# Referring to proved lim of the function.

1. Oct 16, 2012

### A Dhingra

hello..

(My professor solved this question in class but i have a few doubts in it,and this is probably the method used by Sherbert and Beartle in intro to real analysis.)
Prove that lim x→c x2 =c2.

Solution:

Consider
|x2 - c2| < ε --1)
|x-c||x+c|< ε

Suppose
| x-c|<1

so, |x| ≤ |c| +1
|x+c|≤ 2|c|+1

so,
|x-c||x+c| ≤ (2|c|+1)|x-c|< ε --2)

then,

|x-c| < ε/(2|c| +1)

hence
δ = inf{1, ε/(2|c|+1)}
Hence
|x-c| <δ $\Rightarrow$ |x2 - c2| < ε
(This was so far as done in class.)

Now here we assumed the first statement, while this doesn't necessarily mean 2)...
We know, |x-c||x+c|< ε
then putting a constraint on |x-c|, then
finding its max(|x+c|)
How do we show that the product of these two is still less then ε.

Next thing i could not understand was, for a specific value of ε, there exists an upper limit for the value of δ, then how can we randomly select δ to be one and then revise the value of δ ?

Thanks for any help!

2. Oct 16, 2012

### arildno

Re: refrerring to proved lim of the function.

Note that this:
"then,

|x-c| < ε/(2|c| +1)"
is a REQUIREMENT we see must be fulfilled, in order for the product to be less than epsilon.

Now, it is perfectly true that we gained insight in this requirement, by first simplifying by demanding that |x-c|<1, a DIFFERENT requirement.

How can these two requirements be reconciled?
Note that NEITHER of this requirements requires that delta "must" be close to the upper limit, it can be a lot LESS than either of them, in order for BOTH to be fulfilled.

Thus, the line:
"δ = inf{1, ε/(2|c|+1)}"
says we must choose the LEAST of those two values, in order for both our requirements to hold.

3. Oct 16, 2012

### A Dhingra

Re: refrerring to proved lim of the function.

So it is like moving backwards in the proof!

But it is required for delta to be less than its upper limit.
is this said by the equation, |x-c| < ε/(2|c| +1)?

4. Oct 16, 2012

### arildno

Re: refrerring to proved lim of the function.

Upper limits are not unique!
If you first find out something useful if you let delta be less than 1, and then finds out you can say something even more useful if delta is also less than 0.01, you'd better choose delta less than 0.001 (trivially less than 1 as well)! Right?

5. Oct 16, 2012

### arildno

Re: refrerring to proved lim of the function.

Sure you work yourself "backwards" in a sense, but in a way that does not mean circular argumentation!
You could always start by saying:
1. Let e>0
2. Let d=inf(1,e/(2|c|+1)
3. Then it follows that whenever x<d, we have the e-requirement fulfilled, because (insert the inequalities that follows from this choice of d!)
4.QED

6. Oct 16, 2012

### A Dhingra

Re: refrerring to proved lim of the function.

"Upper limits are not unique!"
The infimum of the upper limits are unique for a specific p and selected e, or just e (uniform continuous function),is that not required here?

7. Oct 16, 2012

### arildno

Re: refrerring to proved lim of the function.

Nope.
Because a more nuanced approach might get away with a generally "looser" infimum than the the one you have in the proof (this is OFTEN the case in more complicated estimates; a clever guy might increase the value of the "needed" upper limit significantly relatively to the upper limit utilized in som other, less refined proof!).
For the proof's VALIDITY, only the sufficiency of the chosen infimum of the two indicated values is required.
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For example, we cannot rule out automatically, that there cannot exist "a", 0<p<1, so that it is sufficient, for very large |c|, that the delta value e/(a|c|^p+1) is sufficient to ensure what we want, rather than using e/(2|c|+1) as our infimum.

In this case, it isn't really interesting if we theoretically can get away with such an estimate, but if an estimate of this is a sub-estimate in some larger estimate, it MIGHT be that what we want to prove requires that we prove the latter estimate's validity.

Last edited: Oct 16, 2012
8. Oct 17, 2012

### A Dhingra

Re: refrerring to proved lim of the function.

I didn't get this example.

Let me just elaborate what I meant by saying that "The infimum of the upper limits are unique for a specific p and selected e, or just e (uniform continuous function)". If we plot f(x) on y axis and x on x-axis, and select the point p on x-axis and locate f(x) as x tends to p and select any random ε on the y-axis, and get the values of corresponding x, then there will be a maximum value of δ which can satisfy this condition values greater than that may take us out of the ε region on y-axis. And hence I was asking if this thing has been taken care of in the proof or not?

And if possible, can you please explain the example in a bit simpler way..

9. Oct 22, 2012

### A Dhingra

hi..
I have one more thing to ask.
Can this ε - δ method be used to prove that the limit does not exist at a fixed point?