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Math Help - L1 and L2 limits of f. Prove L1=L2

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    L1 and L2 limits of f. Prove L1=L2

    Suppose f: D ---->R with x_0 an accumulation point of D. Assume L1 and L2 are limits of f at x_0. Prove L1=L2.

    My thought is to assume L1 is not equal to L2, but I'm stuck here.
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    Quote Originally Posted by kathrynmath View Post
    Suppose f: D ---->R with x_0 an accumulation point of D. Assume L1 and L2 are limits of f at x_0. Prove L1=L2.

    My thought is to assume L1 is not equal to L2, but I'm stuck here.
    Ok, since they are not equal, we have |L_1 - L_2| > 0

    Now choose \epsilon = \frac {|L_1 - L_2|}3

    Then |L_1 - L_2| = |L_1 - L_2 + f(x) - f(x)| = \dots

    now continue and try to get a contradiction on the magnitude of |L_1 - L_2|
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    Quote Originally Posted by kathrynmath View Post
    Suppose f: D ---->R with x_0 an accumulation point of D. Assume L1 and L2 are limits of f at x_0. Prove L1=L2.

    My thought is to assume L1 is not equal to L2, but I'm stuck here.
    Here is an outline of proof you need to fill in more details to make it formal.
    |L_1 - L_2| = |(f(x) - L_1) - (f(x) - L_2)| \leq |f(x) - L_1| + |f(x) - L_2| < \tfrac{\epsilon}{2} + \tfrac{\epsilon}{2} = \epsilon

    Since this is for any \epsilon > 0 it forces |L_1 - L_2| = 0 \implies L_1 = L_2.
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