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Math Help - function with greater value has greater limit?

  1. #1
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    function with greater value has greater limit?

    I'm a little bit confused about these similar problems in my analysis text book, and I hope to get some little help. It says, true or false:

    1. with x approaches to infinity suppose that lim f(x) = L and lim g(x) = M and f(x)<= g(x), then L<=M
    2. with x approaches to infinity suppose that lim f(x) = L and lim g(x) = M and f(x)< g(x), then L< M.
    for all x in D
    I know that the second one is wrong and one counter example I can think of is: f(x) = 2/|x| and g(x) = 3/|x| so for all non zero x, f(x) < g(x) but their limits are both zero. Is it just simple as that or I miss something?

    Also for the first one, I think it might be wrong but I cannot think of a counter example. I'm much appreciated if anyone could give me some hints. Thanks in advance.
    Last edited by EmmWalfer; September 29th 2010 at 01:40 AM. Reason: typos
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  2. #2
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    I think you are right about the second proposition.

    The first one is true. To see that, you only need to use the definition of limit and the condition that f(x)\leq g(x).
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  3. #3
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    I came up with this. I wonder if it's correct?
    for every e > 0 and e is very very small then there exists m1 in N such that for every x>= m1 we have: |f(x) - L| <e rewrite as e> f(x) - L> -e (1)
    similarly there exists m2 in N such that for every x>=m2 we have |g(x) - M| <e rewrite as -e< g(x)-M <e (2)
    subtract (1) from (2) we have:
    -2e < g(x) - f(x) + L- M) < 2e.
    we only care about the right part of the inequation above so:
    L-M < 2e +f(x) -g(x)
    since f(x) - g(x) <= 0, there always exists e positive but very very small so that 2e + f(x) - g(x) <= 0
    so L-M <= 0
    so L <= M as desired.

    Do you think i got it wrong somewhere?
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  4. #4
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    I think you are close to the solution, but there are a few problems with your argument. You say that f(x)-g(x)\leq0\Rightarrow 2\varepsilon+f(x)-g(x)\leq 0 which is not true ( 2\varepsilon>0!) and even if it was
    L-M<2\varepsilon+f(x)-g(x)\leq 0\Rightarrow L-M\leq0 is not.
    This requires more careful treatment. An easy way I see involves using the method of contradiction.
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  5. #5
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    thanks very much for your response. contradiction? I wonder if this works?

    from L-M < 2e +f(x) -g(x) and f(x) - g(x) <= 0
    >> L-M < 2e + f(x) -g(x) <= 2e
    >> L-M < 2e

    but it must be for all e > 0 so
    L-M <= 0
    >> L <= M
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  6. #6
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    Quote Originally Posted by EmmWalfer View Post
    L-M < 2e

    but it must be for all e > 0 so
    L-M <= 0
    Can you prove formally your statement?
    By using contradiction I meant this: let's assume that L>M. What then, can you finish?
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  7. #7
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    L-M < 2e and this must be for ALL e>0 including some e very very small and very very close to 0. In order for this to be always true in case e gets really really small, and really really close to 0, L-M must be be less than or equal to 0

    so L <= M

    I don't know how to work it out with the contradiction as you said:

    if L> M but f(x) <= g(x) so f(x) - L < g(x) - M >> f(x) - g(x) - L + M < 0
    but by definition: | f(x) - L| < e
    |g(x) - M| < e
    >> |f(x) - L - M + g(x)| < 2e

    >> -2e < f(x) - g(x) + L - M < 2e since 2e > 0
    >> -2e < f(x) - g(x) + L - M < 0
    >> 0 < f(x) - g(x) + L - M + 2e < 2e
    when f(x) = g(x) >> L-M + 2e < 2e
    but L-M > 0 since L >M >> contradiction

    so L<= M.

    That's all i could think of using contradiction
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  8. #8
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    L-M < 2e and this must be for ALL e>0 including some e very very small and very very close to 0. In order for this to be always true in case e gets really really small, and really really close to 0, L-M must be be less than or equal to 0

    so L <= M
    I don't think that's what a formal argument means.

    >> -2e < f(x) - g(x) + L - M < 2e since 2e > 0
    >> -2e < f(x) - g(x) + L - M < 0
    This is not true, you are making the same mistake.

    What I am trying to suggest is you start at the place you had L-M<2\varepsilon, assume that L>M and show that this leads to a contradiction.
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  9. #9
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    ok I tried again. Thanks for being patient with me:

    L>M >> L-M> 0
    >> -2e < 0 < L-M < 2e >> |L-M| < 2e but if so then L = M so contradiction!

    ???
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  10. #10
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    No problem. I think it's ok now. For the sake of completeness you should verify why your last statement is true though.

    My idea was to do it like this:
    L>M\Rightarrow\exists \delta>0:\ L=M+\delta\Rightarrow \frac{1}{2}(L-M)=\frac{1}{2}\delta<\varepsilon, \forall \varepsilon>0. This is a contradiction since you can choose \varepsilon=\frac{1}{3}\delta<\frac{1}{2}\delta.
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