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Homework Help: Evaluating limit question

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following limit
    lim x->0 f(x) = 2, x rational, -2, x irrational

    2. Relevant equations

    3. The attempt at a solution
    I tried to draw the graph of this question but couldn't. Do you think I have a function that oscillates between 2 and -2 violently?
    My best guess is that the limit doesn't exist but I want to have better explanation than mine.
    Last edited: Oct 15, 2011
  2. jcsd
  3. Oct 15, 2011 #2


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    That looks like a slightly modified Dirichlet function to me. Between any 2 rational numbers is there an irrational number? And between any irrational number is there a rational number?

    Also, what is this limit to? Limit as x goes to what?
  4. Oct 15, 2011 #3
    my apology. x->0.
    There's no irrational number between 2 rational numbers. Vice versa no?
  5. Oct 15, 2011 #4


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    Does your last statement make sense?

    Between 1 and 2 is there an irrational number?
  6. Oct 16, 2011 #5
    When you suspect that a limit does not exist, you can make use of the sequential criterion for limits. The criterion loosely states a limit [itex]\lim_{x→c}f(x)=L[/itex] exists if and only if for every sequence [itex](x_n)[/itex] that converges to [itex]c[/itex] it follows that [itex]\lim_{n→∞}f(x_n)=L[/itex].

    So if you want to prove that the limit does not exist you can do so by showing that there exist two sequences [itex](x_n)[/itex] and [itex](y_n)[/itex] that converge to [itex]c[/itex] and satisfy [itex]\lim_{n→∞}f(x_n)\neq \lim_{n→∞}f(y_n)[/itex].

    Does this make sense to you? Can you construct two sequences that would work in your case?
  7. Oct 17, 2011 #6
    Sorry. Could you explain it easier?
  8. Oct 17, 2011 #7


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    For a limit to exist, every time we make a sequence approach 0, the corresponding function values must approach the same number.

    For example:

    If I choose a sequence of rational numbers: 1,1/2,1/4,...

    We get: f(1),f(1/2),f(1/4),...

    So we get 2,2,2,... and this approaches 2.

    However, if we look at e,e/2,e/4,...

    We get: f(e),f(e/2),f(e/4),...

    So we get -2,-2,-2,... and we get -2.

    Since these are two different values, the limit does not exist.
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