Is it possible for $x$ to appear in the definition of $\delta$ in an $\epsilon-\delta$ proof of limit? Logic tells me it is not (it would be circular, since the allowed interval for $x$ is itself defined by $\delta$), but I don't know where is the error in my workings. I'll explain...
My doubt arised when studying the $\epsilon-\delta$ definition for non-linear polynomial functions, say a quadratic function, when you want to prove that the limit equals the value of the function at any given value of $x$. This could be the case when checking continuity, for example. As I've seen so far, the normal way of solving the proof for quadratic polynomials consists in converting the epsilon inequality like so:
$$|f(x) - L| < \epsilon = -\epsilon < f(x) - L < \epsilon$$
Then equaling the middle part with de lower side of the $\delta$ inequality and finally choosing the minimum of the two possibilities you get, taken into account you have to assure all values of $x$ will have an image inside the $\epsilon$ interval.
But!
As I've recently realised, the polynomial you get after subtracting the limit to the function when both equal each other, as this nullifies to 0, is always factorable. Plus, one of the two factors you get already corresponds to the lower side of the delta inequality. Here is the formal proof of this.
Given:
$$\lim_{x \to n} ax^2 + bx + c = an^2 + bn + c$$
The epsilon inequality of the proof would be:
$$|ax^2 + bx + c - (an^2 + bn + c)| < \epsilon$$
And working from that:
$$|ax^2 + bx + c - an^2 - bn - c| < \epsilon$$
$$|ax^2 + bx - an^2 - bn| < \epsilon$$
$$|a(x^2 - n^2) + b(x - n)| < \epsilon$$
$$|a(x + n)(x - n) + b(x - n)| < \epsilon$$
$$|(x - n)(a(x + n) + b)| < \epsilon$$
$$|x - n||a(x + n) + b| < \epsilon$$
$$|x - n| < \frac{\epsilon}{|a(x + n) + b|}$$
And there it is! You'll always have the lower side of the delta inequality on the left. Thus, you could define $\delta$ as:
$$\delta \le \frac{\epsilon}{|a(x + n) + b|}$$
Et voilà! Here you have $x$ defining delta.
Now, obviously something weird is going on here... isn't it? I highly suspect my math is flawed at some point, possibly when handling absolute values. Add to that I have no idea how to connect this method with the one that has to choose between two possible values of $\delta$.
Any help with this mess would be much appreciated!
 A: When you want to check the continuity at some point $x$, the very first thing you do is fix $x$. From that point on, $x$ is a constant, like $2$ or $3$. Saying that you're unhappy to see $x$ in $\delta$ in such a case is like saying that you're unhappy to see $2$ or $3$.
Now in your situation, you are correct that there is a problem, given that $x$ is a variable that tends to $n$, while $n$ is the fixed one, so it makes no sense to have $x$ part of the definition of $\delta$. 
However, you can easily get rid of the $x$-depending right-hand side but to do that you need to fix the causality in your succession of lines, i.e. insert $\Rightarrow$, $\Leftarrow$, $\Leftrightarrow$ where relevant. It is very important to do this all the time, otherwise you forget what you want to prove and you end up writing down things that make no sense.
So let's do that, what you need is $|x-n|<\delta\implies |f(x)-f(n)|<\varepsilon$. So we need $$|x - n| < \frac{\epsilon}{|a(x + n) + b|}\Longleftarrow |x - n| < \delta$$
And for that it is enough to have $\delta$ less than all values assumed by the function $\frac{\epsilon}{|a(x + n) + b|}$ when $x$ covers a close neighbourhood of $n$. You can easily check that no matter what $a$ and $b$ are, there is such a $\delta$.
