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Math Help - Intro to Real Analysis help: Proving limit is zero.

  1. #1
    Member ilikedmath's Avatar
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    Exclamation Intro to Real Analysis help: Proving limit is zero.

    Problem states:

    Let be a sequence in with , and n.
    Prove that = 0.

    ---------------------------------------------
    Note: The following had to be proven previously, so I assume it will help in completing the above proof: "Let and be sequences of real numbers. If is bounded and = 0,
    then = 0."

    ----------------------------------------------

    My ideas so far:
    Since { x_n} → 0, I know x_n is bounded, and by the theorem if I can show sin (1/ x_n) → 0 then I'm done.

    Or, since x_n → 0, if I can show sin (1/ x_n) is bounded, then I'm done.

    My question now is, which of the two approaches is correct, and how would I go about doing that?

    I thought I could do this:
    I know sin (n) is bounded by [-1, 1] so,
    -1 ≤ sin (n) ≤ 1 so dividing all by n gives,
    -1/n ≤ sin (n)/n ≤ 1/n
    I know both -1/n and 1/n converge to zero, so by squeeze theorem,
    sin (n) / n converges to 0 too. But my classmate pointed out that
    sin (n) / n is not the same as sin (1/ x_n). So I can't do that. :| Therefore I am stuck.

    Any help, suggestions, corrections, ideas are greatly appreciated.
    Thank you for your time!

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  2. #2
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    \left( {\forall x} \right)\left[ {\left| {\sin (x)} \right| \leqslant 1} \right]
    So the sequence \left( {\sin \left( {\frac{1}{x_n}} \right)} \right) is bounded.
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  3. #3
    Member ilikedmath's Avatar
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    Quote Originally Posted by Plato View Post
    \left( {\forall x} \right)\left[ {\left| {\sin (x)} \right| \leqslant 1} \right]
    So the sequence \left( {\sin \left( {\frac{1}{x_n}} \right)} \right) is bounded.
    How can I get the x in \sin (x) to be 1/ x_n to show that sin (1/ x_n) is bounded? I'm not quite "seeing" how to get to that conclusion right away since sin(x) is bounded by 1.
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    This is true \left[ {\left| {\sin (?)} \right| \leqslant 1} \right] no matter what you choose to put in place of the ?.
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  5. #5
    Member ilikedmath's Avatar
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    Quote Originally Posted by Plato View Post
    This is true \left[ {\left| {\sin (?)} \right| \leqslant 1} \right] no matter what you choose to put in place of the ?.
    Okay, thanks! I got it now. So that makes the proof pretty short then, or would I have to prove somehow that sin (x) is bounded?
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