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Math Help - Big Oh

  1. #1
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    Big Oh

    I have to show all the steps showing that (3n-8-4n^3)/(2n-1) is O(n^2). I'm not really sure where to begin since we were never taught how to deal with fractions with Big Oh
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  2. #2
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    Quote Originally Posted by BlakeRobertsonMD View Post
    I have to show all the steps showing that (3n-8-4n^3)/(2n-1) is O(n^2). I'm not really sure where to begin since we were never taught how to deal with fractions with Big Oh
    Polynomial long division leads to:

    -7/(2n - 1) - 2n^2 - n + 1 -> O(n^2)
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  3. #3
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    If \lim_{x\to\infty}\frac{f(x)}{g(x)} < \infty , then f(x) is O(g(x)).

    Now, you can use a L'Hopital's rule to check  \lim_{n\to\infty}\frac {(3n-8-4n^3)/(2n-1)}{n^2} satisfies the above condition.
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