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Proving from first principles that a(n)^2 -> 4 if a(n) -> 2.

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Let an → 2. Prove from first principles (i.e. give a direct ε-N proof) that an2 → 4.

    2. Relevant equations


    3. The attempt at a solution
    I have tried considering |an-2|2 and considering that |an2-4| = |(an+2)(an-2)| but I could not get either of these methods to work. Would someone be able to point me in the right direction?
    I was thinking maybe one of the sequences is a subsequence of the other?

    I was also wondering, is this true generally that if a sequence tends to L then the sequence squared will tend to L2?

    Thanks,
    Tommy
     
  2. jcsd
  3. Feb 28, 2015 #2

    HallsofIvy

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    You are given that, for [itex]\epsilon> 0[/itex], there exist N such that is n> N, [itex]|a_n-2|< \epsilon[/itex] (since [itex]a_n[/itex] goes to 2).

    Now, you want [itex]|a_n^2- 4|= |a_n+ 2||a_n- 2|< \epsilon[/itex]. If you can find an upper bound, A, on |a_n+ 2| then you can say that [itex]|a_n+ 2||a_n- 2|< A|a_n- 2|[/itex] so that if [itex]A|a_n- 2|< \epsilon[/itex], which is the same as [itex]|x_n- 2|< \epsilon/A[/itex], then [itex]|a_n+ 2||a_n- 2|< A|a_n- 2|< \epsilon[/itex].

    Now, if [itex]a_n[/itex] is close to 2, say, [itex]|a_n- 2|< 1[/itex], what can you say about [itex]|a_n+ 2|[/itex]?
     
    Last edited by a moderator: Feb 28, 2015
  4. Feb 28, 2015 #3

    RUber

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    Start with something like :
    There exists a delta such that if |an-2|<delta, then (an-2)^2 < epsilon.
    Then notice that (an+2) = (an-2)+4.
    I think those are the main techniques in this proof.
     
  5. Feb 28, 2015 #4
    Thanks HallsofIvyy I think I will be able to do it now. Is it O.K for me to say that for suitably large n, |an-2|<ε<1 ⇒ |an+2|<3 so 3 can be considered an upper bound on an when n is suitably large??
     
  6. Feb 28, 2015 #5

    HallsofIvy

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    Yes, that will work.
     
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