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Math Help - Limit Analysis Question

  1. #1
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    Limit Analysis Question

    Hey.

    I have played around with this question for my analysis class and I would like some help:

    Given: L= lim f1(x), M=lim f2(x)

    Show that if f1<=f2 for all x in some interval (a,b), then L<=M
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  2. #2
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    It's probably easiest to do this by contradiction. Suppose that L>M and let \varepsilon = \tfrac12(L-M). For x sufficiently close to some limit point c (the limit point isn't specified in the question), it will be true that |f_1(x)-L|< \varepsilon and |f_2(x)-M|< \varepsilon. From that, and the fact that f_1(x)\leqslant f_2(x), you should be able to use the triangle inequality to get a contradiction.
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  3. #3
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    I am still struggling here. I can't find a way to make the triangle inequality work for this because if I use it on |f_1(x)-L|< \varepsilon or |f_2(x)-M|< \varepsilon I know nothing about the relationship between |f_1(x)| + |-L| and epsilon.
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  4. #4
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    Quote Originally Posted by Uriah View Post
    I am still struggling here. I can't find a way to make the triangle inequality work for this because if I use it on |f_1(x)-L|< \varepsilon or |f_2(x)-M|< \varepsilon I know nothing about the relationship between |f_1(x)| + |-L| and epsilon.
    Okay, |f_1(x)-L|< \varepsilon is equivalent to L-\varepsilon<f_1(x)<L+\varepsilon. Similarly, M-\varepsilon<f_2(x)<M+ \varepsilon. From those, you can extract L-\varepsilon<f_1(x)\leqslant f_2(x)<M+ \varepsilon. Can you finish it from there?
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  5. #5
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    Duh. I can't believe I didn't look at it like that. Thanks!
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