3| < (\delta +6)\delta$$ " I don't understand this equation at all, what it represents or how they achieved it. In the previous examples my end goal was to get $$|f(x) - L | < \epsilon$$ if $$0<|x-a|<\delta$$

In the previous examples my end goal was to get $$|f(x) - L | < \epsilon$$ if $$0<|x-a|<\delta$$

That is still your end goal!
In your case, you want to show that given any $$\epsilon>0$$ you can find a sufficiently small $$\delta-$$ neighboorhood about 3, so that
$$|x^{2}-9|<\epsilon$$ whenever you've got $$|x-3|<\delta$$

How do I proceed proving this??

Here is one way:
1.Suppose that $$|x-3|<\delta$$
What is then an upper bound of the expression $$|x^{2}-9|$$??
Let us for simplicity say we can calculate this upper bound; this will depend on the actual value of $$\delta$$, so that our first task can be said to be:
Find a function $$F(\delta)$$ so that we always have:
$$|x^{2}-9|<F(\delta)$$ whenever $$|x-3|<\delta$$

2. Why is this smart?
Well, what is left to you now to reach your end goal is:
Given $$\epsilon>0$$, how small must I choose $$\delta>0$$ so that $$F(\delta)<\epsilon$$?
Let us call a particular delta-value achieving this $$\delta^{*}$$

Note that this will be sufficient; since by combining 1.and 2., we have:
$$|x^{2}-9|<F(\delta^{*})<\epsilon$$ whenever $$|x-3|<\delta^{*}$$

Does this clarify a bit?

$$\delta^*=\delta^2 ?$$
Im sorry i don't understand what you have done...
You are looking for the upper bound of $$|x^{2}-9|$$ ?
Im not sure how this relates to $$|x+3||x-3| \le (|x-3| + 6) |x-3| < (\delta +6)\delta$$

1. $$|x^{2}-9|=|x+3||x-3|$$
Agreed on that?

2. Do you also agree to, that for any x we have:
$$|x+3|=|(x-3)+6|\leq|x-3|+6$$

3. Combining 1. and 2., do you agree that we have:
$$|x^{2}-9|=|x+3||x-3|\leq(|x-3|+6)|x-3|=|x-3|^{2}+6|x-3|$$

4. Thus, if $$|x-3|<\delta$$ do you agree that we have the following upper bound on $$|x^{2}-9|$$:
$$|x^{2}-9|\leq|x-3|^{2}+6|x-3|<\delta^{2}+6\delta$$??

5. Set $$F(\delta)=\delta^{2}+6\delta$$
Are you following so far?

I can't follow you on the second half of #4...

I see that $$|x^2 -9|\le|x-3|^2+6|x-3|$$
but where did $$\delta^2 +6\delta$$ come from?

you take $$|x-3| < \delta$$and multiply each side by $$(|x-3|+6)$$?

Well, if $$|x-3|<\delta$$ then we obviously must have $$|x-3|^{2}<\delta^{2}$$ right?

so you square each side then multiply each side by +6?

I think I can make it a little bit clear for you.
For all positive a, b, c, d such that a < b, and c < d, you have ac < bd.
First, |x - 3| is a non-negative number. Lim x -> 3, means that the 'x' is closed to 3 (but not x = 3). But sice, you just need the 'x' closed to 3, So |x - 3| is a positive number. $\delta$ is also a posituve number.
There fore : $|x - 3| < \delta$ <=> $|x - 3||x - 3| < \delta \delta$
<=> $|x - 3|^2 < \delta^2$
And you will also have $6 |x - 3| < 6 \delta$ 6 is a positive number.
And you have for all a, b, c, d, such that a < b, c < d : a + c < b + d.
So $|x - 3|^2 + 6 |x - 3| < \delta^2 + 6\delta$
Get it?
Viet Dao,

Ok I see what you are doing Thx for all your help, I am really struggling with this whole concept...

So now I know that $$|x-3|^2 +6|x-3| <\epsilon$$ and $$|x-3|^2 +6|x-3| <\delta^2 +6\delta$$

Now I need to some how use this to prove $$\lim_{x\rightarrow 3\\} x^2 = 9$$

The next part in the book is very confusing to me...

"Let us restrict our attention to positive values of $$\delta$$ such that $$(\delta +6)\delta \le \epsilon$$."
Then it shows $$(\delta +6)\delta \le 7\delta$$ and $$7\delta \le \epsilon$$
Finally it shows $$\delta = min(\epsilon/7,1)$$
... and that's the end of the proof

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If we make $|x - 3| < 1$ so we can restrict x to not be too big and in consequence $|x + 3|$ won't be too big.

