Why is l x - 3 l useful in showing the limit of f(x) = x^2 near 3?

[BloodyFrozen]

Foundations of "Formal Limits"

In Spivak's 4th Edition, one examples says:

"The function f(x) = x² is a little more interesting. Presumably, we should be able to show that f(x) approaches 9 near 3. This means that we need to be small enough to show how to ensure the inequality

l x² - 9 l < ε

for any given positive number ε by requiring l x - 3 l to be small enough. The obvious first step is to write

l x² - 9 l = l x - 3 l* l x + 3 l,

which gives use the useful l x - 3 l factor. Unlike the situation with the previous examples, however, the extra factor is l x + 3 l, which isn't a convenient constant like 3 or 3,000,000. But the only crucial thing is to make sure that we can say something about how big l x + 3 l is. ............

MY question is why is l x - 3 l useful and not l x + 3 l?

Also, I don't really understand this. Anyone care to explain? (I don't think this is ε-δ because δ hasn't been introduced yet.)

so confusing...

 

[Unit]

Spivak is outlining the general idea behind the formal definition of a limit. I'm not entirely sure if the following is what you're looking for, but ...

|x - 3| is useful because |x - 3| < δ. This means that x is within δ of 3. This is important, because we can relate the |x - 3| with δ, and create an interval around x: (3-δ, 3+δ).

|x + 3| is not so useful because it doesn't seem immediately obvious how to relate it to δ. However, with a bit of careful manipulation, you can arrive at 6 - δ < x + 3 < 6 + δ, (given that 0 < |x - 3| < δ). From here, so long as δ is sufficiently small (i.e. δ < 1), we can say that |x + 3| < 6 + δ < 6 + 1 = 7.

We are now able to write the product |x - 3||x + 3| entirely in terms of δ. This is important! Since |x - 3| < δ and |x + 3| < 7, we get that |x^2 - 9| < 7δ. But what is δ?

|x^2 - 9| represents the "vertical difference" between the function evaluated at x-values near 3, and the desired vertical height, 9. To show that the function approaches 9 as x approaches 3, we need to show that the difference |x^2 - 9| gets arbitrarily small. This is where ε comes in. Taking any ε > 0, we need to find a small-enough interval (3-δ, 3+δ) such that for all x inside this interval, |x^2 - 9| is less than ε. To find this small-enough interval, we need to pick an appropriate δ.

It is good to pick δ in terms of ε. We have already shown that |x^2 - 9| < 7δ. For any ε > 0, pick δ to be min{1, ε/7}. Then we know that |x^2 - 9| < 7δ < 7(ε/7) = ε.

I hope this helps!

 

[BloodyFrozen]

How did you get 6-delta<x+3<6+delta?

Also, in this (or any) example, what is delta exactly as Spivak has not mentioned it yet.

 

[BloodyFrozen]

Spivak 4th Edition, Chapter 5 Limits

Can someone explain to me what Spivak is trying to say about Limits?

I know the epsilon-delta definition, but he doesn't use delta only epsilon.

I'm trying to get a good idea on what he's trying to say because it seems like he uses this kind of definition quite often.

 

[micromass]

Spivak does use the epsilon-delta definition, but he introduces it quite late (on p.84 in my edition). All the pages before that are just a motivation for the definition and showing that it works like we expect it to work.

In the beginning of the chapter, he introduces a provisional definition:

The function f approaches the limit l near a, if we can make f(x) as close as we like to l by requiring that x be sufficiently close to, but unequal to, a.​

This is a very informal definition which is only meant to give you some intuition on the subject. But this actually is an epsilon-delta definition, as you will see later.

If you already know epsilon-delta, then it might be a good exercise for you to rephrase all the intuitive things that he does in formal language...

Does this answer your question a bit?

 

[Stephen Tashi]

BloodyFrozen,

Perhaps dealing with the "simplified" examples of the definition is actually more confusing than dealing with the actual definition.

MY question is why is l x - 3 l useful and not l x + 3 l?

The number |x -3 | indicates how close x is to 3. It is small when x is close to 3. The value 3 is where we wish to find the limit of the function. The number | x + 3 | is small when x is near -3, but in this probem, we aren't trying to find out what happens to the function when x is near -3.

 

[BloodyFrozen]

Hey micromass

Well, my problem is partly on that.

On one of his problems, he shows that f(x)=x² approaches 9 near 3.

He then proceeds the show that

l x²-9 l < ε

He says that we need to require l x - 3 l to be small enoguh for any given ε.

