Just wondering if I can get some feedback regarding this proof, and whether or not there is a way to do it without limits.
My Solution:
$(n + \log(n))(n+8)$ is not $O(n)$
Suppose by contradiction that $(n + \log(n))(n+8)$ is O(n). Then there exists c > 0 and $n_0 \geq 1$ such that:
$(n + log(n))(n+8) \le c \cdot n $
$n^2 + 8n + nlog(n) + 8log(n) \le c \cdot n$
$\frac{(n^2 + 8n + nlog(n) + 8log(n))}{n} \le c$
$n + 8 + log(n) + \frac{8log(n))}{n} \le c$
$\lim \limits_{n\to \infty} (n + 8 + log(n) + \frac{8log(n))}{n} ) = \infty $
The left side approaches infinity and therefore will never be less than a constant. This leads to a contradiction. $(n + \log(n))(n+8)$ is not $O(n)$
Yes you are correct indeed
$$ (n + \log n)(n+8)=n^2+8n+n\log n + 8 \log n=O(n^2)$$