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Math Help - Finding the Limit

  1. #1
    Junior Member
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    Oct 2009
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    Finding the Limit

    My problem is this:



    I've tried this multiple times, and I've gotten (1/16) or -(1/16).

    Using L'Hospital's Rule, the derivative of the top fuction was:

    cos(pi*e^x^8), which I calculated to be -1 after I put 0 in.

    The bottom derivative turned out to be:

    -16/(16*x^2+8*x+1)

    I plugged in 0, and I got -16.

    My homework still tells me that I'm wrong.
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  2. #2
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    Quote Originally Posted by Kimmy2 View Post
    My problem is this:



    I've tried this multiple times, and I've gotten (1/16) or -(1/16).

    Using L'Hospital's Rule, the derivative of the top fuction was:

    cos(pi*e^x^8), which I calculated to be -1 after I put 0 in.

    The bottom derivative turned out to be:

    -16/(16*x^2+8*x+1)

    I plugged in 0, and I got -16.

    My homework still tells me that I'm wrong.

    \left(x\cos^5\left(\pi e^{x^8}\right)\right)'=\cos^5\left(\pi e^{x^8}\right)-40x^8e^{x^8}\pi \cos^4\left(\pi e^{x^8}\right)\sin\left(\pi e^{x^8}\right) and this thing goes to -1 when x goes to zero.

    And the bottom's derivative is nothing close to what you say: \left(\ln(1+4x)\right)'=\frac{4}{1+4x}\xrightarrow [x\to 0] {}4 , so the limit is in fact -\frac{1}{4}

    Tonio
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