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Math Help - Disproving the existance of a limit at infinity

  1. #1
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    Disproving the existance of a limit at infinity

    Hello

    I'm given the function x*sin(x), and asked whether it has a limit of some sort as x goes to negative infinity. In other words does it converge to a number or go off to positive infinity or negative infinity.

    Based on the graph of the function my intuition is that it doesn't have a limit of any sort since it is constantly fluctuating between very large and very small values.

    I think i've managed to prove that it doesn't converge to any number at negative infinity, because on any interval can always supply an x1 = -pi/2 -n*pi for which the value is -x1 and an x2 = -(3/2) * pi -n*pi for which the value is x2, i.e. the function doesn't get closer to any single number.

    But I'm having a lot of trouble with the infinite limits, proving that the function doesn't go off to positive or negative infinity...help would be much appreciated
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  2. #2
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    Quote Originally Posted by moses View Post
    Hello

    I'm given the function x*sin(x), and asked whether it has a limit of some sort as x goes to negative infinity. In other words does it converge to a number or go off to positive infinity or negative infinity.

    Based on the graph of the function my intuition is that it doesn't have a limit of any sort since it is constantly fluctuating between very large and very small values.

    I think i've managed to prove that it doesn't converge to any number at negative infinity, because on any interval can always supply an x1 = -pi/2 -n*pi for which the value is -x1 and an x2 = -(3/2) * pi -n*pi for which the value is x2, i.e. the function doesn't get closer to any single number.

    But I'm having a lot of trouble with the infinite limits, proving that the function doesn't go off to positive or negative infinity...help would be much appreciated


    Take the sequences \displaystyle{\{-n\pi\}\,,\,\left\{-(4n-1)\frac{\pi}{2}\right\} . We know that if \lim\limits_{x\to -\infty}x\sin x exists, then it is

    the same no matter how we choose to make x\to -\infty , so do this through the above two seq's.

    Tonio
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  3. #3
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    I'm afraid I haven't learned anything about sequences yet...is there another way?
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  4. #4
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    Quote Originally Posted by moses View Post
    I'm afraid I haven't learned anything about sequences yet...is there another way?
    I am truly puzzled by this question.
    If you can do it for x\to -\infty it is exactly the same for x\to \infty.
    After all (-\infty,0]~\&~[0,\infty) are 'copies' of one another as for as the sine function is concerned.
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  5. #5
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    Quote Originally Posted by Plato View Post
    I am truly puzzled by this question.
    If you can do it for x\to -\infty it is exactly the same for x\to \infty.
    After all (-\infty,0]~\&~[0,\infty) are 'copies' of one another as for as the sine function is concerned.


    It probably is a witty question: prove that \lim\limits_{x\to -\infty}x\sin x=\lim\limits_{x\to\infty}x\sin x , since the

    function x\sin x is even...

    Anyway, I can't see right now a straightforward way to prove what the OP wants with resourcing to sequences.

    Tonio
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  6. #6
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    Okay, I see I phrased the question pretty unclearly

    Basically what I meant was, how can I prove that the limit of xsinx as x goes to negative infinity is not positive infinity, and also that the limit of xsinx as x goes to negative infinity is not negative infinity?
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