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Convergence and limits analysis problem

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume that (xn) and (yn) are sequences, both of which
    converge to L. Assume further that (zn) is a sequence satisfying
    xn <or= zn <or= yn

    for all n in N. Prove that (zn) also converges to L

    2. Relevant equations



    3. The attempt at a solution

    Let xn and yn be sequences that converge to L and let zn converge to L'
    The fact that xn is <or= to zn for all n in N implies that L<or=L'
    The fact that zn is <or= to yn for all n in N implies that L'<or=L'
    Meaning L<or=L'<or=L
    Therefore, L=L'

    Would this be correct? Do I need to show that Zn converges before I assume it has a limit?
     
  2. jcsd
  3. Sep 26, 2009 #2
    Yes. You have not shown that zn converges at all; we can't start by assuming it converges, unless we later plan to append a proof for the case where it is not assumed to converge, in which case the latter proof is stronger and replaces the former!
    Also, I'm not sure what theorems you have already proven, but you may need to show the proof for the second line's inequality limit implication.
     
  4. Sep 26, 2009 #3
    we've already proven the theorem that if
    lima<or= limb for all n in N then a<b so I assumed I could just use that theorem.

    How would you reccomend I go about proving that zn converges? Should I do an induction proof? Because the sequence is bounded, so all I'd have to prove would be that it is either increasing or decreasing, right?
     
  5. Sep 26, 2009 #4
    i'm sorry i mean we have already proven that
    if lima<or=limb for all n in N then a<or=b
     
  6. Sep 26, 2009 #5
    Unfortunately, the existence of lim b is part of the assumption for that theorem, so it cannot be used directly until we show zn does indeed converge. This is actually easier than it looks and need only uses the definition of the limit of a sequence.
    Consider the two inequalities:
    xn - L <= zn - L <= yn - L
    L - xn >= L - zn >= L - yn
    which are perfectly valid pointwise. Then add in the definition of the limit of the sequences x and y tending to L, and show that this gives the definition of z converging to L as well.
     
  7. Sep 26, 2009 #6
    Alright how does this look?

    assume xn<=zn<=yn
    Meaning, xn-L<=zn-L<=yn-L

    Since L is the limit of xn and yn, that means:
    for e>0 there exists N1,N2 in N so that:

    |xn-L|<e and |yn-L|<e
    which is equivalent to:
    -e<xn-L<e and -e<yn-L<e
    since -e<xn-L and yn-L<e
    that means:
    -e<xn-L<=zn-L<=yn-L<e
    and -e<zn-L<e
    meaning for e>0 there exists an N in N so that
    |zn-L|<e
    and therefore, L is the limit of zn
     
  8. Sep 26, 2009 #7
    Excellent work. Very straightforward. :smile:
     
  9. Sep 26, 2009 #8
    thanks so much for the help :D
     
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