Prove that there exists an integer greater than x such that any polynomial $f(x)$ will be strictly non-negative and get large? Hi I am taking a number theory class and so far I have been proving modular congruences, modular arithmetic, and prime properties. There is this theorem that came up in the textbook and apparently it does not involve any modular arithmetic. The theorem is as follows   

Suppose $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ is a polynomial of degree $n>0$. Then there exists an integer $k$ such that if $x>k$ then $f(x)>0$. 

I feel like this would come up in real analysis, but I have not come that far in my studies. I have an idea of applying induction and using the ceiling function somehow, but I have no idea how to start off this proof. Any help will do and thank you

One should note that the field of interest here (presumably) is $\mathbb{R}$, so that $f(x)$ is a real polynomial. Notice first that for $a_n<0$, this fails completely, because we can just choose $f(x)=-x$ and it does not have the property.
If we assume that $a_n>0$, then write
$$ \frac{f(x)}{x^n}=a_n+a_{n-1}x^{-1}+\cdots +a_1 x^{1-n}+a_0 x^{-n}$$
so that $\lim_{x\to\infty} \frac{f(x)}{x^n}=a_n>0$. This implies that $\lim_{x\to\infty}f(x)=\infty$, so that $f(x)>0$ for $x\ge N$ for some $N\in \mathbb{R}$. 
