I'm looking for an explicit formulation for $\delta$ in the $\epsilon-\delta$ formulation of the limit for a polynomial $p(x) = \sum_{n=0}^N a_nx^n$.
For example, in the the specific linear case $p(x) = a_1x+a_0$, for any $\epsilon > 0$ choosing $\delta = \frac{\epsilon}{|a_1|}$ means that $|x-c|<\delta$ implies $|p(x)-p(c)|<\epsilon$. I expect that in general, $\delta$ would depend on $\epsilon, c$, and the coefficients $a_0, \dots a_N$
I've started by looking at the expression: $$|p(x)-p(c)| = |a_N(x^N-c^N)+a_{N-1}(x^{N-1}-c^{N-1}) + \dots + a_1(x-c) + a_0|$$ and followed my intuition to factor this as: $$|(x-c)\cdot[a_N(x^{N-1}+x^{N-2}c+x^{N-3}c^2+\dots+c^{N-1})+\frac{a_{N-1}}{c}(x^{N-2}c+x^{N-3}c^2+\dots+c^{N-1})+\frac{a_{N-2}}{c^2}(x^{N-3}c^2+\dots+c^{N-1})+\dots+\frac{a_1}{c^{N-1}}(c^{N-1})]|$$
But my ideas fizzle out at this point. There should be a way to express a bound for the bracketed sum, possibly using a restriction like $|x-c|<1$, but I don't see what it is. Any insights?
A separate (and easier) argument is needed for $c=0$. So assume $c\ne 0$. We will use the Triangle Inequality repeatedly without explicit mention.
First insist that $\delta\lt |c|/2$. Then if $|x-c|\lt \delta$, we will have $|x|\lt 3|c|/2$. Let $m=\max(3|c|/2,1)$.
Now note that $\frac{x^q-c^q}{x-c}=x^{q-1}+cx^{q-2}+\cdots +c^{q-1}$. We have $q$ terms, each with absolute value $\le m^{q-1}$. The sum has absolute value $\le Nm^{N-1}$.
Let $A$ be an upper bound on the $|a_i|$. Then each term $a_i\frac{x^i-c^i}{x-c}$ has absolute value $\le ANm^{N-1}$. There are $N$ of them, so the absolute value of their sum is $\le AN^2 m^{N-1}$. Thus $|P(x)-P(c)|\le |x-c|AN^2 m^{N-1}$ if $|x-c|\lt |c|/2$.
Now there is no problem in choosing $\delta$ that ensures that if $|x-c|\lt \delta$, then $|P(x)-P(c)|\lt \epsilon$.