Prove that a polynomial diverges to infinity. I would like to prove the following statement:
Let $P$ be a polynomial of degree $n$ where $n$ is an odd natural number and $x$ $\in$ $\mathbb{R}$. $P(x)=a_{0}+a_{1}x+ ... + a_{n}x^{n}$
If $a_{n} > 0$, then $\lim_{x\to\infty}P(x)=\infty$
I am thinking of three ideas in order to prove the statement above.
$1$. $\lim_{x\to\infty}P(x)=\infty$ if for every $M>0$, there exists $K \in \mathbb{R}$ such that if $x \in \mathbb{R}$ with $x>K$, then $P(x)>M$.
$2$. $P$ is continuous on $\mathbb{R}$.
$3$. Contradiction: $a_{n} > 0$ but $\lim_{x\to\infty}P(x)\neq\infty$
Although I'm thinking the ideas above might help me prove the statement, but I cannot prove it. Could anyone help me with this? 
 A: The fact that $n$ is odd is irrelevant. Rewrite the polynomial, over $(0,\infty)$ which is not restrictive, as
$$
x^n\left(a_n+\frac{a_{n-1}}{x}+\dots+\frac{a_0}{x^n}\right)
$$
Now it should be known that
$$
\lim_{x\to\infty}\left(a_n+\frac{a_{n-1}}{x}+\dots+\frac{a_0}{x^n}\right)
=a_n
$$
because each summand, except for the first one, has limit $0$.
Hence there is $x_0$ such that, for $x>x_0$,
$$
a_n+\frac{a_{n-1}}{x}+\dots+\frac{a_0}{x^n}>\frac{a_n}{2}
$$
Finally, for $x>x_0$,
$$
a_nx^n+\dots+a_0>\frac{a_n}{2}x^n
$$
and you should be able to finish.
The same strategy proves that if


*

*$\lim_{x\to c}f(x)=\infty$

*$\lim_{x\to c}g(x)=l>0$


then
$$
\lim_{x\to c}f(x)g(x)=\infty
$$
(here $c$ can be any real, $\infty$ or $-\infty$; $l$ can also be $\infty$).
A: Let $A=\max\{|a_0|,|a_1|,\ldots,|a_{n-1}|\}$. Then, for $x\ge1$, we have
$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\ge a_nx^n-nAx^{n-1}=(a_nx-nA)x^{n-1}\ge a_nx-nA$$
It's clear (and/or easy to show) that $a_nx-nA\to\infty$ as $x\to\infty$ if $a_n\gt0$.
A: A small intuition behind the solution: Just somehow use the fact that $a_nx^n$ will dominate all the other monomials.
It should be enough to prove that a polynomial like this is eventually positive. For this use something like $$\exists m>0: x\geq m \implies a_nx^n>\frac{a_ix^i}{n}$$After this, as $a_n>0,\ P(x)>\frac{a_n}{2}x^n$ eventually . I'm sure you can prove the rest.
A: Here's a pretty simple induction proof, done backwards (since it makes more sense that way):
Inductive Step: We can easily show that $\lim_{x\rightarrow\infty}P(x)=\infty$ if we can show that $\lim_{x\rightarrow\infty}P'(x)>0$. So, let's go about doing that. We differentiate, and get $P'(x)=nx^{n-1}+(n-1)x^{n-2}+(n-2)x^{n-3}...$. We don't seem to be any closer to proving the original statement, or our revised version. However, notice that if $\lim_{x\rightarrow\infty}P'(x)=\infty$, then $\lim_{x\rightarrow\infty}P'(x)>0$. So, we've now brought the whole problem down one derivative, by showing that if the $P'(x)$ case is true, then the $P'(x)$ must also be true.
Initial Case:
Notice also that in the case $n=1$, the statement is trivial to prove. So, if we can bring the problem all the way down to $n=1$, then we will proven the $n$th case. We can easily repeat the procedure already demonstrated, in total $n-1$ times, since $P^{n-1}(x)$ will be of degree $1$, which by the Principle of Math Induction means that the $n$th case is true.
Q.E.D.
