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Math Help - Infinite limits

  1. #1
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    Infinite limits

    Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)



    how do i solve this problem.
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  2. #2
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    Hello

    1)Expand the bottom out.
    2)Divide both the top and bottom of the fraction by  u^4

    And you should arrive at:

     \lim_{u \to \infty} = \frac{3 + \frac{3}{u^4}}{2 - \frac{17}{u^2} + \frac{8}{u^4}}

    You know that \lim_{u \to \infty} \frac{1}{u} = 0 right? So anything like  \lim_{u \to \infty} \frac{3}{u^4} you can express as  3 \lim_{u \to \infty} \frac{1}{u^4} or even  3 \lim_{u \to \infty} \frac{1}{u} . \lim_{u \to \infty} \frac{1}{u}. \lim_{u \to \infty} \frac{1}{u} . \lim_{u \to \infty} \frac{1}{u} = 0

    So the limit in your question is  \frac{3 + 0}{2 - 0 - 0}

    =  \frac{3}{2}
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  3. #3
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    Quote Originally Posted by plstevens View Post
    Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)



    how do i solve this problem.
    Did you mean
    \lim_{u \to \infty} \frac{3u^4 + 3}{(u^2 - 8)(2u^2 - 1)}

    There are a variety of ways. I think the clearest at this stage of the game would be to expand the denominator.
    \lim_{u \to \infty} \frac{3u^4 + 3}{(u^2 - 8)(2u^2 - 1)}

    = \lim_{u \to \infty} \frac{3u^4 + 3}{2u^4 - 17u^2 + 8}

    The next step is standard: Divide the numerator and denominator by the highest power appearing in the problem, in this case u^4:
    = \lim_{u \to \infty} \frac{3 + \frac{3}{u^4}}{2 - \frac{17}{u^2} + \frac{8}{u^4}}

    Now take your limit.

    -Dan
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