01 Jul 16
6 edits
Originally posted by twhiteheadI might be missing something or even stupid but you have just confused me.
Sin(x)/x
[edit]It doesn't have an infinite limit, but I don't think your form deals with infinite limits does it? The limit in your form is L is it not?
Can you give me a specific example of a limit L value and an x value which is such that there exists no real z such that z>x ∧ |f(z) − L| < |f(x) − L|
01 Jul 16
Originally posted by humyThe limit L is zero.
I might be missing something or even stupid but you have just confused me. Can you give me a specific example of a limit L value and an x value which is such that there exists no real z such that z>x ∧ |f(z) − L| < |f(x) − L|
Take x as Pi.
f(x) = 0
There is no z such that z>x ∧ |f(z) − L| < |f(x) − L| (which simplifies to z>pi ∧ |f(z)| < 0 )
01 Jul 16
Originally posted by twhiteheadoh I see. Obviously cannot have |f(z)| < 0 .
The limit L is zero.
Take x as Pi.
f(x) = 0
There is no z such that z>x ∧ |f(z) − L| < |f(x) − L| (which simplifies to z>pi ∧ |f(z)| < 0 )
I will revise my formula and then came back here later.
01 Jul 16
Originally posted by humyThe Wikipedia version doesn't have that problem which is why they do it that way (and is the standard definition). You would do well just to stick with that. It doesn't use infinity, so it shouldn't concern you on that count.
oh I see. Obviously cannot have |f(z)| < 0 .
I will revise my formula and then came back here later.
01 Jul 16
In English rather than mathematical notation, the Wikipedia version says:
For any interval around the limit, no matter how small, there is always a point after which the function never goes outside that interval.
Your version says:
For any point in the function, there is always another point after which the function is closer to the limit than at that point. Clearly for any function that crosses the limit, this is not true.
I believe |tan(x)| + x shows similar behaviour with infinity as tan repeatedly goes to infinity.
01 Jul 16
4 edits
Originally posted by twhiteheadYes, but, at least for me, I think it would be a useful maths exercise to occasionally try and come up independently with a solution of my own rather than just always stick to the short-term easy option of sticking to a standard solution:
The Wikipedia version doesn't have that problem which is why they do it that way (and is the standard definition). You would do well just to stick with that. It doesn't use infinity, so it shouldn't concern you on that count.
exercises my brain so that when there is no standard/known solution to a problem, which is the situation I often encounter with my unusual line of research, I had made myself as good as possible at coming up with a solution for myself.
I have done a similar mental exercise several times in the past for solving certain problems when I already had just seen the standard/known solution! -I do it anyway just to improve my ability for solving problems with no known solution.
01 Jul 16
5 edits
Originally posted by humySo, if I am at LAST thinking about this correctly, the problem with that formula:
oh I see. Obviously cannot have |f(z)| < 0 .
I will revise my formula and then came back here later.
lim {x→∞} f(x) = L ⇒ ∀x ∈ ℝ ∃z ∈ ℝ : z>x ∧ |f(z) − L| < |f(x) − L| ( FOUL! )
( " {...} " indicates subscript )
is that, if f(x) is an oscillating function, as x increases, f(x) may 'pass through' f(x) = L not once but many times and thus, if we make the value of x such that f(x) = L where x is such that that L can be reached yet again by increasing x by a sufficient non-zero value yet again, then we cannot have;
|f(z) − L| < |f(x) − L| as the closest we can get to that is |f(z) − L| = |f(x) − L| .
Now, if I am at last thinking about this correctly, a fix for that would be to simply replace that "<" with "≤" so that we can allow that closest approach of |f(z) − L| = |f(x) − L| and that means changing the formula to;
lim {x→∞} f(x) = L ⇒ ∀x ∈ ℝ ∃z ∈ ℝ : z>x ∧ |f(z) − L| ≤ |f(x) − L|
For similar reason, I got it a bit wrong for;
lim {x→p} f(x) = L ⇒ ∀x ∈ ℝ{≠p} ∃z ∈ ℝ : |z − p| < |x − p| ∧ |f(z) − L| < |f(x) − L| ( FOUL! )
and that should be;
lim {x→p} f(x) = L ⇒ ∀x ∈ ℝ{≠p} ∃z ∈ ℝ : |z − p| < |x − p| ∧ |f(z) − L| ≤ |f(x) − L|
Anyone:
Have I at last got both of those above formulas exactly correct?