Remember

$$|x - 3| < \delta$$

therefore, our delta can be less or equal to 1.

$$\delta \le 1$$

so

$$\delta + 6 \le 7$$

$$\delta(\delta + 6) \le 7 \delta$$

Then it follows the proof of your book.

So $$\mid x+3\mid\mid x-3\mid < \epsilon$$ if $$0< \mid x-3 \mid < \delta$$ is satisfied if x is 1 or 7 units from a(3)?

What makes the delta less than or equal to one? I know we need to approach 1 closely, but 1 seems kind of arbitrary...

In my opinion the books complicates itself.

to prove $$\lim_{x \rightarrow 3} x^2 = 9$$

there must be a $$\epsilon > 0$$ and $$\delta > 0$$ so, for each x, if $$|x - 3| < \delta$$, then $$|x^2 - 9| < \epsilon$$

Now,

$$|x^2 - 9| < \epsilon$$

Factorizing

$$|x + 3| |x - 3| < \epsilon$$

The problem is with making $|x + 3|$ small, so we need to find a bound so we can "tweak" $|x - 3|$.

Now, let's make $|x - 3| < 1$, so x is not too big, and therefore $|x + 3|$ is not too big.

Now,

$$|x| - 3 \le |x - 3| < 1 (1)$$

such as

$$|x| < 1 + 3$$

therefore,

$$|x + 3| \le |x| + 3 < 2(3) + 1$$

We can see

$$|x+3||x-3| < 7|x-3|$$

and we need to make

$$7|x-3| < \epsilon$$

so

$$|x-3| < \frac{\epsilon}{7} (2)$$

and therefore

$$|x-3| < min(1, \frac{\epsilon}{7})$$

This means $\delta = min(1, \frac{\epsilon}{7})$ so we can guarantee both inequalities (2) and (1) work.

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We are only interested in the values of x close to 3, so we can suppose x is in a distance of 1 to 3, this means $|x - 3| < 1$, yes 1 looks rather arbitrary we could have picked any other positive number different from 1, and it will had worked the same.

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DieCommie, if you still don't understand, this thread explains delta-epsilon proofs quite well in my opinion.

Thx for trying to help me out man

DieCommie:
Your basic problem seems to be that you don't understand what is meant by the logical structure IF "A" THEN "B" ("A" implies "B")

Here's how you must learn to think in a particular case:
If a number "a" is less than a number "b", this implies a lot of other statements.
For example:
If a<b, THEN a+1<b+1
Or:
If a<b and a,b are both non-negative numbers, then $$a^{2}<b^{2}$$

Do you understand this?

Yes I understand that.

I don't see how that proves the limit is 9...

So, do you now get that IF $$|x-3|<\delta$$, THEN
$$|x-3|^{2}+6|x-3|<\delta^{2}+6\delta$$?

Yes I see that. Now I need to relate $$|x-3|^{2}+6|x-3|<\delta^{2}+6\delta$$ to $$\epsilon$$ somehow?

Precisely!
What we've actually managed to show now, is that given $$|x-3|<\delta$$ we have the upper bound:
$$|x^{2}-9|<\delta^{2}+6\delta$$
(You can think of my F-function as $$F(\delta)=\delta^{2}+6\delta$$)

Note that as yet, we have placed NO restrictions upon $$\delta$$!

Such restrictions will appear when we want to deduce
$$|x^{2}-9|<\epsilon$$ whenever $$|x-3|<\delta$$

that is, in the final step we need in order to prove that $$\lim_{x\to3}x^{2}=9$$

are you following thus far?

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Now, as long as we keep in mind which restrictions we place on $$\delta$$, we may choose them in order to simplify our work.
Let us therefore assume $$\delta<\delta_{1}=1$$
It then follows that:
$$\delta^{2}+6\delta<\delta\delta_{1}+6\delta=\delta*1+6\delta=7\delta$$
Thus, we have gained:
$$|x^{2}-9|<7\delta$$
whenever: $$|x-3|<\delta,\delta<\delta_{1}$$

Now, let $$\epsilon>0$$ be an arbitrary number.
As long as $$\delta<\frac{\epsilon}{7}$$, we have:
$$7\delta<\epsilon$$

Thus, we have gained:
$$|x^{2}-9|<\epsilon$$
whenever: $$|x-3|<\delta,\delta<\delta_{1}=1,\delta<\frac{\epsilon}{7}$$

Thus, if $$\delta$$ is chosen to be the MINIMUM value of $$1,\frac{\epsilon}{7}$$, we are finished!

