Why doesn't derivative difference quotient violate the epsilon-delta definition of a limit? So the difference quotient is defined as:
$$\lim \limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
So if we take a function such as $f(x)=x^2$ and go through the simplification, we get 
$$\lim \limits_{h \to 0} 2x+h $$
We say $h$ is zero, and that makes sense because it becomes negligible. But here's what I don't understand: $\delta$-$\epsilon$ says (in casual terms) we can make $h$ as close as we want to zero and $2x+h$ will be sufficiently close to $2x$. But isn't one of the constraints of delta epsilon that $|\delta - c| > 0$? So how can we say $h$ is exactly zero if this constraint must be met? In other words, we always say the value of the limit at the point of interest does not necessarily equal the value of the function - but here it seems like we're saying that they are by making $h$ exactly zero, not just sufficiently close. Why can we do this?

We can do this because the function $g(h)=2x+h$ (for some fixed value of $x$) is continuous at $h=0$.  This means that $\lim_{h\to 0}g(h)=g(0)$, so we can evaluate the limit by just plugging in $h=0$.
Of course, this is ultimately circular: we only know that $g(h)$ is continuous at $h=0$ because we can evaluate the limit by using the $\epsilon$-$\delta$ definition and find that it coincides with the value of $g(0)$.  So you should think of the "plug in $h=0$" rule as only a heuristic; the true rigorous way to compute limits is by using the $\epsilon$-$\delta$ definition.  However, if by previous work you have done you know that the expression you get is continuous in $h$ (e.g., in this case, if you have proven that all polynomial functions are continuous), then you can compute the limit by plugging in $h=0$.
