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Math Help - Help with L'Hospital's Rule question

  1. #1
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    Help with L'Hospital's Rule question

    Hey guys,

    I was just wondering if you could help me out with a question. I think I've wittled it down enough but I'm having troubles simplifying it and I'm not sure if I've done one step right.

    Here's the question:

    Evaluate the limit.



    This is what I have gotten so far:

    lim x --> 0 = sin(8x) / -tan(4x)^-2 (because, 1/1/cot(4x) = tan(4x)^-1, and the derivative of that is what I have written)

    Took the deriv. again because I get 0/0
    lim x --> 0 = 8cos(8x) / -4sec^2(4x)^-2 << this is the step I'm not sure about, with sec. I don't think that's the right derivative.

    Any help's appreciated. Thanks!
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  2. #2
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    Quote Originally Posted by dark-ryder341 View Post
    Hey guys,

    I was just wondering if you could help me out with a question. I think I've wittled it down enough but I'm having troubles simplifying it and I'm not sure if I've done one step right.

    Here's the question:

    Evaluate the limit.



    This is what I have gotten so far:

    lim x --> 0 = sin(8x) / -tan(4x)^-2 (because, 1/1/cot(4x) = tan(4x)^-1, and the derivative of that is what I have written)

    Took the deriv. again because I get 0/0
    lim x --> 0 = 8cos(8x) / -4sec^2(4x)^-2 << this is the step I'm not sure about, with sec. I don't think that's the right derivative.

    Any help's appreciated. Thanks!
    \lim_{x \to 0}\cot{(4x)}\sin{(8x)} = \lim_{x \to 0}\frac{\sin{(8x)}}{\tan{(4x)}}

    Since this tends to the indeterminate \frac{0}{0}, we can use L'Hospital.

     = \lim_{x \to 0}\frac{\frac{d}{dx}[\sin{(8x)}]}{\frac{d}{dx}[\tan{(4x)}]}

     = \lim_{x \to 0}\frac{8\cos{(8x)}}{4\sec^2{(4x)}}

     = \lim_{x \to 0}2\cos{(4x)}\cos^2{(8x)}

     = 2\cos{0}\cos^2{0}

     = 2.
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  3. #3
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    another way : just simplify the term \sin(8x)\cot(4x).
    you'll get the same result without using the L'hospital rule.
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