Proceeding,

l x^2 - 9l = l x - 3 l*l x +3 l, {I get it up to this part}

He shows that l x - 3 l is the useful factor.

Then says we need to say something about how big l x + 3 l is.
so...

l x - 3 l < 1 or 2 < x < 4 then,
5 < x + 3 < 7 ------> this means

l x^2 - 9l = l x - 3 l*l x +3 l < 7*l x - 3 l


Why does he say l x - 3 l is useful and not l x + 3 l?

5 < x + 3 < 7 ----> At here he added 3 to both sides right?

l x^2 - 9l = l x - 3 l*l x +3 l < 7*l x - 3 l ---------> Don't get this part...

Reminder, he does nothing with delta yet.

 

[BloodyFrozen]

BloodyFrozen,

Perhaps dealing with the "simplified" examples of the definition is actually more confusing than dealing with the actual definition.



The number |x -3 | indicates how close x is to 3. It is small when x is close to 3. The value 3 is where we wish to find the limit of the function. The number | x + 3 | is small when x is near -3, but in this probem, we aren't trying to find out what happens to the function when x is near -3.

Oh, so l x - 3 l = 0

So we are trying to find as f(x) approaches 3?

What is Unit saying about using l x + 3 l then ?

 

[I like Serena]

This problem is also currently in progress here: https://www.physicsforums.com/showthread.php?t=507802

 

[BloodyFrozen]

This problem is also currently in progress here: https://www.physicsforums.com/showthread.php?t=507802

Yeah.. I totally forgot I had this one until today. I guess one can be locked/deleted.

 

[I like Serena]

Yeah.. I totally forgot I had this one until today. I guess one can be locked/deleted.

Not likely, looking at the timestamps. I don't mind, especially since you're still new to the forum.

But I do consider it impolite to Unit, Steven Tashi, and micromass, who are all doing their utmost to help you.
Being a homework helper myself, I know how it feels if you try your utmost to help someone, only to discover he just tuned out (without serious effort) and asked someone else, and that without me even knowing.

It's not a big deal, just don't do it again, please.

 

[Stephen Tashi]

Oh, so l x - 3 l = 0

No. The idea of limit is very subtle. "the limit of f(x) as x approaches 3" doesn't depend on what f(x) is when x is exactly 3. Notice the problem involves [itex] 0 < |x-3| < \delta [/itex], not simply [itex] |x - 3| < \delta [/itex].

For the functions encountered in elementary algebra, it often happens that "the limit of f(x) as x approaches a" is the number f(a). Such a function is called "continuous". But there are functions which are not continuous.

So we are trying to find as f(x) approaches 3?

We are trying to prove "the limit of f(x) as x approaches 3" is 9.

What is Unit saying about using l x + 3 l then ?

I'll let Unit explain that. The general form of limit proofs is that you pretend that someone gave you an [itex] \epsilon [/itex] and you must prove you can find a [itex] \delta [/itex]. To prove that, you often do steps that amount to solving for [itex] \delta [/itex] as a function of [itex] \epsilon [/itex]. Since we are dealing with inequalities instead of equations, you'll see quite a variety of ways of solving for [itex] \delta [/itex]. These will include verbal arguments as well as symbolic manipulations. You can't rely on learning one particular method for doing it.

 

[Unit]

How did you get 6-delta<x+3<6+delta?

Also, in this (or any) example, what is delta exactly as Spivak has not mentioned it yet.

I see micromass has also explained in https://www.physicsforums.com/showthread.php?t=507889 what Spivak is trying to do. Micromass is correct: it is just an intuitive idea of a limit.

As for your questions:

If I am given that |x - 3| < δ, it means that x is inside (3 - δ, 3 + δ). That's a tiny little interval centered around 3. If you're not familiar with how |a| < b implies -b < a < b, then the following approach will explain it more clearly: since x is in (3 - δ, 3 + δ), 3 - δ < x < 3 + δ. With me so far? :)

Now, we are interested in saying something about the factor |x + 3|. Well, since we have 3 - δ < x < 3 + δ, we can surely say something about x + 3 (without absolutes). Add 3 to each part of the inequality: 3 - δ + 3 < x + 3 < 3 + δ + 3. This simplifies to 6 - δ < x + 3 < 6 + δ. Since δ is small, we can say δ < 1, and write the inequality as 6 - 1 < 6 - δ < x + 3 < 6 + δ < 6 + 1, or, 5 < x + 3 < 7. Since (x + 3) > 5, (x + 3) is positive, so (x + 3) = |x + 3|.

