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Math Help - Limit help please!

  1. #1
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    Limit help please!

    How is it possible that lim(x to infinity) sin(x/2)/x = 1/2? I thought it was zero.....? After all, the maximum value of the numerator is 1 in absolute value while the denominator approaches infinity, thus making the expression approaching zero.

    So what's up guys? What am I missing?
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  2. #2
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    Quote Originally Posted by Kaitosan View Post
    How is it possible that lim(x to infinity) sin(x/2)/x = 1/2? I thought it was zero.....? After all, the maximum value of the numerator is 1 in absolute value while the denominator approaches infinity, thus making the expression approaching zero.

    So what's up guys? What am I missing?
    \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{x} = \frac{1}{2}

    not x \to \infty

    must be a misprint.
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  3. #3
    MHF Contributor matheagle's Avatar
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    \lim_{x\to\infty}{\sin x\over x}=0 while \lim_{x\to 0}{\sin x\over x}=1

    giving you \lim_{x\to 0}{\sin(x/2)\over x}=1/2

    SO which one do you want?
    Last edited by matheagle; March 27th 2009 at 10:55 PM.
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  4. #4
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    Guys! I didn't misprint.

    It's lim(x to infinity) sin(x/2)/x. The book must be wrong then.
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  5. #5
    MHF Contributor matheagle's Avatar
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    Then  {-1\over x} \le {\sin (x/2)\over x} \le {1\over x} and since  {1\over x}\to 0 as  x\to \infty the limit is zero.
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  6. #6
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    Talking

    If the limit is being taken as x goes to infinity, then you are correct.

    I suspect that there is a typo in the book, and that the limit was supposed to have been "as x tends toward zero". If so, then the book's answer would have been correct.

    Ask your instructor for correction or clarification.
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