# Prove that an odd degree polynomial must converge to $\pm\infty$ when $x\to\pm\infty$

While what I'm trying to prove is related to this topic but alas I cannot just assume that given: $p(x)$ where $\deg(p(x))=2n+1$ where $n\in\mathbb{N}$ that the only two cases are:

• $\lim \limits_{x\to -\infty}f(x)=-\infty$ and $\lim \limits_{x\to \infty}f(x)=\infty$

or

• $\lim \limits_{x\to -\infty}f(x)=\infty$ and $\lim \limits_{x\to \infty}f(x)=-\infty$

I've tried proving that given that

$x^{2n+1}$ and $\sum \limits_{k=0}^{2n}{a_{k}x^k}$, $n\in\mathbb{N}$ and $a_k>1$

that

$\exists x_1\in\mathbb{R} , \forall x>x_1$

$x^{2n+1} > \sum \limits_{k=0}^{2n}{a_{k}x^k}$

And I think I proved it but it's an ugly inductive proof: for $n=1$

$x^{3} > a_2x^2+a_1x+a_0$

then I've tried choosing $x_1 = (a_2a_1a_0)^2$

then assuming $x^{2n+1} > \sum \limits_{k=0}^{2n}{a_{k}x^k}$ proving that $x^{2n+3} > \sum \limits_{k=0}^{2n+2}{a_{k}x^k}$

Which is the same as : $x^{2n+3} > \sum \limits_{k=0}^{2n}{a_{k}x^k} + a_{2n+1}x^{2n+1}+ a_{2n+2}x^{2n+2}$ and then replace $\sum \limits_{k=0}^{2n}{a_{k}x^k}$ with $x^{2n+1}$

meaning : $x^{2n+3} > x^{2n+1} + a_{2n+1}x^{2n+1}+ a_{2n+2}x^{2n+2}$ and assuming $x\neq0$ dividing by $x^{2n+1}$ will mean that it's sufficient to show that:

$x^{2} > 1 + a_{2n+1} + a_{2n+2}x$ which is too fairly easy to show.

The problem is I don't know whether showing this is really sufficient ( I intended to use that to show that if for example the coefficient of the highest degree variable is positive that $\forall M\in\mathbb{R} \exists X_1 , x>X_1 \Rightarrow f(x)>M$ ).

1. Whether it makes sense at all because it feels like I have holes in my proof.
2. Is there a more mathematically elegant way to prove this not using derivatives.

Thanks and I deeply apologize if I broke any rule, I've really tried not to.

-

$$\lim_{x \to +\infty} 1 + 2x + x^2 = \lim_{x \to +\infty} x^2 \left(\frac{1}{x^2} + \frac{2}{x} + 1 \right)$$
$$\lim_{x \to 1} 1 - x - x^2 + x^3 = \lim_{x \to 1} (x-1)^2 \left((x-1) + 2 \right)$$
So you propose to write that given the previously mentioned $p(x)$ I can always describe it as $a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + \cdots + a_{2}x^{2} + a_{0}$ and then just factor it into $\lim\limits_{x \to +\infty} x^{2n+1} \left(\frac{a_0}{x^{2n+1}} + \cdots + \frac{a_{2n}}{x} + a_{2n+1} \right)$ ? And then using limit properties show that this limit is indeed $\pm\infty$ depending on the sign of $a_{2n+1}$ when $x\to\pm\infty$ ? – Scis Jan 12 '13 at 9:50