01 Jul 16
Originally posted by humyNo, you are still not thinking about it correctly.
[b]Now, if I am at last thinking about this correctly, a fix for that would be to simply replace that " < " with " ≤ " so that we can allow that closest approach of |f(z) − L| = |f(x) − L| and that means changing the formula to;
lim {x→∞} f(x) = L ⇒ ∀x ∈ ℝ ∃z ∈ ℝ : z>x ∧ |f(z) − L| ≤ |f(x) − L|
I could come up with a function that crosses the limit only once and thereafter approaches it gradually. That would violate your definition.
01 Jul 16
1 edit
Even worse, your new definition can be used to show that the line x=2 converges to 3.
lim {x→∞} f(x) = L ⇒ ∀x ∈ ℝ ∃z ∈ ℝ : z>x ∧ |f(z) − L| ≤ |f(x) − L|
Take L=3
∀x ∈ ℝ , |f(x) − L| = |2-3|=1
and
|f(z) − L| =1 ∀ z ∈ ℝ
Therefore for any x, there will always exist z ∃z ∈ ℝ : z>x ∧ |f(z) − L| ≤ |f(x) − L|
01 Jul 16
4 edits
Originally posted by twhiteheadArr yes you are right.
No, you are still not thinking about it correctly.
I could come up with a function that crosses the limit only once and thereafter approaches it gradually. That would violate your definition.
Just one more fix then (I hope) by adding the constraint of f(x) ≠ L ;
lim{ x→∞ } f(x) = L ⇒ ∀x ∈ ℝ{ f(x)≠L } ∃z ∈ ℝ : z>x ∧ |f(z) − L| < |f(x) − L|
and similarly:
lim{ x→p } f(x) = L ⇒ ∀x ∈ ℝ{ x≠p, f(x)≠L } ∃z ∈ ℝ : |z − p| < |x − p| ∧ |f(z) − L| < |f(x) − L|
and note I have gone back to not using " ≤ ".
Have I got it right now?
01 Jul 16
Originally posted by humyNo, still not right. First, a typo, you included x≠p, which doesn't apply to this case.
Arr yes you are right.
Just one more fix then (I hope);
lim{ x→∞ } f(x) = L ⇒ ∀x ∈ ℝ{ x≠p, f(x)≠L } ∃z ∈ ℝ : z>x ∧ |f(z) − L| < |f(x) − L|
Have I got it right now?
Secondly, your definition works for the sine function.
Let f(x)=sin(x)
Let L=0
It is true that
∀x ∈ ℝ{ f(x)≠L } ∃z ∈ ℝ : z>x ∧ |f(z) − L| < |f(x) − L|
substituting
∀x ∈ ℝ{ sin(x)≠ 0} ∃z ∈ ℝ : z>x ∧ |sin(z)| < |sin(x)|
If that is not obviously true, then find the first z>x for which sin(z)=0, thus we have 0<|sin(x)| and we know |sin(x)|≠ 0 so it must be true.
01 Jul 16
12 edits
Originally posted by twhiteheadI probably should check over this more but I have came up with:
No, still not right. First, a typo, you included x≠p, which doesn't apply to this case.
Secondly, your definition works for the sine function.
Let f(x)=sin(x)
Let L=0
It is true that
∀x ∈ ℝ{ f(x)≠L } ∃z ∈ ℝ : z>x ∧ |f(z) − L| < |f(x) − L|
substituting
∀x ∈ ℝ{ sin(x)≠ 0} ∃z ∈ ℝ : z>x ∧ |sin(z)| < |sin(x)|
If that is not obviously true, th ...[text shortened]... irst z>x for which sin(z)=0, thus we have 0<|sin(x)| and we know |sin(x)|≠ 0 so it must be true.
lim { x→∞ } f(x) = L
⇒
∀x ∈ ℝ { f(x)≠L } : ∃a ∈ ℝ : a>x ∧ ¬∃ b ∈ ℝ { b>a } : |f(b) − L| ≥ |f(x) − L|
01 Jul 16
1 edit
Originally posted by humyMultiple issues, but an easy one is that it doesn't work for f(x)=2 as your f(x)≠L is a non-starter.
I probably should check over this more but I have came up with:
lim { x→∞ } f(x) = L
⇒
∀x ∈ ℝ { f(x)≠L } : ∃a ∈ ℝ : a>x ∧ ¬∃ b ∈ ℝ { b>a } : |f(b) − L| ≥ |f(x) − L|