That is, we have shown:
$$|x^{2}-9|<\epsilon$$
whenever $$|x-3|<\delta, \delta<min(1,\frac{\epsilon}{7})$$

Note that the restriction on $$\delta$$ is now dependent upon $$\epsilon$$, and that whenever we know $$\epsilon$$ we can readily determine an acceptable $$\delta$$-value.

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Now, the first thing you should do, is to ascertain that what I have done is VALID.

Thank you for sticking around and helping me.

I have gone through the steps, and I see that how it is all valid.

I do have a few particulars I still don't understand, as you predicted.

1)you choose $$\delta < \delta_1 = 1$$ arbitrarily.. Is that enable you to solve the delta expression in terms of epsilon?

2)When you did this $$\delta^2 + 6x < \delta\delta_1 + 6\delta = 7\delta$$ why did you choose to replace only one of the deltas with 1? And why did you choose to replace that particular delta?

3)How do we know this is all true if $$[|x^2 - 9|<\epsilon$$? Simply because it was stated in the begining?

4)The maxium values $$\delta$$ can be are $$1$$or$$\epsilon/7$$ right? I don't understand why one has epsilon in it and one does not. Why is this and what does it imply?

Thx again for helping me through this. Our teacher just glanced over it and didnt even test us on it, but I still really want to know. It is late and I am tired so I hope this made sense.

What part didnt I read as it was presented?

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DieCommie said:
Thank you for sticking around and helping me.

I have gone through the steps, and I see that how it is all valid.

I do have a few particulars I still don't understand, as you predicted.
1)you choose $$\delta < \delta_1 = 1$$ arbitrarily.. Is that enable you to solve the delta expression in terms of epsilon?
I did it because it would simplify my work further on; it certainly wasn't a NECESSARY step.
2)When you did this $$\delta^2 + 6x < \delta\delta_1 + 6\delta = 7\delta$$ why did you choose to replace only one of the deltas with 1? And why did you choose to replace that particular delta?
Again, a VALID simplification trick.[/QUOTE]
3)How do we know this is all true if $$[|x^2 - 9|<\epsilon$$? Simply because it was stated in the begining?
OOOPS!
Here you've misunderstood the issue!
If we CHOOSE $$\delta$$ to fulfill $$\delta<min(1,\frac{\epsilon}{7})$$, then $$|x-3|<\delta$$ IMPLIES that $$|x^{2}-9|<\epsilon$$
This is what we're after, not the other way around!
The "restrictions" (or conditions) I mentioned are to be understood in the following manner:
"What restrictions/conditions must be placed upon $$\delta$$ so that $$|x-3|<\delta$$ IMPLIES $$|x^{2}-9|<\epsilon$$?"
4)The maxium values $$\delta$$ can be are $$1$$or$$\epsilon/7$$ right? I don't understand why one has epsilon in it and one does not. Why is this and what does it imply?
Read again, in particular my comment on 3), and if you still have problems, let's tackle those.

Remember that to be able to make valid and smart simplification tricks is a matter of experience (you'll develop a skill in this through practice).

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## 1. What is Epsilon Delta?

Epsilon Delta is a mathematical concept used in calculus to define limits. It involves using the symbols epsilon (ε) and delta (δ) to represent small numbers and their relationship to a given function or sequence.

## 2. How is Epsilon Delta used in calculus?

Epsilon Delta is used to formally define the concept of a limit in calculus. It allows us to determine the behavior of a function or sequence as its input approaches a certain value, known as the limit point.

## 3. What is the Triangle Inequality?

The Triangle Inequality is a mathematical principle that states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In other words, it ensures that the shortest distance between two points is a straight line.

## 4. How is the Triangle Inequality related to Epsilon Delta?

The Triangle Inequality is often used in conjunction with Epsilon Delta to prove the existence of limits in calculus. It helps to show that the difference between a function's output and its limit can be made arbitrarily small by choosing an appropriate input value.

## 5. What are some real-world applications of Epsilon Delta and the Triangle Inequality?

Epsilon Delta and the Triangle Inequality have various applications in fields such as physics, engineering, and economics. For example, they can be used to analyze the behavior of moving objects, design optimal routes for transportation, and model supply and demand in markets.

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