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 $8^5/n^5$?
 A: 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 $O(n^5\log n)$. 
A: You have $$0\le 25\log n + 15 + \frac{8^5}{n^5}\le c\log n\;,$$
and you want to choose $c>0$ and $n_0$ so that this will be true for all $n\ge n_0$. First you could notice that if $n\ge 8$, then $\frac{8^5}{n^5}\le 1$, so for $n\ge 8$ we have
$$0\le 25\log n + 15 + \frac{8^5}{n^5}\le 25\log n+16\le c\log n\;,$$
provided that we can choose $c$ properly. Factor out the $25$: we’d like to get
$$25\left(\log n+\frac{16}{25}\right)\le c\log n\;.$$
Suppose that $n$ is big enough so that $\log n\ge\frac{16}{25}$; then
$$\log n+\frac{16}{25}\le 2\log n\;,\tag{1}$$
and therefore $$25\left(\log n+\frac{16}{25}\right)\le 50\log n\;,$$
so we can take $c=50$. How big does $n$ have to be for this to work? $\log n\ge\frac{16}{25}$ if and only if $n\ge e^{16/25}$, so taking $n\ge e$ will make $(1)$ true. We already required $n$ to be at least $8$, so we already know that $n\ge e$, and we can set $n_0=8$ and $c=50$.
