1. ## 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

2. Originally Posted by BlakeRobertsonMD
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
3. 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.