How to prove an odd-degree polynomial starts and ends at values of different sign? Consider an odd-degree polynomial. How to prove that it starts a value that has different sign from its end value?
Or 
$$
\lim_{x\to -\infty} f(x) \lim_{x\to+\infty } f(x)<0
$$
Please don't use calculus (limit, derivative, integral) in the proof.
 A: Fix a polynomial $f(x) \in \mathbb{R}[x]$ of odd degree.  Then $f$ has an even number of non-real roots (because roots appear in conjugate pairs), hence an odd number of real roots (counting multiplicities).
Suppose that $f(x)$ has a root of multiplicity $m$ at $x_0$.  It's easy to show that $f$ changes sign at $x_0$ if and only if $m$ is odd.  As well, $f$ may only change sign at these points, by continuity.  It follows that $f$ has an odd number of sign changes (once again, by a parity argument).
But this is precisely to say that
$$\lim_{x \to \infty} f(x) \qquad \text{and} \qquad \lim_{x \to -\infty} f(x)$$
have different signs (of course, both will be infinite).
A: If your polynomial is odd it is of the form : $a_0 + a_1x + \ldots a_nx^n$ where $n$ is odd. Now we can write this as $x^n (a_0/x^n + \ldots a_n)$. Now for $|x|$ sufficiently large  we see that we can make $(a_0/x^n + \ldots a_{n-1}/x)$ arbitrarily small i.e smaller than the leading coefficient $a_n$ thus we see that taking $\lim x \to \infty$ we must have $x^n (a_0/x^n + \ldots a_n) = + \infty$ and since the degree $n$ is odd we get $\lim x \to -\infty (a_0/x^n + \ldots a_n) = - \infty $.  As we for each $x$ are multiplying something negative $x^n$ with something positive $(a_0/x^n + \ldots a_n)$ 
