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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- Nov 9th 2008, 01:25 PM #1

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- Nov 9th 2008, 06:24 PM #2

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- Nov 9th 2008, 07:11 PM #3

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If $\displaystyle \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 $\displaystyle \lim_{n\to\infty}\frac {(3n-8-4n^3)/(2n-1)}{n^2}$ satisfies the above condition.