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Math Help - L'Hopitals rule... getting right answer, but it should be negative not positive. Why?

  1. #1
    Member nautica17's Avatar
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    L'Hopitals rule... getting right answer, but it should be negative not positive. Why?

    Okay, so I'm working with L'Hopitals rule and indeterminate forms.. and I ran across this problem in my homework. I get the right answer, only that it is positive instead of negative and I can't figure out why it's negative.

    The problem:
    limit (sec(3x)cos(5x)) x-> (pi/2) from the left

    since that is in infinity * 0 form... I did:

    = cos(5x) / (1/sec(3x)) = 0/0

    Great... I can now use L'Hopitals...

    = -5sin(5x) / -3sin(3x) = 5/3

    Negatives cancel right?? So now where did I go wrong? The answer should be negative according to my textbook, Wolfram Alpha, and MS Math.
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  2. #2
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    Hello, nautica17!

    There is a simple fact you overlooked . . . (I almost did!)


    \lim_{x\to\frac{\pi}{2}^-} \sec(3x)\cos(5x)
    Your work is correct . . .

    Using L'Hopital, the problem becomes: . \lim_{x\to\frac{\pi}{2}^-}\frac{5\sin(5x)}{3\sin(3x)}


    As x approches \tfrac{\pi}{2}, the fraction approaches: . \frac{5\sin\left(\frac{5\pi}{2}\right)}{3\sin\left  (\frac{3\pi}{2}\right)} \;=\; \frac{5(1)}{3({\color{red}-1})} \;=\;-\frac{5}{3}

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  3. #3
    Member nautica17's Avatar
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    Thank-you! That's so obvious now! I've been doing to many of these tonight. :|
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