serious trouble finding big O

Original statement:

$\displaystyle f(n)=(n^3+4log_2 n)/(n+4)$

I'm sure the n+4 in both the denominator and numerator have something to do with it, but if i make $\displaystyle n$in the denominator a $\displaystyle n^3$, wouldn't that then make it smaller the the LHS?

I'm suppose to prove

$\displaystyle f(n)=0(n^2)$

which means i need the inequality to make the RHS bigger than the LHS

thanks mate, just on one line...

Thanks. All makes sense except that one line starting with "Then as... we have:", hate to be a noob, but can you please explain what you mean with the arrows and stuff thanks.

Beautiful CB, just filling in some details for him

That notation means as n approaches infinity, $\displaystyle \frac{4log_2(n)}{n}$ becomes arbitrarily close to 0. It is read the limit of $\displaystyle \frac{4log_2(n)}{n}$ as n approaches infinity is 0.

So given any $\displaystyle \epsilon >0$, there exists an integer N for which, $\displaystyle |\frac{4log_2(n)}{n}|<\epsilon$ for all n>N.

In particular, if $\displaystyle \epsilon=1$, there exists an N, such that for all n>N, $\displaystyle \frac{4log_2(n)}{n}<1$, this means we clearly have the identity $\displaystyle \frac{4log_2(n)}{n}<1<n^2\Rightarrow n^2+\frac{4log_2(n)}{n}<n^2+n^2=2n^2$.

Thus $\displaystyle |\frac{x^2+\frac{4log_2(n)}{n}}{n^2}|\leq 2 $ for all n>N. Which is the definition of $\displaystyle f(n)=O(n^2)$