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Homework Help: Prove that the limit of [x]+[-x] at infinity doesn't exist.

  1. Jun 24, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem is to prove that the limit of [x]+[-x] at infinity does not exist.

    3. The attempt at a solution

    I used the argument that the function [x]+[-x] is equivalent to the function f such that it gives 0 for all integers and gives -1 otherwise. therefore because the function f oscillates between 0 and -1 at infinity the limit can not exist. then I realized that my argument might be not so convincing for the professor. I then tried to use the following logical argument: assuming the limit exists and is equal to L, then there exists a positive number N that for any epsilon greater than zero when x>N we can conclude: |f(x) - L| < epsilon. therefore If I show that this assumption leads to a contradiction I have solved the problem. I said let's take x to be greater than N, f(x) is either 0 or -1 by definition. if f(x) is 0 then |0-L| < epislon which says L is between -epsilon and +epsilon, now since epsilon is an arbitrary number, let's take it to be equal to L/2. then it says that -L/2 < L < L/2 which is false.
    Now if f(x) is -1, then |-1-L|=|L+1|<epislon. which says L is between -epsilon-1 and +epislon-1. now since epsilon is again an arbitrary number, let's assume that epsilon is equal to L. then it tells us that -L-1<L<L-1 and there is no such L that satisfies this inequality. therefore in both cases we've shown that the limit does not exist.

    am I right? Can I write it to the professor as the answer?
    Thanks in advance.
     
  2. jcsd
  3. Jun 24, 2011 #2
    Hi AdrianZ! :smile:

    Your argument does not work for L=0. Indeed, you took epsilon to be equal to L/2. But if L is 0, then you've taken epsilon=0, which is not allowed. So you'll have to do something different for L=0.

    In any case, I found your argument a bit weird (not that's it's wrong, it's just uncommon). You've basically fixed x and then said that there is no corresponding L. A more structured argument takes the following approach (which I'll leave to you to complete):

    Assume the limit L exists, there are three possibilities for L:
    • L=0. But this can't be because...
    • L=1. But this can't be because...
    • L is not 0 or 1. But this can't be because...
     
  4. Jun 24, 2011 #3
    Yes You're right. my argument doesn't work for L<0 as well. I was just thinking about this a few seconds ago before you reply.

    I found your argument convincing, I mean you've assumed that there are 3 possibilities which is a right thing to do. Now let me try to work with your suggested argument.

    well, assuming L=0 then we should have |f(x) - 0| < epsilon which says |f(x)| < 0 must be true for every f(x) depending on the epsilon that I choose. am I right? now If f(x) = -1 then it says that |-1|<0 which is a contradiction.

    the second assumption too is easy to prove. but the third one is a little bit tricky. could you give me a clue of how to prove the 3rd one?

    my first argument can be corrected if I take epsilon to be equal to |L/2| and then prove L=0 separately. am I right?
     
  5. Jun 24, 2011 #4
    No, it says that |f(x)|<epsilon for each f(x). Now what happens if you choose epsilon=1/2?

    Yes, I believe so.
     
  6. Jun 24, 2011 #5
    well, sorry for being so confused at the beginning. I modified my argument this way, It turned out to be very similar to what micromass suggested:

    assuming that the limit exists and is equal to L but not equal to 0 or -1, f(x) must be either 0 or -1. if f(x)=0 then |0-L|<epsilon which means |L|<epsilon. Now take the epsilon to be |L|/2 and we have come to a contradiction.
    Now if f(x)= -1 then we have |-1-L|=|L+1|<epsilon. assuming that epsilon is equal to |L+1| leads to |L+1|<|L+1| which is false.
    Now, assume that L=0. it suffices to show that there is one value for f(x) which leads us to contradiction. take f(x) = -1. then we'll have: |-1 - 0| < epislon which says that for any epsilon less than 1 the argument will fail which is a contradiction.
    Now, assume that L=-1. it suffices to show that there is one value for f(x) which leads us to contradiction. take f(x) = 0. then we'll have |0 - (-1)| < epsilon which says that epsilon is greater than 1 which means any epsilon less than one would fail which is a contradiction.

    Now, Is this one a correct argument?
     
  7. Jun 24, 2011 #6
    Yes, that sounds good!
     
  8. Jun 24, 2011 #7
    Thank you micromass for your help. I do appreciate it xP
     
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