1. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 09:161 edit
    Originally posted by humy
    I like a simple example of such an oscillating function with an infinite limit so I can analyze this.
    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?
  2. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 09:223 edits
    Sin(x)+x/2
    has an infinite limit and doesn't conform to your definition. [edit] Sorry, your definition does work for this one, taking L to be infinity.
  3. Joined
    06 Mar '12
    Moves
    642
    01 Jul '16 13:526 edits
    Originally posted by twhitehead
    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?
    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|
  4. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 14:05
    Originally posted by humy
    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|
    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 )
  5. Joined
    06 Mar '12
    Moves
    642
    01 Jul '16 14:19
    Originally posted by twhitehead
    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 )
    oh I see. Obviously cannot have |f(z)| < 0 .
    I will revise my formula and then came back here later.
  6. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 14:25
    Originally posted by humy
    oh I see. Obviously cannot have |f(z)| < 0 .
    I will revise my formula and then came back here later.
    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.
  7. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 14:33
    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.
  8. Joined
    06 Mar '12
    Moves
    642
    01 Jul '16 16:514 edits
    Originally posted by twhitehead
    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.
    Yes, 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:
    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.
  9. Joined
    06 Mar '12
    Moves
    642
    01 Jul '16 16:535 edits
    Originally posted by humy
    oh I see. Obviously cannot have |f(z)| < 0 .
    I will revise my formula and then came back here later.
    So, if I am at LAST thinking about this correctly, the problem with that formula:

    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?
  10. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 17:03
    Originally posted by humy
    [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|
    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.
  11. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 17:481 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|
  12. Joined
    06 Mar '12
    Moves
    642
    01 Jul '16 17:494 edits
    Originally posted by twhitehead
    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.
    Arr yes you are right.
    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?
  13. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 17:59
    Originally posted by humy
    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?
    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, 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.
  14. Joined
    06 Mar '12
    Moves
    642
    01 Jul '16 18:4012 edits
    Originally posted by twhitehead
    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.
    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|
  15. Cape Town
    Joined
    14 Apr '05
    Moves
    52945
    01 Jul '16 20:391 edit
    Originally posted by humy
    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|
    Multiple issues, but an easy one is that it doesn't work for f(x)=2 as your f(x)≠L is a non-starter.
Back to Top

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.I Agree