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Math Help - Calculation of a limit

  1. #1
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    Calculation of a limit

    Hey

    I'm not managing to calculate the following limit:

    The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)

    L'hopital's rule doesn't work because the limit of the quotient of the derivatives doesn't exist, and I haven't managed to find any trig identities that work. I understand intuitively that the limits is 1, since cosx and sinx are bounded and x isn't. But how can I get the function into a form where it's obvious?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by moses View Post
    Hey

    I'm not managing to calculate the following limit:

    The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)

    L'hopital's rule doesn't work because the limit of the quotient of the derivatives doesn't exist, and I haven't managed to find any trig identities that work...
    Hint: Divide numerator and denominator by x.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by moses View Post
    Hey

    I'm not managing to calculate the following limit:

    The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)

    L'hopital's rule doesn't work because the limit of the quotient of the derivatives doesn't exist, and I haven't managed to find any trig identities that work. I understand intuitively that the limits is 1, since cosx and sinx are bounded and x isn't. But how can I get the function into a form where it's obvious?
    Your 'intuition' is good because is...



    Kind regards

    \chi \sigma
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  4. #4
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    L'Hopital's rule does not apply because the limit is not of the form 0/0 or the variations on that. As x goes to 0, x+ cos(x)/2 goes to 1/2, not 0. Since the denominator does go to 0, the limit of the fraction does not exist.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by moses View Post
    The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)
    Quote Originally Posted by HallsofIvy View Post
    L'Hopital's rule does not apply because the limit is not of the form 0/0 or the variations on that. As x goes to 0, x+ cos(x)/2 goes to 1/2, not 0. Since the denominator does go to 0, the limit of the fraction does not exist.
    The limit in the OP is as x goes to infinity, not 0.

    -Dan
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  6. #6
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    Okay thanks
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