Your other question:

Delta is what makes the small little interval around your desired point (in this case, 3). Delta is "how close x is to 3". Delta is all up to you: you need to pick an appropriate delta (in terms of epsilon) that will satisfy any given epsilon.

 

[Unit]

l x - 3 l < 1 or 2 < x < 4 then,
5 < x + 3 < 7 ------> this means

l x^2 - 9l = l x - 3 l*l x +3 l < 7*l x - 3 l

Why does he say l x - 3 l is useful and not l x + 3 l?

5 < x + 3 < 7 ----> At here he added 3 to both sides right?

l x^2 - 9l = l x - 3 l*l x +3 l < 7*l x - 3 l ---------> Don't get this part...

Reminder, he does nothing with delta yet.

So, no deltas involved yet. Perfectly fine. He has said that |x - 3| < 1. (Note: This "1" is like delta.) Anyway. If |x - 3| < 1, then 2 < x < 4. So adding 3 to all sides (you said "both sides"), we get 5 < x + 3 < 7. Like I explained in the other thread, (x + 3) > 0, so |x + 3| = (x + 3).

For the part you don't get, it's just the multiplicative property of inequalities. You should look over that section in your text. If I have c < d and a number b > 0, then bc < bd. So here, I have |x + 3| < 7, and a number |x - 3| > 0. So, |x - 3|*|x + 3| < |x - 3|*7.

There are countless tangible examples of the multiplicative property if you just try any numbers: if 2 < 3 and 5 > 0, then 2*5 < 3*5 i.e. 10 < 15.

 

[BloodyFrozen]

Ok, thanks to everyone:)

 

[sponsoredwalk]

Why does he say l x - 3 l is useful and not l x + 3 l?

5 < x + 3 < 7 ----> At here he added 3 to both sides right?

l x^2 - 9l = l x - 3 l*l x +3 l < 7*l x - 3 l ---------> Don't get this part...

Reminder, he does nothing with delta yet.

Far better way to think about this part is as follows:


|x² - 9| = |(x + 3)(x - 3)| = |x + 3||x - 3|
Now take the |x + 3| part & manipulate it so that
|x + 3| = |x + 3 +(3 - 3)| = |x - 3 + 6| ≤ |x - 3| + |6|
So if we stipulate that as long as |x - 3| < δ we can have
|x + 3| ≤|x - 3| + |6| < δ + |6|.

So going back to the original:
|x² - 9| = |(x + 3)(x - 3)| = |x + 3||x - 3| ≤ (δ + |6|)|x - 3| < ε
Therefore if we let |x - 3| < δ = [ε / (δ + |6|)]

you get:

|x - 3| < δ ---> |x² - 9| = |(x + 3)(x - 3)| = |x + 3||x - 3| < (δ + |6|)|x - 3| < (δ + |6|)[ε / (δ + |6|)] = ε.

http://math.ucalgary.ca/~jling/calculus/examples/limitdefinition.pdf

-----------------

Just curious, I set δ = [ε / (δ + |6|)], does anybody have a good reason why this is
a bad idea? It seems circular to include 2 δ' but I don't see how it is & actually I think it's
a good idea! I don't see why people should magically pick δ = 1 as was done above when
people chose 7 as a safe bound, but I could be wrong! Thanks!

 

[Unit]

|x² - 9| = |(x + 3)(x - 3)| = |x + 3||x - 3|
Now take the |x + 3| part & manipulate it so that
|x + 3| = |x + 3 +(3 - 3)| = |x - 3 + 6| ≤ |x - 3| + |6|
So if we stipulate that as long as |x - 3| < δ we can have
|x + 3| ≤|x - 3| + |6| < δ + |6|.

So going back to the original:
|x² - 9| = |(x + 3)(x - 3)| = |x + 3||x - 3| ≤ (δ + |6|)|x - 3| < ε
Therefore if we let |x - 3| < δ = [ε / (δ + |6|)]

you get:

|x - 3| < δ ---> |x² - 9| = |(x + 3)(x - 3)| = |x + 3||x - 3| < (δ + |6|)|x - 3| < (δ + |6|)[ε / (δ + |6|)] = ε.

-----------------

Just curious, I set δ = [ε / (δ + |6|)], does anybody have a good reason why this is
a bad idea? It seems circular to include 2 δ' but I don't see how it is & actually I think it's
a good idea! I don't see why people should magically pick δ = 1 as was done above when
people chose 7 as a safe bound, but I could be wrong! Thanks!

