Elementary proof without limits? Consider the following statement:
A polynomial $p:\mathbb R \to \mathbb R$ of odd degree has at least one real root.
I was trying to prove it and thought of the following proof idea: If $p$ has degree $2n + 1$ then $\lim_{x \to \infty} p(x) = \lim_{x \to \infty} x^{2n +1} = \infty$ and $\lim_{x \to -\infty} p(x) = \lim_{x \to -\infty} x^{2n +1} = -\infty$ therefore $p$ changes sign and the intermediate value theorem can be applied. 
But, I was wondering if there are proofs without limits? 
 A: If $p(x) = a_{n} x^{n} + \ldots + a_0 = a_{n} x^{n} \left( 1 + \ldots + \frac{a_0}{a_nx^n} \right)$, then $p\left(-n(|a_0/a_n|+1)\ldots(|a_{n-1}/a_n|+1)\right)$ and $p\left(n(|a_0/a_n|+1)\ldots(|a_{n-1}/a_n|+1)\right)$ have different signs.
A: Consider the roots of $p$ in $\mathbb C$.
Check, that if $x+iy$ is a root, then also the complex conjugate $x-iy$ is a root of $p$. Thus, the non-real roots of $p$ come in pairs, so the number of non-real roots is even. However, in $\mathbb C$, $p$ has an odd number of roots, because its degree is odd.
A: Yes and no. ;-)
It's not restrictive to assume the polynomial is monic. The result you use can be proved in this way: write
$$
x^{2n+1}+a_{2n}x^{2n}+\dots+a_0=
x^{2n+1}\left(1+\frac{a_{2n}}{x}+\dots+\frac{a_0}{x^{2n+1}}\right)
$$
and observe that all the terms between the parentheses have limit $0$ when $x\to\infty$ or $x\to-\infty$. This means that you can find $M>0$ such that, for $|x|>M$,
$$
\left|\frac{a_{k}}{x^{2n+1-k}}\right|<\frac{1}{2n+1}
$$
so
$$
1+\frac{a_{2n}}{x}+\dots+\frac{a_0}{x^{2n+1}}>0
$$
Now it's sufficient to determine such $M$: take $M>1$ such that
$$
\frac{M}{2n+1}>\max\{|a_0|,|a_1|,\dots,|a_{2n}|\}
$$
and you're done.
However, you'll be using continuity when applying the intermediate value theorem, so, in some sense, limits are still involved in the proof.
