Correct this improper definition of a limit

  1. 1. The problem statement, all variables and given/known data
    Eddy wrote on his midterm exam that the definition of the limit is the
    following: The sequence {an} converges to the real number L if there
    exists an N ∈ Natural numbers so that for every [itex]\epsilon[/itex] > 0 we have |an − L| < [itex]\epsilon[/itex] for all
    n > N. Show Eddy why he is wrong by demonstrating that if this were
    the definition of the limit then it would not be true that lim n→∞ 1/n = 0.
    (Hint: What does it mean if |a − b| < [itex]\epsilon[/itex] for every [itex]\epsilon[/itex] > 0?)


    2. Relevant equations
    |a-b| <ε means that ||a|-|b|| < ε from the reverse triangle inequality


    3. The attempt at a solution
    I know it has to do with the fact that the actual definition of a limit has "for every ε > 0, there exists an N [itex]\in[/itex] Natural numbers S.T. ...." so, Eddy reversed that part of the definition. I just haven't been able to quite see the difference of the two. A little push in the right direction would be greatly appreciated. I like figuring these out on my own, so no full on answers, please.
     
  2. jcsd
  3. micromass

    micromass 18,444
    Staff Emeritus
    Science Advisor

    So there exists a certain N.

    Let [itex]\varepsilon = |1/N|[/tex]

    and try to prove that 1/n does not converge to 0 with this definition.
     
  4. I'm not quite sure I follow. I ended up answering it this way: Eddy's definition implies there is a single natural number, N, such that for all n>N |1/n|< every epsilon greater than zero. Which is not true. For every epsilon you give me, I can find an N such that 1/n is less than that epsilon for all n>N, but if you pick a newer, smaller epsilon, my N has to be larger, and since the natural numbers are unbounded, we can do this forever. But, it's a different N for each new epsilon, not one single N like eddy implied. Does that make sense?
     
  5. vela

    vela 12,444
    Staff Emeritus
    Science Advisor
    Homework Helper

    That's right.
     
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