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Math Help - Aqueeze

  1. #1
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    Aqueeze

    I dont understand the idea and proof behind Squeeze Theorem
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by HELLMACH View Post
    I dont understand the idea and proof behind Squeeze Theorem
    Hmm, now let me recall the lecture I had myself

    Lets say we want to know what the limit \lim_{x \rightarrow 0} approaches for the function x \ \sin(x)

    Now we know \sin(x) oscillates between 1 and -1

    So we can say:

    -1 < \sin(x) < 1 (We know the value of the limit must be somewhere between these two values.)

    But the original question asks for \lim_{x \rightarrow 0} x \ \sin(x)

    So multiply throughout the equation with x

    -x < x \ \sin(x) < x

    We are given that x approaches 0

    0\leq \lim_{x\to0}( x \sin x )\leq 0

    Now observe, the value must be 0. It lies between 0 and... well uhm... 0 So it must be 0 itself.

    And so we have calculated the limit!


    (The proof behind it will be in your textbook, go through it again now that I've shown you the idea behind the theorem.)
    Last edited by janvdl; July 13th 2008 at 02:09 PM.
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by janvdl View Post
    We are given that x approaches 0

    0 < x \ \sin(x) < 0
    I think it'd make more sense to write  0\leq \lim_{x\to0}( x \sin x )\leq 0  .
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi


    I think it'd make more sense to write  0\leq \lim_{x\to0}( x \sin x )\leq 0  .
    Tonight is not my night for doing math...

    Thank You.
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