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 $a_{n} &gt; 0$, 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 $M&gt;0$, there exists $K \in \mathbb{R}$ such that if $x \in \mathbb{R}$ with $x&gt;K$, then $P(x)&gt;M$.

$2$. $P$ is continuous on $\mathbb{R}$.

$3$. Contradiction: $a_{n} &gt; 0$ 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?
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 $x&gt;x_0$,
$$
a_n+\frac{a_{n-1}}{x}+\dots+\frac{a_0}{x^n}&gt;\frac{a_n}{2}
$$
Finally, for $x&gt;x_0$,
$$
a_nx^n+\dots+a_0&gt;\frac{a_n}{2}x^n
$$
and you should be able to finish.

The same strategy proves that if


$\lim_{x\to c}f(x)=\infty$
$\lim_{x\to c}g(x)=l&gt;0$


then
$$
\lim_{x\to c}f(x)g(x)=\infty
$$
(here $c$ can be any real, $\infty$ or $-\infty$; $l$ can also be $\infty$).
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 $\lim_{x\rightarrow\infty}P'(x)&gt;0$. 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 $\lim_{x\rightarrow\infty}P'(x)&gt;0$. 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.