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Math Help - A limit

  1. #1
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    A limit

    Got any leads on calculating:
    \lim{x\to\frac{\pi}{2}} : \frac{cotx}{2x-\pi}

    Sorry for the bad latex, but I am trying you know.
    Last edited by a4swe; October 8th 2005 at 04:44 AM.
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  2. #2
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    Lim(X-->pi/2)[cotX / (2x -pi)]
    = cot(pi/2) / 2*pi/2 -pi
    = 0/0
    Indeterminate.
    Use the L'Hopital's Rule: Lim(x->a)[f(x)/g(x)] = Lim(x->a)[f'(x)/g'(x)].

    = Lim(X->pi/2)[-csc^2(X) / 2
    = -csc^2(pi/2) / 2
    = -[1/sin(pi/2)]^2 /2
    = -[1/1]^2 / 2
    = -1/2 ------------answer.
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  3. #3
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    That's true, thanks.

    But is there a (there is) way to solve it without using L'Hopital's Rule?
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  4. #4
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    Quote Originally Posted by a4swe
    Got any leads on calculating:
    \lim{x\to\frac{\pi}{2}} : \frac{cotx}{2x-\pi}

    Sorry for the bad latex, but I am trying you know.
    Here's the Latex for reference.

    \lim_{x\to\frac{\pi}{2}} \frac{\cot(x)}{2x-\pi}

    And I don't see a way to do this without L'Hopitals either.
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  5. #5
    hpe
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    Using the substitution x - \pi/2 \to x,we get
    \lim_{x\to\frac{\pi}{2}} \frac{\cot(x)}{2x-\pi} = \lim_{x \to 0}\frac{\cot(x + \pi/2)}{2x} = \lim_{x \to 0}\frac{\cos(x+\pi/2)}{2x\sin(x+\pi/2)} which becomes \lim_{x \to 0}\frac{-\sin(x)}{2x\cos(x)} = -\frac{1}{2}.
    The less L'Hopital, the better.
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