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Thread: Proof that lim sup (sin x) = 1

  1. #1
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    Proof that lim sup (sin x) = 1

    Please tell me if the following is sufficient to prove that lim sup (sin x) = 1. I have marked steps that I might need to explain--i.e., prove as well--with a **. My question is, at an Analysis level, would this proof suffice?

    Part A: Show lim sup (sin x) >= 1.
    **1. Observe that sin(2*n*pi + x) = sin x for any integer n.
    2. sin (pi/2) = 1.
    3. x = pi/2 => sin(2*n*pi + pi/2) = 1 for any integer n.
    4. => lim sup (sin x) >= 1.

    Part B: Show lim sup (sin x) >= 1.
    **5. The domain of sin x is [-1, 1].
    6. => lim sup (sin x) <= 1.

    7. Combining statements 4 and 6, lim sup (sin x) = 1.


    Again, at the Analysis level, would I need to prove steps 1 and 5, or is what I have up here sufficient?
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  2. #2
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    Re: Proof that lim sup (sin x) = 1

    In "B" you meant "Show lim sup(sin x)<= 1".
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    Re: Proof that lim sup (sin x) = 1

    Oops, yeah.
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    Re: Proof that lim sup (sin x) = 1

    Hi,
    What you've proved is 1=supremum { lim sup (sin(xn) : xn is a sequence}. Is this what is meant by lim sup sin(x)? I quickly googled the term lim sup and found only the standard idea for sequences ( or collections of sets). Whatever, when you use a non-standard term, you ought to define what is meant.
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  5. #5
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    Re: Proof that lim sup (sin x) = 1

    lim sup is a standard term. It's basically the supremum of the sequence as the sequence goes to infinity.

    Apparently it and its counterpart, lim inf, are fundamental concepts in number theory.

    Anyway, good to know that this proof suffices. I will be using this as a lemma for a later proof.
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