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Math Help - Quick limit question

  1. #1
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    Quick limit question

    So I've boiled down a problem to the point that I've got ln|x-1|-ln|x+1| evaluated at B=-2 and A=-infinity

    Boiling it down further, I get ln|(-3)| - ln|-1|-(limits as X->-inf for both of the terms)

    This gets me to LN(3)-0-(infinity-infinity). Is that valid? It says the answer is LN(3) but can I really subtract infinity from infinity and get zero?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Wolvenmoon View Post
    So I've boiled down a problem to the point that I've got ln|x-1|-ln|x+1| evaluated at B=-2 and A=-infinity

    Boiling it down further, I get ln|(-3)| - ln|-1|-(limits as X->-inf for both of the terms)

    This gets me to LN(3)-0-(infinity-infinity). Is that valid? It says the answer is LN(3) but can I really subtract infinity from infinity and get zero?
    Post the original question.

    CB
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  3. #3
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    Original question is the integral from -infinity to -2 of 2dx/(x^2-1)
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  4. #4
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    Quote Originally Posted by Wolvenmoon View Post
    Original question is the integral from -infinity to -2 of 2dx/(x^2-1)
    \ln|x-1|-\ln|x+1|=\ln\left|\frac{x-1}{x+1}\right|
    As x\to -\infty that limit is 0.
    So your answer is just \ln(3).
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