Proof (without calculus) that any polynomial with a positive leading coefficient is positive?

Question

Theorem:

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

I have been asked to show this is true without any use of calculus. I factored my given polynomial to

$$f(x)=x^n\left(a_n+a_{n-1}\frac1x+a_{n-2}\frac1{x^2}+\ldots+a_1\frac1{x^{n-1}}+a_0\frac1{x^n}\right)$$ My particular struggle is with showing that $a_{n-1}\frac1x+a_{n-2}\frac1{x^2}+\ldots+a_1\frac1{x^{n-1}}+a_0\frac1{x^n}$ converges to $0$ without using a limit.

Any thoughts/hints? Thank you!

Answer 1

$$f(x)=x^n\left(a_n+a_{n-1}\frac1x+a_{n-2}\frac1{x^2}+\ldots+a_1\frac1{x^{n-1}}+a_0\frac1{x^n}\right)$$

Hint:   for any $\,x \gt \max\left( 1, \dfrac{|a_{n-1}|+|a_{n-2}|+\ldots+|a_0|}{a_n}\right)\,$:

$$ \left|a_{n-1}\frac1x+a_{n-2}\frac1{x^2}+\ldots+a_1\frac1{x^{n-1}}+a_0\frac1{x^n}\right| \le \frac{1}{x}\big(|a_{n-1}|+|a_{n-2}|+\ldots+|a_0|\big) \lt a_n $$

Answer 2

If the claim is false then either $f(x)<0$ for all $x>k$ for some sufficiently large $k$ or $f(x)$ changes sign infinitely often. The latter cannot occur since a polynomial of degree $n$ has at most $n$ distinct zeros and therefore cannot change sign more than $n$ times. In the former case we would have for all $x>k$ $$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0<0\\\Rightarrow a_{n-1}x^{n-1}+...+a_1x+a_0<0\hspace{0.2cm}\text{and}\hspace{0.2cm}a_nx^n<|a_{n-1}x^{n-1}+...+a_1x+a_0|\leqslant Cx^{n-1}$$ where $C:=|a_0|+...+|a_{n-1}|\geqslant 0$. But $a_nx^n<Cx^{n-1}$ implies $x<C/a_n$ which contradicts the fact that $x>k$ is unbounded.