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Math Help - Improper Integration

  1. #1
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    Improper Integration



    Here is my answer.



    I know this is correct, but I'm confused about one thing. When I take the limit as t approaches 4 from the left, do I approach 4 from the left on \frac{1}{x^2 - 3x - 4} or on ln|x-4|?
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  2. #2
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    If you approached on \frac{1}{x-4}, the over all limit would not exist.
    Last edited by dwsmith; April 17th 2010 at 09:05 PM. Reason: added a quantifer
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    If you approached on \frac{1}{x-4}, the over all limit would not exist.
    Aren't we approaching ln|x-4|?
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  4. #4
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    Quote Originally Posted by temaire View Post
    Aren't we approaching
    Quote Originally Posted by temaire View Post
    ln|x-4|?


    That limit doesn't exist. It only exist on the original equation. You can't evaluate this
    ln|x-4| at 4 since one side is +\infty and the other side is -\infty
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    [font=Arial]

    That limit doesn't exist. It only exist on the original equation. You can't evaluate this
    ln|x-4| at 4 since one side is +\infty and the other side is -\infty
    Actually, aren't both sides negative? This is the graph from x=3 to x=5.

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  6. #6
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    I understand now. You can approach ln|x-4| from the left.
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  7. #7
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    Quote Originally Posted by temaire View Post
    Actually, aren't both sides negative? This is the graph from x=3 to x=5.


    That is what I was telling you. That limit exist at 4 and the ln function you were asking about DNE
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