Uniqueness of limit proof

[Alpharup]

Spivak proves that limit of function f (x) as x approaches a is always unique.
ie...If lim f (x) =l
x-> a
and lim f (x) =m
x-> a
Then l=m.
This definition means that limit of function can't approach two different values.
He takes definition of both the limits.
He says for first limit, we can find some δ1> 0 for every ε> 0
Such that if 0 <|x-a|<δ1, then |f (x)-l|<ε
and also some δ2> 0 for every ε> 0
Such that 0 <|x-a|<δ2, then |f (x)-m|<ε
He then proves by saying a delta which works for one definition may not work for another and since both in qualities must be satisfied, he takes δ=min (δ1, δ2)
My questions here are as follows:
1. Why did he take only one ε in both definitions but 2 δs in both defintions?
Shouldn't there be ε1 and ε2?
2. He then says since
0 <|x-a|<δ1 and
0 <|x-a|<δ2 are both true, so
0 <|x-a|<min (δ1, δ2) By definition.
My question is what if δ1 is exclusively satisfied by first sentence alone and not second sentence, though δ1 <δ2?
Also what if δ2 is satisfied by second sentence alone and not the first one, even if δ2 <δ1.
How can we guarantee that atlest one of δ1 or δ2 satisfy both the statements, leave alone the prospect of taking minimum?
To be clear, say, if δ2>δ1, so δ=δ1 which is the minimum value. Then how can we prove that such a δ also satisfies second statement?
Sorry, if my questions sounds silly.

 

[mfb]

Shouldn't there be ε1 and ε2?

The limit definition says "for every ε". You are free to choose the same one for both.
You need two δ because all you get is "there exists some δ", it does not have to be the same.

My question is what if δ1 is exclusively satisfied by first sentence alone and not second sentence, though δ1 <δ2?

Then 0 <|x-a|<min (δ1, δ2) is true because it is equivalent to 0 <|x-a|<δ1.

Also what if δ2 is satisfied by second sentence alone and not the first one, even if δ2 <δ1.

]Then 0 <|x-a|<min (δ1, δ2) is true because it is equivalent to 0 <|x-a|<δ2.

Then how can we prove that such a δ also satisfies second statement?

δ does not satisfy anything. Apart from δ=min (δ1, δ2) of course, that's how it got defined.

 

[Alpharup]

1.Do you mean to say that, since ε1 and ε2 ∈(0,∞) ( This is open interval), so both can be taken as ε?
2. Also, do you mean to say that x and a are common to inequality equations in both the definitions and so, we can find a δ( either δ1 or δ2)? That's why, have you mentioned min(δ1, δ2) satisfied both equations?

 

[mfb]

1.Do you mean to say that, since ε1 and ε2 ∈(0,∞) ( This is open interval), so both can be taken as ε?

You don't "take" anything. You pick an arbitrary value and you call it ε.

2. Also, do you mean to say that x and a are common to inequality equations in both the definitions and so, we can find a δ( either δ1 or δ2)?

I don't understand that question.

 

[Alpharup]

what I asked is,
since
0 <|x-a|<δ1( for first definition) and
since 0 <|x-a|<δ2(for second definition),
Can we definitely say that both inequalities are true provided, in these inequalities, "x","a" are both the same?

 

[mfb]

We only consider x where both inequalities are satisfied.
a is the same, sure.

 

[Alpharup]

Please say whether my understanding is right. There is one definition of limit. And there are two friends: Ron and John.
Both use same definition of limit.
Ron: As x approaches a, f (x) approaches l.
John: As x approaches a, f (x) approaches m.
Both Ron and John use the same epsilon-delta definition of limit and we are the judge. Since, "for every ε> 0" means we can take any ε. But we, as judge, take one value of ε and analyze it.
Ron: Now, I am able to find δ1 for this ε.
John: Now, I am able to find δ2. for this ε.
We should judge, who is right. For us, it is immaterial. We just take minimum of these two δs. This is because, both can be considered as true. If we question the validity of either of δs, then we are doubting either Ron or John. This, may undermine the fact that both have used the same definition and both are right.

 

[Mark44]

Please say whether my understanding is right. There is one definition of limit. And there are two friends: Ron and John.
Both use same definition of limit.
Ron: As x approaches a, f (x) approaches l.
John: As x approaches a, f (x) approaches m.

Unstated, but should be stated is that l $\ne$ m.

Both Ron and John use the same epsilon-delta definition of limit and we are the judge. Since, "for every ε> 0" means we can take any ε. But we, as judge, take one value of ε and analyze it.
Ron: Now, I am able to find δ1 for this ε.
John: Now, I am able to find δ2. for this ε.
We should judge, who is right. For us, it is immaterial. We just take minimum of these two δs. This is because, both can be considered as true. If we question the validity of either of δs, then we are doubting either Ron or John. This, may undermine the fact that both have used the same definition and both are right.

