Formal definition for limits at a point (2 Viewers)

[mathgeekof2050]

Dear BOS members,

I would like any immediate assistance on a problem currently at hand, please. Any form of help, which is relevant is much appreciated.

Formal definition for limits at a point:

Let f be a function defined on an open interval D containing c. Let L be a real number. We can assert that

lim (x->c) f(x) = L​

if for every ϵ > 0, there exists δ > 0 such that for all x in D and

0 < |x - c| < δ​

we have,

|f(x) - L| < ϵ​

The problem is understanding it. Say for example, we have to prove

lim (x->5) (3x - 3) = 12​

using this definition

How do I do it? Do I prove that 0 < |x - c| < δ implies |f(x) - L| < ϵ or the other way around?


Regards,
anon

 

[mathgeekof2050]

OR what is the most logical, neatest way of laying out such a proof?

 

[Shadowdude]

"Let f be a function defined on an open interval D containing c. Let L be a real number. We can assert that

lim (x->c) f(x) = L


if for every ϵ > 0, there exists δ > 0 such that for all x in D and

0 < |x - c| < δ
we have,

|f(x) - L| < ϵ "


Your proof should follow the definition.

So, see the bolded.

Your proof thus should start:

Let ϵ > 0...

 

[mathgeekof2050]

Is the aim to find a value for delta in terms of epsilon such that is satisfies this 0 < |x - c| < δ implies |f(x) - L| < ϵ condition?

 

[Carrotsticks]

Is the aim to find a value for delta in terms of epsilon such that is satisfies this 0 < |x - c| < δ implies |f(x) - L| < ϵ condition?

Yes.

So in a sense, we can make epsilon as small as we like and by doing so, we drag delta to also be as small as we like.

 

[Shadowdude]

Is the aim to find a value for delta in terms of epsilon such that is satisfies this 0 < |x - c| < δ implies |f(x) - L| < ϵ condition?

yes

 

[mathgeekof2050]

Thanks for your input Shadowdude.

I will try it out:

Let ϵ > 0, there exists δ > 0 such that for all x in reals and 0 < |x - 5| < δ we have, |(3x-3) - 12| < ϵ

Simplifying and collecting like terms, we have

|3x - 15| < ϵ

Dividing by 3 on each side

|x - 5| < ϵ/3

Now, setting δ = ϵ/3

Thus, the limit lim (x->5) (3x - 3) = 12 is true for every ϵ > 0, there exists δ > 0 such that for all x in reals and 0 < |x - 5| < ϵ/3, we have |f(x) - 12| < ϵ

 

[Carrotsticks]

 

[mathgeekof2050]

Thanks for your input Carrotsticks, I follow on what you are saying.

 

[mathgeekof2050]

So since all steps are reversible, then |x - 5| < delta implies |f(x) - L| < epsilon?

 

[mathgeekof2050]

Adding to the above (Carrotstick's working out), it is displaying |f(x) - L| < ϵ implies 0 < |x - c| < δ. So is there an 'iff' implied in the formal definition?

 

[Shadowdude]

Thanks for your input Shadowdude.

I will try it out:

Let ϵ > 0, there exists δ > 0 such that for all x in reals and 0 < |x - 5| < δ we have, |(3x-3) - 12| < ϵ

Simplifying and collecting like terms, we have

|3x - 15| < ϵ

Dividing by 3 on each side

|x - 5| < ϵ/3

Now, setting δ = ϵ/3

Thus, the limit lim (x->5) (3x - 3) = 12 is true for every ϵ > 0, there exists δ > 0 such that for all x in reals and 0 < |x - 5| < ϵ/3, we have |f(x) - 12| < ϵ

see with your proof, you're starting at the end and then go back again


The proof's structure, should be:

Let ϵ > 0. Choose δ = ϵ/3. Suppose x is in D, and that 0 < |x - c| < δ. Then: |f(x) - L| < ϵ.

Follows the definition exactly.

Not "Let ϵ > 0." and then "|f(x) - L| < ϵ." and then go backwards to "0 < |x - c| < δ. " and then "Choose δ = ϵ/3".

 

[mathgeekof2050]

see with your proof, you're starting at the end and then go back again


The proof's structure, should be:

Let ϵ > 0. Choose δ = ϵ/3. Suppose x is in D, and that 0 < |x - c| < δ. Then: |f(x) - L| < ϵ.

Follows the definition exactly.

Not "Let ϵ > 0." and then "|f(x) - L| < ϵ." and then go backwards to "0 < |x - c| < δ. " and then "Choose δ = ϵ/3".

How do we know to choose δ = ϵ/3? It just came without evidence.

 

[Shadowdude]

How do we know to choose δ = ϵ/3? It just came without evidence.

You do off the paper work to show it.


And then in your proof, you'll show that choosing that particular value will make it work out. So you'll go through the working you did off-paper, but in reverse.

 

[mathgeekof2050]

You do off the paper work to show it.


And then in your proof, you'll show that choosing that particular value will make it work out. So you'll go through the working you did off-paper, but in reverse.

Wow, that's pretty weird. Thank you anyways.

 

[Carrotsticks]

I'll use D to denote delta and E to denote epsilon.

I didn't have the concluding line in my proof, my fault.

The whole idea is IF 0<|x-c|< D, THEN |f(x)-L| < E

What we do above by starting backwards is what we generally do to find the relationship between E and D. Once we've done that (we found that D = E/3), we then re-write our statement formally.

IF 0 < |x-c| < E/3, THEN |f(x)-L| < E.