I like your method! There is no real problem with setting δ = [ε / (δ + 6)] except an aesthetic one. Suppose I gave you ε = 1. Can you easily find a δ such that δ = [1 / (δ + 6)]? You can; with the quadratic formula, it's roughly 0.1623. Rigourously, this is fine, but for "prettiness", I'd impose an early restriction that δ be less than 1 and continue your line thus:

|x + 3| ≤|x - 3| + |6| < δ + |6| < 1 + |6| = 7
This way, setting δ < min{1, ε/7}, we have:
|x² - 9| = |(x + 3)(x - 3)| = |x + 3||x - 3| ≤ 7|x - 3| < 7δ < 7(ε/7) = ε.

 

[BloodyFrozen]

Ok, so if we were to do this for lim_x->3 x²+2x-11= 4

We would say...

l x²+2x-15 l < ε and 0 < l x-3 l < δ

l x+5 l*l x-3 l < ε , we assume that l x+5 l < F (any random variable), so that
l x-3 l < ε/F
l x-3 l < 1 as the tolerance (right word?) -----> 2 < x < 4 = 7 < x+5 < 9 = l x+5 l < 9

F = 9

l x-3 l < ε/9 and l x-3 l < 1

So we have (for δ) δ = min{1,ε/9}

For this------ l x-3 l < ε/9 and l x+5 l < 9

From the beginning:

l (x²+2x-11)-4) l = l x²+2x-15 l
= l x+5 l*l x-3 l
< 9*l x-3 l
< 9*l ε/9 l
< ε

This is what we tried to prove?

l (x²+2x-11)-4) l < ε whenever 0 < l x-3 l < min{1,ε/9}

Is this process correct and is this what Spivak is trying to say?

EDIT: Are my signs ( < , > , = ) wrong?

 

[Unit]

Hi, BloodyFrozen! Sorry for not replying sooner.

You've got all the main points down - good. Well done on figuring out all the inequalities beforehand before writing out the complete proof. However, your presentation is very informal. A proper solution would look something more like the following. I have added important notes in italics, and commentary, which you may want to include in your written proofs, in parentheses.

Definition: we say lim_x→a f(x) = L if and only if for all ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ then |f(x) - L| < ε.

Claim:
lim_x→3 x² + 2x - 11 = 4

Proof:
Let ε > 0. This sentence should begin every epsilon-delta proof for limits. It basically says, "do your worst; give me any epsilon you like."
Pick a positive δ < min{1, ε/9}. This guarantees that δ < 1 and δ < ε/9. We are now going to show that this δ is THE required δ, the very one that will satisfy the given ε.
Suppose that 0 < |x - 3| < δ. This must be assumed if we are to proceed.
Note that |x + 5| = |x + 5 + (8 - 8)| = |x - 3 + 8| ≤ |x - 3| + |8| (by the triangle inequality) = |x - 3| + 8 < δ + 8 < 1 + 8 = 9 (since |x - 3| < δ and δ < 1). All your dirty work goes here, expressed very succinctly.
Finally, |(x² + 2x - 11) - 4| = |x² + 2x - 15| = |(x + 5)(x - 3)| = |x + 5||x - 3| < 9δ < 9(ε/9) = ε.
Q.E.D.

Thus we have shown that for any possible positive ε you can think of, I can give you a δ that will always work. I hope this helps!

 

[BloodyFrozen]

Ok, thanks Unit. Are all formal limits proved this way (similarly)?

 

[Unit]

In my experience, yes: proofs using ε-δ are done in this fashion.

Note that proofs using the ε-δ definition directly tend to get difficult for more complicated functions. Serious use of ε-δ is found in proving things like the squeeze theorem and how limits behave under addition, scalar multiplication, and composition of functions. Once these theorems have been proven, limits and matters of continuity can usually be resolved without much hassle.

 

[BloodyFrozen]

Is this very important to learn and perfect at the exact moment, or can it be held back *a little* and be learned later?

 

[Unit]

Practise. As someone who's read Spivak, his stuff doesn't get any easier. If you find yourself not getting it, it is really just a matter of time. You'll get used to ε-δ the more you do it. I'd also suggest working with limits of sequences, sequences diverging to infinity, limits of functions as x approaches infinity, and continuity, all of which have very similar definitions. This way, you'll be exercising the same kind of formal logic but in different settings. Good luck!

 

[BloodyFrozen]

Thanks, appreciate the help.