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Math Help - Sequence Limit Question

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    Super Member Aryth's Avatar
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    Sequence Limit Question

    First it states to give an alternate proof of a theorem that showed that a sequence's limit is unique.

    Choose \epsilon > 0. There is an N_1 such that for n \geq N_1, |a_n - A| < \frac{\epsilon}{2} and there is an N_2 such that for n \geq N_2, |a_n - B| < \frac{\epsilon}{2}. Use the triangle inequality to show that this implies that |A-B| < \epsilon. Argue that A = B.

    Now, I've already shown that the above implies that |A-B| < \epsilon, but I'm having trouble seeing how I can argue that the limits are equal (even though I know they are).
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    Re: Sequence Limit Question

    Quote Originally Posted by Aryth View Post
    First it states to give an alternate proof of a theorem that showed that a sequence's limit is unique.

    Choose \epsilon > 0. There is an N_1 such that for n \geq N_1, |a_n - A| < \frac{\epsilon}{2} and there is an N_2 such that for n \geq N_2, |a_n - B| < \frac{\epsilon}{2}. Use the triangle inequality to show that this implies that |A-B| < \epsilon. Argue that A = B.

    Now, I've already shown that the above implies that |A-B| < \epsilon, but I'm having trouble seeing how I can argue that the limits are equal (even though I know they are).
    Start by supposing that A\ne B.
    Then in what you did above let \epsilon=|A-B|>0 .
    You last line says |A-B|<\epsilon=|A-B| that is a contradiction.
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