# Epsilon Delta and The Triangle Inequality

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 dont understand is how to get this term.. $$(\delta+6)\delta$$ ?

Thanks

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Pyrrhus
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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 im not very good at this...

I dont 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 dont 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 dont 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 dont 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 $$0<|x-a|<\delta$$ arildno
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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 dont understand what you have done...
You are looking for the upper bound of $$|x^{2}-9|$$ ?
Im not sure how this relates to $$|x+3||x-3| \le (|x-3| + 6) |x-3| < (\delta +6)\delta$$

arildno
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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 cant 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 $$(|x-3|+6)$$?

arildno
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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?

VietDao29
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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, im 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 thats the end of the proof Last edited:
Pyrrhus
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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...

Pyrrhus
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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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Pyrrhus
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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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Pyrrhus
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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

arildno
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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 dont see how that proves the limit is 9...

arildno
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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?

arildno
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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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arildno
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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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