Epsilon Delta and The Triangle Inequality

[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

 

[Pyrrhus]

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?

 

[DieCommie]

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 ?

 

[Benny]

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.

 

[DieCommie]

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&lt; |x-3| &lt; \delta, then
|x+3||x-3| \le (|x-3| + 6) |x-3| &lt; (\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 | &lt; \epsilon if 0&lt;|x-a|&lt;\delta

 

[arildno]

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

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

Your task is now:
How do I proceed proving this??

Here is one way:
1.Suppose that |x-3|&lt;\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|&lt;F(\delta) whenever |x-3|&lt;\delta

2. Why is this smart?
Well, what is left to you now to reach your end goal is:
Given \epsilon&gt;0, how small must I choose \delta&gt;0 so that F(\delta)&lt;\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|&lt;F(\delta^{*})&lt;\epsilon whenever |x-3|&lt;\delta^{*}


Does this clarify a bit?

 

[DieCommie]

\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| &lt; (\delta +6)\delta

 

[arildno]

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|&lt;\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|&lt;\delta^{2}+6\delta??

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

 

[DieCommie]

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| &lt; \deltaand multiply each side by (|x-3|+6)?

 

[arildno]

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

 

[DieCommie]

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

 

[VietDao29]

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| &lt; \delta <=> |x - 3||x - 3| &lt; \delta \delta
<=> |x - 3|^2 &lt; \delta^2
And you will also have 6 |x - 3| &lt; 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| &lt; \delta^2 + 6\delta
Get it?
Viet Dao,

 

[DieCommie]

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| &lt;\epsilon and |x-3|^2 +6|x-3| &lt;\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

 

[Pyrrhus]

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

Remember

|x - 3| &lt; \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.

 

[DieCommie]

So \mid x+3\mid\mid x-3\mid &lt; \epsilon if 0&lt; \mid x-3 \mid &lt; \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...

 

[Pyrrhus]

In my opinion the books complicates itself.

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

there must be a \epsilon &gt; 0 and \delta &gt; 0 so, for each x, if |x - 3| &lt; \delta, then |x^2 - 9| &lt; \epsilon

Now,

|x^2 - 9| &lt; \epsilon

Factorizing

|x + 3| |x - 3| &lt; \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| &lt; 1, so x is not too big, and therefore |x + 3| is not too big.

Now,

|x| - 3 \le |x - 3| &lt; 1 (1)

such as

|x| &lt; 1 + 3

therefore,

|x + 3| \le |x| + 3 &lt; 2(3) + 1

We can see

|x+3||x-3| &lt; 7|x-3|

and we need to make

7|x-3| &lt; \epsilon

so

|x-3| &lt; \frac{\epsilon}{7} (2)

and therefore

|x-3| &lt; min(1, \frac{\epsilon}{7})

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

 

[Pyrrhus]

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| &lt; 1, yes 1 looks rather arbitrary we could have picked any other positive number different from 1, and it will had worked the same.

 

[Pyrrhus]

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

 

[DieCommie]

Thx for trying to help me out man

 

[arildno]

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}&lt;b^{2}

Do you understand this?

 

[DieCommie]

Yes I understand that.

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

 

[arildno]

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

 

[DieCommie]

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

 

[arildno]

Precisely!
What we've actually managed to show now, is that given |x-3|&lt;\delta we have the upper bound:
|x^{2}-9|&lt;\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|&lt;\epsilon whenever |x-3|&lt;\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?

 

[arildno]

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&lt;\delta_{1}=1
It then follows that:
\delta^{2}+6\delta&lt;\delta\delta_{1}+6\delta=\delta*1+6\delta=7\delta
Thus, we have gained:
|x^{2}-9|&lt;7\delta
whenever: |x-3|&lt;\delta,\delta&lt;\delta_{1}

Now, let \epsilon&gt;0 be an arbitrary number.
As long as \delta&lt;\frac{\epsilon}{7}, we have:
7\delta&lt;\epsilon

Thus, we have gained:
|x^{2}-9|&lt;\epsilon
whenever: |x-3|&lt;\delta,\delta&lt;\delta_{1}=1,\delta&lt;\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|&lt;\epsilon
whenever |x-3|&lt;\delta, \delta&lt;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.

 

[arildno]

Now, the first thing you should do, is to ascertain that what I have done is VALID.
Secondly, if you are unsure about, or don't understand why I have made the particular choices I made, then ask about that.

 

[DieCommie]

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 &lt; \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 &lt; \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|&lt;\epsilon? Simply because it was stated in the begining?

4)The maxium values \delta can be are 1or\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.

 

[Pyrrhus]

DieCommie, please rest, and read again arildno's reply, it seems you didn't read it as it was presented.

 

[DieCommie]

What part didnt I read as it was presented?

 

[arildno]

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 &lt; \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 &lt; \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|&lt;\epsilon? Simply because it was stated in the begining?

OOOPS!
Here you've misunderstood the issue!
If we CHOOSE \delta to fulfill \delta&lt;min(1,\frac{\epsilon}{7}), then |x-3|&lt;\delta IMPLIES that |x^{2}-9|&lt;\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|&lt;\delta IMPLIES |x^{2}-9|&lt;\epsilon?"

4)The maxium values \delta can be are 1or\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).