Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Negation of Limit

  1. Nov 21, 2005 #1
    so is that a good negation of the defenition of the limit?

    A function f with domain D doesnt not have limit L at a point c in D iff
    not for every number E > 0 there is a corresponding number G >0 such if |F(x) - L| <E then is not the case 0< |x-a|<G
    am i right? wat am i doing wrong?
    thx so much. I made a new post that way ppl wont get confuse with the other post.
    thank u so much
     
  2. jcsd
  3. Nov 21, 2005 #2

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    I am guessing that the statement you want to negate is something like "for each e there is a d such that X(d) implies Y(e)." So the negation should be "for some e there is not any d such that X(d) implies Y(e)."
     
  4. Nov 21, 2005 #3

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    It's easy if you use quantifiers:
    Definition of [itex]\lim_{x\to a}f(x)=L[/itex]:

    [tex]\forall \epsilon>0 \exists \delta>0 : |x-a|<\delta \Rightarrow |f(x)-L|<\epsilon[/tex]

    To negate this, simply use the rules:
    [tex]\neg (\forall x:P) \iff \exists x: \neg P[/tex]
    [tex]\neg (\exists x:P) \iff \forall x : \neg P[/tex]
     
  5. Nov 21, 2005 #4
    is this right?

    is this the right negation of the statement above?

    [tex]\exists\epsilon>0 \forall \delta>0 : |x-a|<\delta \wedge\|f(x)-L|\geq\epsilon[/tex]

    i am also using this fact
    ~(P=>Q) = P^~Q

    how can i illustrate this negation? would it be a function that is not continous at a point such as f(x) = 1/(x+1)?
    thank you very much for al ur help!
     
    Last edited: Nov 22, 2005
  6. Nov 23, 2005 #5

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    I always seem to forget that f(a) isn't important in the definition.
    The function doesn't even have to be defined at the limit point. The correct definition is:

    [tex]\forall \epsilon>0 \exists \delta>0 : 0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon[/tex]

    So change [itex]|x-a|<\varepsilon[/itex] to [itex]0<|x-a|<\varepsilon[/itex], then it's correct.

    This is different from continuity! A function is continuous at a if [itex]\lim \limits_{x\to a}f(x)=f(a)[/itex], which says 3 things:
    1. The limit exists
    2. f(a) is defined
    3. The 2 are equal.

    More precisely, a function is continuous at x=a if
    [tex]\forall \epsilon>0 \exists \delta>0 : |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon[/tex]

    The function 1/(x+1) is perfectly continuous everwhere on its domain. The point f(-1) is not defined so it's no problem. If you define f(-1)=0, then it's not continuous anymore.
     
    Last edited: Nov 23, 2005
  7. Feb 13, 2009 #6
    Is this the right negation of a finite limit?

    [tex]
    \neg(\lim_{x\to a}f(x)=L) \iff \exists \epsilon>0 \forall \delta>0\exists x: 0<|x-a|<\delta \Rightarrow |f(x)-L|\geq\epsilon \vee \neg\exists f(x) [/tex]

    Thanks.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook