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Math Help - Limit of nondecreasing function that is bounded above

  1. #1
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    Limit of nondecreasing function that is bounded above

    Prove that if f(x) is nondecreasing and bounded above on (a,b) then Lim x->a (f(x)) = M where M = least upper bound {f(x)|x \in (a,b)}.

    I know that since f is bounded above, there is a number M so that f(x) \leq M for all x in (a,b). Then if I let L = least upper bound of f(x), I claim that L = lim x->a^- (f(x)). So, let \epsilon > 0. Then L - \epsilon < L. So L - \epsilon is not an upper bound for {f(x)}. So, for some x_0 \in (a,b), f(x_0) > L - \epsilon.

    Now I am not sure how to choose \delta... I am thinking \delta might be b - x_0 but I am not sure.
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  2. #2
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    Nothing wrong with \delta= |a- x_0|!
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    OK, suppose I choose \delta to be a - x_0 (since L = the limit as x -> a from the left). Now I must show that if 0 < a - x_0 < \delta, then |f(x) - L| < \epsilon. Now a - x < \delta \implies a - x < a - x_0, so x > x_0. Since f is nondecreasing, f(x) > f(x_0) > L - \epsilon. Since L = l.u.b.{f(x)}, then f(x) \leq L. Then
    L - \epsilon \leq f(x) \leq L
    L - \epsilon \leq f(x) \leq L \leq L + \epsilon
    L - \epsilon \leq f(x) \leq L + \epsilon
    -\epsilon \leq f(x) - L \leq \epsilon
    |f(x) - L| < \epsilon QED

    Right?
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  4. #4
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    Quote Originally Posted by tarheelborn View Post
    Prove that if f(x) is nondecreasing and bounded above on (a,b) then Lim x->a (f(x)) = M where M = least upper bound {f(x)|x \in (a,b)}.
    Tarheelborn, please reread what you have posted.
    It is not true. I think you mean  \displaystyle \lim _{x \to b} f(x) = M.
    But you cannot even prove that!

    This is true,  \displaystyle \lim _{x \to b^{-}} f(x) = M.
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  5. #5
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    Oops, duh...

    So do I need to change my \delta to relate to b or can I still use a - x_0?
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    If \varepsilon  > 0 then M - \varepsilon  < M = \sup \left\{ {f(x):x \in (a,b)} \right\}.

    So \left( {\exists x_0  \in (a,b)} \right)\left[ {M - \varepsilon  < f(x_0)  \leqslant M \leqslant f(b)} \right].

    Now finish.

    Now finish.
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  7. #7
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    I have to have an epsilon/delta proof. So I think I got goofed by trying to follow the example from the limit as x approaches a from the right. I should have been dealing with b all along. So, if \epsilon > 0, then L - \epsilon < L. So L - \epsilon is not an upper bound... So for some x_0 \in (a,b), f(x_0) > L - \epsilon. Choose \delta to be b - x_0. Now I must show that if 0 < b - x < \delta, then |f(x) - L| < \epsilon.
    b - x < \delta ==> b - x > b - x_0 so x > x_0. Since f is nondecreasing,
    L - \epsilon \leq f(x_0) \leq f(x). Since L = l.u.b.{f(x)}, certainly f(x) \leq L. Now I have
    L - \epsilon \leq f(x) \leq L
    L - \epsilon \leq f(x) \leq L \leq L + \epsilon
    -\epsilon \leq f(x) - L \leq \epsilon
    |f(x) - L| < \epsilon QED

    Does that get it? Thank you for your help!
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