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Math Help - uniformly continuous and boundeness

  1. #1
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    uniformly continuous and boundeness

    Let a be in R. Suppose f is a real valued function on [a, infinity) satisfying that lim as x -> infinity of f(x) is L, where L is in R.
    I need to show:

    f is bounded on [a, infinity)

    here is what I have so far:
    if f is bounded on [a, infinity), then there exists a constant M s.t. |f(x)| <= M for all x in [a, infinity)

    From the fact that lim f(x) = L, I can say that given epsilon > 0, there is a delta s.t |f(x) - L| < epsilon.

    if L=0, then |f(x)| < epsilon. I can set M <= to epsilon, but I don't think this last part makes any sense because there is no way I can claim that L =0

    or what if I say that since |f(x) - L| < epsilon, then |f(x) - L + L| < epsilon + |L|. If I let M = epsilon + |L|, then it would work, wouldn't it?
    Please provide me with some hints

    By the way, the second part of the problem is to show f is uniformly continuous on [a, infinity)
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  2. #2
    Senior Member Tinyboss's Avatar
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    I assume you're given that f is continuous?

    Edit: Or you have to show existence of an a such that f is bounded on [a,inf)? Use the definition of a limit at infinity, i.e. there is an X such that for x>X, |f(x)-L|<epsilon.
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  3. #3
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    yes, it is a continuous function
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  4. #4
    Senior Member Tinyboss's Avatar
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    Okay, find some b>a (as above) so that f is bounded on [b,inf). And f is bounded on [a,b] because it's continuous on a compact set.
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  5. #5
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    how do I prove that f is bounded on the [a,b]. I understand that if I show f is bounded on [a, b], since it is closed, then it is compact. the f is continuous on the compact set.
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  6. #6
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by inthequestofproofs View Post
    how do I prove that f is bounded on the [a,b]. I understand that if I show f is bounded on [a, b], since it is closed, then it is compact. the f is continuous on the compact set.
    Closed intervals in R are compact. Continuous functions are bounded on compact sets. Since f is continuous, f is bounded on [a,b].
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