Prove boundedness in the polynomial 
Prove using first principles: $42n^8 + 16n^5 + 5n\log(n) + 2017 \in O(n^{10})$

I am not sure where to begin and what "first principles" exactly means?
 A: Proving this is not difficult, once you understand intuitively why the statement holds. In general, you have $n^m \in O(n^k)$ for $k \geq m$ (note that you only need to bound the first term by a multiple of the second for all $n > n_0$ for some suitable $n_0$, in this case e.g. $n_0 = 1$; assuming that n is supposed to be an integer, which was not explicitly mentioned above, this doesn't matter, but it would be important if we wanted to prove the asymptotics on R). Furthermore, it is immediately clear that for $f(n), g(n) \in O(h(n))$ you have $f(n)+g(n) \in O(h(n))$ (by adding the constants) and $\lambda f(n) \in O(h(n))$ (by scaling the constant).
Therefore, the term to worry about is the logarithmic term. Proving that $n \log n \in O(n^{10})$ means to find an $n_0$ and a $\lambda$ such that, for $n > n_0$, $n \log n$ is bounded by $\lambda n^{10}$, or equivalently, $$\log n \leq \lambda n^{9}$$
Now imagine the graph of the logarithm function: It is very flat towards $+\infty$, whereas $n^9$ is very steep. I don't know if it can be considered a first principle that $\log n \in O(n^\alpha)\,\forall \alpha 
> 0$. It can be seen e.g. by noting $$\forall x > 1: \log x = \int_1^x \frac{dt}{t} < \int_1^x \frac{dt}{t^{1-\alpha}} = \frac 1 \alpha ( x^{\alpha} - 1)$$
A: 
$$\lim_{n \rightarrow 0}\frac{ 42n^8 + 16n^5 + 5n\log(n) + 2017}{n^{10}}=0$$

Thus for $\epsilon=\frac{1}{2}$ exists $n_0 \in \mathbb{N}$ such that  $42n^8 + 16n^5 + 5n\log(n) + 2017 <\frac{1}{2}n^{10} \leq Cn^{10}$ forall  $n \geq n_0$ and for some $C\geq\frac{1}{2}$ forall  $n \geq n_0$
Thus  $42n^8 + 16n^5 + 5n\log(n) + 2017 \in O(n^{10})$
