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Math Help - Limit Proof

  1. #1
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    Limit Proof

    Define f: R R by f(x) = x if x is rational and f(x) = 0 if x is irrational. Prove that f has a limit at c iff c = 0.
    This has me totally stumped...

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  2. #2
    Super Member Gamma's Avatar
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    well if c is 0 it is pretty easy application of definitions because both of the piecewise functions converge to 0 at 0, so you can just choose the smaller of the two deltas to get it within any given epsilon.

    For the other direction, I suggest a proof by contradiction. That is is say c is not 0. Anywhere else on the real line it is pretty clear that if you took epsilon to be say |c|/3 there is no way you can choose a delta small enough to ensure this function is continuous there. I think |c|/2 works actually, but better safe than sorry in my opinion.

    Oh I should mention it is important that both the rationals and irrationals are dense, so any delta ball will contain points from both pieces.
    Last edited by Gamma; April 26th 2009 at 11:01 PM. Reason: add last line about dense subsets
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  3. #3
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    Quote Originally Posted by bearej50 View Post
    Define f: R R by f(x) = x if x is rational and f(x) = 0 if x is irrational. Prove that f has a limit at c iff c = 0.








    This has me totally stumped...
    Assume \lim_{x\rightarrow}{f(x)} = m and c\neq 0

    Then we have two cases :

    case I : m=0

    In this case we have for all ε>0 ,there exists a δ>0 and such that :

    for all x :if xεD(f) and 0<|x-c|<δ ,then |f(x)|<ε.......................................... ..........1

    BUT in the interval (c-δ,c+δ) -{c} x can take both rational and irrational values and so we will have :

    |f(x)| = |x| and |f(x)|=0 and if we choose ε<|x| we will have a contradiction



    case II : m\neq 0.

    In this case we have for all ε>0 ,there exists a δ>0 and such that :

    for all x : if xεD(f) and 0<|x-c|<δ ,then |f(x)-m|<ε.............................................. ................................2

    Αnd in this case again in the interval (c-δ,c+δ)-{c} x can take both rational and irrational values.

    Hence |f(x)-m|=|x-m| and |f(x)-m|= |m| and if we put ε =|m| we will end with a contradiction

    Conversely:

    Let c=0 we must show : \lim_{x\rightarrow 0}{f(x)} =m

    AGAIN we have two cases :

    case I : m=0

    In this case given ε>0 we must find a δ>0 and such that :

    for all x:if 0<|x|<δ and xεD(f) ,then |f(x)|<ε.......................................... ...........................3

    Again when x, rational |f(x)| =|x| and if we choose δ<ε ,then \lim_{x\rightarrow 0}{f(x)} = 0 and hence the limit exists ,

    while in the case where ,x is irrational :

    |f(x)|=0< ε and the limit again exists.

    HENCE \lim_{x\rightarrow 0}{f(x)} =0 and the limit exists

    Thus we do not have to consider the case where m\neq 0

    Note the above problem is based on the following theorem:

    LET z be a real No. Given ε>0 .there exists a rational No, a such that |a-c|<ε.

    Τhe same theorem holds when ,a is irrational
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  4. #4
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    ^ Thank you for being so helpful. I would +rep you but they wont let me
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bearej50 View Post
    ^ Thank you for being so helpful. I would +rep you but they wont let me
    just clicking the thanks button is fine. chances are your reputation isn't high enough to give a lot of point through +rep anyway.
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