I think that you are missing the point here. Both of them can't be right.

Let's look at a concrete example: $\lim_{x \to 1} 2x$. Suppose Ron claims that this limit is 2, and John claims that it is 3. At least one of them has to be wrong.

I will act as the judge and skeptic. I choose ε = 1/2. (Note that this is (3 - 2)/2.

Ron makes this calculation: |2x - 2| < ε = 1/2, so |x - 1| < 1/4. John says, "I take δ = 1/4. So if x is anywhere in the interval (.75, 1.25), 2x will be in the interval (1.5, 2.5). I.e, 2x will be within 1/2 unit of 2." So far, so good.

John notes that |2x - 3| < ε = 1/2, so 2x is in the interval (2.5, 3.5). This means that x is in the interval (1.25, 1.75). Note that 1 is not even in this interval. John is unable to come up with a δ for which every x ∈ (1 - δ, 1 + δ) implies that 2x ∈ (2x - ε, 2x + ε). Right off the bat, John is skunked.

On the other hand, I can continue testing Ron's asserted limit by giving him smaller and smaller values for ε. He will always be able to find a suitable δ simply by taking δ = ε/2 (due to the simple nature of the function we're using here). Things only get worse for John; when I give him smaller values for ε, he still cannot come up with a δ that works.

In the end, because Ron is able to make 2x arbitrarily close to 2 by taking x suitably close to 1, I concede that Ron's claim, $\lim_{x \to 1} 2x = 2$ is correct, and that John's claim is incorrect.

 

[Alpharup]

Sorry, I should have been more elaborate. As you told, I should have assumed l is not equal to m( according to Ron and John). Both Ron and John use the same definition and they declare that respective value of limit of function f as x approaches a are l and m, respectively. Both of them argue that their l and m are true answers( they both can find some δ for every ε). But, me as a judge should prove them that 'l not equal to m' is a contradiction...ie...l = m. If I am not able to prove, this epsilon-delta definition is useless.. Or in other words, the limit approaches two different values at a point.

 

[Samy_A]

Sorry, I should have been more elaborate. As you told, I should have assumed l is not equal to m( according to Ron and John). Both Ron and John use the same definition and they declare that respective value of limit of function f as x approaches a are l and m, respectively. Both of them argue that their l and m are true answers( they both can find some δ for every ε). But, me as a judge should prove them that 'l not equal to m' is a contradiction...ie...l = m. If I am not able to prove, this epsilon-delta definition is useless.. Or in other words, the limit approaches two different values at a point.

But you can prove one of them wrong. That is what Spivak did in his book and Mark44 in his example.

You can formalize it in an $\epsilon \ \delta$ argument as follows:

Assume $\displaystyle \lim_{x\rightarrow a} f(x)=l$ and also $\displaystyle \lim_{x\rightarrow a} f(x)=m$.
Take any $\epsilon >0$.
By the definition of limit we have:
(1) $\exists \delta_1>0 :\ 0<|x-a| \lt \delta_1 \Rightarrow |f(x)-l| \lt \frac{\epsilon}{2}$
(2) $\exists \delta_2>0 :\ 0<|x-a| \lt \delta_2 \Rightarrow |f(x)-m| \lt \frac{\epsilon}{2}$
Set $\delta=\min(\delta_1,\delta_2)>0$.
Now pick any $x$ satisfying $0<|x-a|<\delta$.
By (1), we have $|f(x)-l| \lt \frac{\epsilon}{2}$
By (2), we have $|f(x)-m| \lt \frac{\epsilon}{2}$
Then $|m-l|=|m-f(x)+f(x)-l| \leq |m-f(x)|+|f(x)-l| <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

So, for each $\epsilon >0$, we have $|l-m|<\epsilon$.
In words, $|l-m|$ is a non negative number that is smaller than each positive number, however small that number is. Conclusion: $|l-m|=0$ or $l=m$.

 

[mfb]

( they both can find some δ for every ε)

They cannot. That is the whole point of the unique limit. And you can use the proof to show that they cannot both do that.

 

[Alpharup]

But, why can't they find some δ for every epsilon? Why can't both δ1 and δ2 can't get satisfied at same time.
Spivak has told, if A <B always and A <C always, then it is definitely true that A <min (B, C). But since, both definitions are true, we being the judge, shouldn't it work for both δ1 and δ2?
@Samy_A Yes, Mark44 has proven that both Ron and John can't be right at some time. But he has taken l=2 and m=3. But if they both use the same definition of limit for every ε, shouldn't they get the same limit value?
Also, I get a feeling of whether my intuition of this proof is wrong? What I thought we should prove is if any person uses the epsilon delta definition of limit for a function approaching a number 'a', he should always get the same value such that, for every smaller or bigger ε> 0, he can find some δ.
Should I change my perception on what to prove?

 

[mfb]

But, why can't they find some δ for every epsilon?

See Mark's post (#8) for an example. They both use the same definition of a limit, the numbers 2 and 3 are the different limits they claim. Sure, they should get the same limit because there is just one limit, but that's what you want to prove.

Why can't both δ1 and δ2 can't get satisfied at same time.

That question does not make sense. "Can 4 be satisfied?"

Also, I get a feeling of whether my intuition of this proof is wrong? What I thought we should prove is if any person uses the epsilon delta definition of limit for a function approaching a number 'a', he should always get the same value such that, for every smaller or bigger ε> 0, he can find some δ.
Should I change my perception on what to prove?

That is correct, but the logic flow in the proof is a bit different: "imagine someone would claim to get a different limit, find a way to prove him wrong".

 

[Mark44]

But, why can't they find some δ for every epsilon?

Take a closer look at the example I posted in post #6. In that example I show that John cannot possibly come up with a δ that works.

 

[Samy_A]

@Samy_A Yes, Mark44 has proven that both Ron and John can't be right at some time. But he has taken l=2 and m=3. But if they both use the same definition of limit for every ε, shouldn't they get the same limit value?

Yes, of course they should get the same value. Mark44 showed with an example that if they have two different values, one of these has to be wrong.

Also, I get a feeling of whether my intuition of this proof is wrong? What I thought we should prove is if any person uses the epsilon delta definition of limit for a function approaching a number 'a', he should always get the same value such that, for every smaller or bigger ε> 0, he can find some δ.
Should I change my perception on what to prove?

I posted the proof in detail. What part of the proof don't you understand?

 

[Alpharup]

See Mark's post (#8)That is correct, but the logic flow in the proof is a bit different: "imagine someone would claim to get a different limit, find a way to prove him wrong".

Now, I think am getting it.You mean to say that l is 2 and m is 3. One guy is lying. But initially, what I thought was that I should prove that l and m obtained by Ron and John are both equal( by using the same definition) Now I realize that one guy is wrong and one guy is right. That is, if l=2 is the right value, then it must be right value. If some other guy proves it as m=3, then the definition is useless. Is this right?

 

[Alpharup]

But you can prove one of them wrong. That is what Spivak did in his book and Mark44 in his example.

You can formalize it in an $\epsilon \ \delta$ argument as follows:

Assume $\displaystyle \lim_{x\rightarrow a} f(x)=l$ and also $\displaystyle \lim_{x\rightarrow a} f(x)=m$.
Take any $\epsilon >0$.
By the definition of limit we have:
(1) $\exists \delta_1>0 :\ 0<|x-a| \lt \delta_1 \Rightarrow |f(x)-l| \lt \frac{\epsilon}{2}$
(2) $\exists \delta_2>0 :\ 0<|x-a| \lt \delta_2 \Rightarrow |f(x)-m| \lt \frac{\epsilon}{2}$
Set $\delta=\min(\delta_1,\delta_2)>0$.
Now pick any $x$ satisfying $0<|x-a|<\delta$

what about δ between δ1 and δ2...like δ=(δ1+δ2)/2? can we prove such a δ is invalid for the condition "for all x satisfying, 0 <|x-a|<δ"?

 

[Mark44]

Now, I think am getting it.You mean to say that l is 2 and m is 3. One guy is lying. But initially, what I thought was that I should prove that l and m obtained by Ron and John are both equal( by using the same definition) Now I realize that one guy is wrong and one guy is right. That is, if l=2 is the right value, then it must be right value. If some other guy proves it as m=3, then the definition is useless. Is this right?

My point is that, if l = 2 is the correct limit, the other guy CAN'T prove that the limit is 3. He (John) is unable to find a δ that works.

 

[Samy_A]

what about δ between δ1 and δ2...like δ=(δ1+δ2)/2? can we prove such a δ is invalid for the condition "for all x satisfying, 0 <|x-a|<δ"?

It doesn't matter. We need one $\delta>0$ that can be used for both limits. The minimum clearly works. There may be values larger than the minimum that would work: we don't know and actually don't care.
The proof works just fine with the minimum.

 

[Alpharup]

You nailed it @Mark44 but my only doubt is why @Samy_A's mathematical argument (I have quoted in post 17 ) is consistent with your description.

 

[Alpharup]

It doesn't matter. We need one $\delta>0$ that can be used for both limits. The minimum clearly works. There may be values larger than the minimum that would work: we don't know and actually don't care.
The proof works just fine with the minimum.

