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?

Let's review (one of?) the definition of limit. We say the limit of $f$ as $h$ tends to $c$ is $L$ and write $\lim_{h\to c}f(h)=L$ when
$$\forall \epsilon>0, \, \exists \delta >0, \, \forall h\text{ with } 0<|h - c|<\delta, \,|f(h)-L|< \epsilon$$
So, let's use your example and work through the definition. We have $f(h)=2x+h$ and $c=0$. We claim the limit $L$ is $2x$. Indeed, let $\epsilon >0$ and take $\delta=\epsilon$. Clearly, whenever $0<|h-0|=|h|<\delta$, we have $$|f(h)-L|=|2x+h - 2x|=|h|<\delta=\epsilon$$ so that indeed $\lim_{h \to 0}2x+h=2x$.
Notice that throughout this process we never really cared about $f(0)$ or substituted $h$ for $0$ somewhere. That said, there are many theorems on the algebra of limits, that one may use with continuous functions, to ease some of the work.
If somehow you know beforehand that $f$ is continuous on $h$, you can skip the whole epsilon-delta stuff and just plug $h=c$ to get the limit. In this case, $2x+h$ is continuous on $h$, so $h=0$ yields the limit; notice how it agrees with the one we found via epsilon-delta.
