Any polynomial with a positive leading coefficient is positive

Question

I have some questions about this answer: https://math.stackexchange.com/a/2711058/1196218

1. Is $x$ positive (how can you tell?)? If not then im confused about a few things. For example, why would the following be true: $a_nx^n<|a_{n-1}x^{n-1}+...+a_1x+a_0|\leqslant Cx^{n-1}$? I understand that $|a_{n-1}x^{n-1}+...+a_1x+a_0|\leqslant C$ (by triangle inequality), but if $x<0$, you cant multiply the $C$ right? You cant just put $Cx^{n-1}$ on the right side.

2. What exactly is $C$ here? I read that $C$ is often used to represent constant values. Is the letter $C$ here just randomly chosen or does it have some common meaning, related to absolute values or something else? Also I heard something that $C$ is often called constant of integration, does it occur here?

3. I don't understand the last sentence:

   "But $a_nx^n<Cx^{n-1}$ implies $x<C/a_n$ which contradicts the fact that $x>k$ is unbounded."
   What does it mean that $x>k$ is unbounded? Is it just that $x$ can be arbitrary larger than $k$? What does $x<C/a_n$ mean? Why is it a contradiction?

Answer 1

1. i think the answer there forgets to take absolute value of x^n-1
2. It is a common technique that used in real analysis, find something that must be bounded above, C is defined by the answer(by using the notation “:=“)
3. C/an is a number that C is defined as sum of all absolute value of a0 + … . As it is smaller than a number, then it can’t be arbitrary large, hence it is a contradiction. But if you can prove it is greater than whatever number you make, then it is unbounded I dont know whether it can help you

Answer 2

I’ll just add a few notes to @Question’s answer. For the first point, you can do away with the absolute value because the result is supposed to hold from some $k$ onwards. In particular, either $k$ is already greater than $1$, and so you are considering values for which $x^{n-1} > x^k > 1$, for all $k < n-1$. Or if $k$ is smaller than $1$, you can just restrict your considerations to $x > 1$, since your hypothesis will still hold by transitivity (if it holds for all $x > k$, and $1 > k$, it holds for all $x > 1$)

For the third point, the idea is that you have one assumption that says “$P(x)$ is negative for all $x > k$”, meaning that starting from some point it will be negative for all positive values of $x$, while at the same time you just proved “$P(x)$ is negative only if $x < \frac{C}{a_n}$”, which is some positive real number. So by your first assumption, if you take $x > \frac{C}{a_n}$, $P(x)$ has to be negative, and by the second statement, it has to be positive, resulting in a contradiction.

Also, to spell it out explicitly $C$ has nothing to do with integrals.