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Definition of a limit of a sequence

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the definition of a limit to prove that lim [(1+an)-1] = 1/2 if lim an = 1.

    2. Relevant equations

    ([itex]\forall[/itex][itex]\epsilon[/itex]>0)([itex]\exists[/itex]N[itex]\in[/itex]N)(n[itex]\geq[/itex]N [itex]\Rightarrow[/itex]|an-L|<[itex]\epsilon[/itex])

    3. The attempt at a solution

    Let [itex]\epsilon[/itex] be arbitrary. Since lim an exists, [itex]\exists[/itex]N[itex]\in[/itex]N such than |an-1|<[itex]\epsilon[/itex]'.

    My professor helped me a bit, but once we started comparing two different epsilons, I couldn't follow him anymore. He said to choose [itex]\epsilon[/itex]'< 1/2 since 1/2 < an, but I don't understand why we can say that the sequence is greater than or equal to 1/2 since we only know the value of its limit.

    Any help would be appreciated, I've always had a hard time with the rigorous definitions.
     
  2. jcsd
  3. Oct 13, 2011 #2

    SammyS

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    Welcome to PF.

    Let me try to read his mind.
    Since lim an exists, ∃N∈ such than |an-1| < ϵ'.

    Letting ϵ' = 1/2, since it is arbitrary, gives: |an-1| < 1/2  → -1/2 < an-1 < 1/2

    Therefore, 1/2 < an < 3/2.  It's the 1/2 < an you're interested in.


    If an > 1/2, then 2(an + 1) > 3  → [itex]\displaystyle \frac{1}{2(a_n+1)}<\frac{1}{3}[/itex]
    That should be handy in working with |(1+an)-1 - 1/2|, when you write that as one fraction with a common denominator.
     
  4. Oct 14, 2011 #3
    So then, we have showed that 1/(1+a_n) < epsilon'/3 < epsilon', but is this the end of the proof? I guess my question is, when does the other epsilon come in? or is epsilon' just our other epsilon in disguise? Thank you so much! I could use a mind-reader for this professor :)
     
  5. Oct 14, 2011 #4
    Also, if you wouldn't mind--(this homework is due in an hour, so this is more for my understanding)--in general, how do we know how to choose epsilon so that it will give us the result we are looking for? Is there a reason choosing epsilon'=1/2 was particularly useful for this problem?
     
  6. Oct 14, 2011 #5

    SammyS

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    No, that's not the complete proof.

    Let ε' = min(ε, 1/2)

    Then you have N∊ such that for n > N, |an-1| < ε' ≤ 1/2
    This leads to 1/(2|an+1|) < 1/3.

    Also, for n > N, |an-1| < ε' ≤ ε

    Put them together, & you get (the desired quantity) < ε/3 < ε​
     
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