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Math Help - Trig limit

  1. #1
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    Trig limit

    Unfortunately, I can't use any advanced techniques to find this limit because our class hasn't "learned" l'hopital's rule or anything. The most I think I can use are the lim(sinx)/x and lim(1-cosx)/x rules. Here is the problem:

    lim as x approaches pi/2 of: (cosx)/(x - pi/2)

    Thanks.
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  2. #2
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    Quote Originally Posted by uberbandgeek6 View Post
    Unfortunately, I can't use any advanced techniques to find this limit because our class hasn't "learned" l'hopital's rule or anything. The most I think I can use are the lim(sinx)/x and lim(1-cosx)/x rules. Here is the problem:

    lim as x approaches pi/2 of: (cosx)/(x - pi/2)

    Thanks.
    Hint: \cos x = \sin \left( \tfrac{\pi}{2} - x \right)
    Hint: Now use substitution t= \tfrac{\pi}{2} - x
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    alternatively, just let t = x - \frac {\pi}2

    your limit becomes \lim_{t \to 0} \frac {\cos (t + \pi / 2)}t

    now use the addition formula for cosine. the result will be pleasantly surprising
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  4. #4
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    Quote Originally Posted by uberbandgeek6 View Post
    Unfortunately, I can't use any advanced techniques to find this limit because our class hasn't "learned" l'hopital's rule or anything. The most I think I can use are the lim(sinx)/x and lim(1-cosx)/x rules. Here is the problem:

    lim as x approaches pi/2 of: (cosx)/(x - pi/2)

    Thanks.
    Or just use L'Hopital's rule:
    \lim_{x \to \pi/2} \frac{cos(x)}{x - \frac{\pi}{2}} = \lim_{x \to \pi/2} \frac{-sin(x)}{1} = -1

    -Dan
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Or just use L'Hopital's rule:
    \lim_{x \to \pi/2} \frac{cos(x)}{x - \frac{\pi}{2}} = \lim_{x \to \pi/2} \frac{-sin(x)}{1} = -1

    -Dan
    well, the poster did say they couldn't use L'Hopital's
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    well, the poster did say they couldn't use L'Hopital's
    I mentioned it for the sake of simplicity. Yeah, that's why I did it. I certainly read the OP's first post and caught that. (Yeah, right!)

    Ah well. Thanks for the catch.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    I mentioned it for the sake of simplicity. Yeah, that's why I did it. I certainly read the OP's first post and caught that. (Yeah, right!)

    Ah well. Thanks for the catch.

    -Dan
    don't worry about it. i always read the OP's first post and catch stuff like that (Yeah, right!) too
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  8. #8
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    You can also use the definition of the derivative:

    \lim_{x \to \frac{\pi}{2}} \frac{\cos{x}-0}{x-\frac{\pi}{2}}=(\cos{x})'~\text{at}~x = \frac{\pi}{2}

    EDIT: that is if you covered derivatives.
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