Showing $f(n)$ has upperbound of $O(g(n))$ I am currently enrolled in an algorithms course and was learning about upper, lower and tight bounds of functions. I am confused on how to show that a function $f(n) = O(g(n))$ for some $n > n_0$ and $c$. The definition for upper bound is: there exists positive constants $c$ and $n_0$ such that $0 \le f(n) \le c \cdot g(n)$ for all $n \ge n_0$. I am currently stuck on a homework question similar to this: > $$T(n) = 25 n^5 \log(n) + 15n^5 + 8^5$$ Show that $T(n)$ has an upper bound of $O(n^5 \log(n))$ So this is what I have done so far: $$ 0 \le 25n^5 \log(n) + 15n^5 + 8^5 \le c n^5 \log(n) $$ divide everything by $n^5$ $$ 0 \le 25\log(n) + 15 + 8^5/n^5 \le c\log(n) $$ Now I am stuck, how do I deal with $8^5/n^5$?

There is a more simple-minded approach.

Note that for $n\ge 3$, $15n^5\lt 15n^5\log n$. (Here I am assuming that by $\log$ you mean the natural logarithm. If it is the base $10$ logarithm, then $n$ has to be a bit larger.)

Note also that for $n\ge 8$, we have $8^5\lt n^5\log n$.

So if $n\gt 8$, your function is less than $41n^5\log n$. It follows that your function is $O(n^5\log n)$.