TITLE: Can we find a general $\delta$ to prove the continuity of polynomials?
QUESTION [6 upvotes]: Polynomials are continuous functions. In other words, for all $\epsilon > 0$ and all $a$, there is some $\delta > 0$ such that if $|x-a|<\delta$, $|P(x)-P(a)|<\epsilon$ where $P(x)$ is a function of the form 
$$P(x) = \sum_{i=0}^n a_i x^i$$
The proof of this is usually done via limit theorems and mathematical induction, which circumvents the difficulty of finding $\delta$. 
My question is, is it possible to prove the continuity of polynomials by explicitly finding a $\delta$ (a closed form), in terms of $n$, $a$, $a_i$ and $\epsilon$ where we have $\delta = \min(....)$.
If this is too ambitious, are there special cases (with, say, quadratics)?

REPLY [5 votes]: Assume $|x - a| \leq \delta \leq 1$. We have 
$$
\begin{align*}
|P(x) - P(a)| &= \left|\sum_{i = 1}^n a_i(x - a)(x^{i-1} + x^{i-2}a + \dots + a^{i-1})\right| \\
&\leq \sum_{i = 1}^n i|a_i||x - a| \max(|x|^{i-1}, |a|^{i-1}) \\
&\leq |x-a|\max(|x|^{n-1},|a|^{n-1},1)\sum_{i=1}^n i|a_i| \\
&\leq \delta (|a| + 1)^{n-1}n^2 \max_{i \geq 1}(|a_i|).
\end{align*}$$
So (assuming $P(x)$ is non-constant) you can choose $\delta = \min(1,\epsilon/M)$, where $M = (|a|+1)^{n-1}n^2 \max_{i \geq 1}(|a_i|)$.
What I wrote works, but it might be possible to do this a bit more cleanly by using $|P(x) - P(a)| \leq |x-a| \max_{y \in I_{a,x}} |P'(y)|$, where $I_{a,x}$ is either $[a,x]$ or $[x,a]$.