Brilliant one. Now I get it. In our consideration, Ron is right and John is plain wrong. Ron can always find some δ for every ε. But John can't find the same. Ron can find δ like 1 or 0.1 or 0.00001 for any ε like 100, 10000( For any ε> 0, the some smallest of smallest δ can be found). But John can't do the same. If l=2, John can't attack this value by saying m=3. Thus, our definition is attack proof.

 

[Mark44]

You nailed it @Mark44 but my only doubt is why @Samy_A's mathematical argument (I have quoted in post 17 ) is consistent with your description.

What Samy_A is laying out is the usual proof by contradiction that the limit of a function is unique. The proof starts by assuming that if $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} f(x) = M$, and then concludes that L = M, a contradiction, thereby establishing the uniqueness of the limit of a function.

What I said is that if L and M are different, then one of them (at least) has to be wrong.

 

[Samy_A]

Brilliant one. Now I get it. In our consideration, Ron is right and John is plain wrong. Ron can always find some δ for every ε. But John can't find the same. Ron can find δ like 1 or 0.1 or 0.00001 for any ε like 100, 10000( For any ε> 0, the some smallest of smallest δ can be found). But John can't do the same. If l=2, John can't attack this value by saying m=3. Thus, our definition is attack proof.

This is correct.
But the emphasis on large ε and small δ is somewhat questionable.

Actually, if ε is large, like 100, 10000, chances are that John can produce some δ to satisfy the limit definition.
It is for small ε that John gets into trouble. Note that in his example, Mark44 chose ε=½. That was small enough to make it impossible for John to produce a δ>0, however small a δ he would try.
Had Mark started with ε=1.1, John could have produced δ=0.04 to satisfy the limit condition. But that of course doesn't mean anything. The limit definition contains ∀ε>0. As we chose smaller and smaller ε, Ron can continue to come up with an appropriate δ (that will indeed in general become smaller as ε gets smaller), but not John, or anyone how claims that the limit is not 2.

 

[zinq]

This statement: "Spivak proves that limit of function f (x) as x approaches a is always unique" cannot be correct.

It is necessary to make some assumption about f(x) being a "nice" function. The most obvious such assumption is that f is continuous, at least at the point a that x is approaching.

Or, one could alternatively assume at least that the limit exists. Any limit that exists is automatically unique. This should not be confused with the concept of a "limit point" in topology.

Otherwise, consider the function defined as f(x) = 0 if x is rational and f(x) = 1 if x is irrational. The limit of f(x) as x —> a does not exist, so it is not unique.

 

[Alpharup]

A small doubt over here. The orginal definition of limit of f (x) approaches l as x approaches a is, "For every ε>0, there is some δ>0, such that for all x, if 0 <|x-a|<δ, then|f (x)-l|<ε" In this definition, why can't we drop the clause "for all x"? Will there be any implications in doing so?

Moreover Apostol defines like, "It means For every ε>0, there is some δ>0, such that |f (x)-l|<ε, whenever 0 <|x-a|<δ". In this definition we can see, he has dropped the clause. Am I right in recommending to drop this clause?

 

[mfb]

In this definition, why can't we drop the clause "for all x"?

If we drop it, we have no idea what x is. Technically it should be "for all x in the set of the real numbers".
Apostol hid it in "whenever 0 <|x-a|<δ".

 

[Alpharup]

But anyways, we are studying about real X right?

 

[mfb]

And we know that because we write it down.
It is often dropped if it is clear from the context, but that is sloppy notation.

 

[Svein]

Reductio ad absurdum proof:

Assume $\lim_{x\rightarrow a}f(x)=l$ and $\lim_{x\rightarrow a}f(x)=m$ and $l\neq m$. This means that $\left\lvert l-m \right\rvert>0$. Therefore we can choose an $\epsilon = \frac{\left\lvert l-m \right\rvert}{3}$. Since the limits exist, we can find δ₁ such that $\left\lvert x-a\right\rvert <\delta_{1}\Rightarrow \left\lvert f(x)-l \right\rvert<\epsilon$ and δ₂ such that $\left\lvert x-a\right\rvert <\delta_{2}\Rightarrow \left\lvert f(x)-m \right\rvert<\epsilon$. Let δ=min(δ₁, δ₂). Then for $\left\lvert x-a\right\rvert <\delta$, we have $\left\lvert l-m\right\rvert= \left\lvert f(x)-m - (f(x)-l)\right\rvert\leq \left\lvert f(x)-m\right\rvert+\left\lvert f(x)-l\right\rvert<\epsilon + \epsilon \leq \frac{2}{3}\left\lvert l-m\right\rvert$ which cannot hold if $\left\lvert l-m\right\rvert>0$. Thus our premise ($l\neq m$) must be wrong, so l = m.
Q.E.D