Proof of a limit of a real function with rational paraphernalia A real-valued function $f(x)$ of a real argument $x$ has a limit $A$ for $x=x_0$ . I prove this by choosing $\epsilon$, and showing that a number $\delta$ exists so that $\left|f(x) – A\right| < \epsilon$ if $\left|x-x_0\right|< \delta$. 
My question is now: if I only have access to rational values for $\epsilon$ and $\delta$, is my proof still valid edit: for all such functions f(x)?  
 A: The key concept in an $\epsilon - \delta$ limit proof is that given any $\epsilon > 0$, there exists $\delta > 0$ such that $\left|f(x) - A\right| < \epsilon$ whenever $\left|x - x_0\right| < \delta$: we must be able to do this for any $\epsilon$, but notice that $\delta$ has no restrictions - we just need to find it. One can observe that if for a particular $\epsilon - \delta$ pair such that the inequalities hold, any smaller $\delta'$ will also work for the same $\epsilon$. Similarly, if we have an $\epsilon - \delta$ pair such that the inequalities hold, any larger $\epsilon'$ will work for the same $\delta$. We can use this and the density of the rational numbers to show that it suffices to prove that for all $\epsilon\in\Bbb Q^+$, there exists a $\delta$ such that the inequalities hold.
Suppose you know that for all $\epsilon\in\Bbb Q^+$ you can find a $\delta_{\epsilon}$ such that the inequalities hold. Now you want to show they hold for all irrational $r\in\Bbb R^+$ as well. Then let $r$ be an arbitrary positive irrational and choose a rational number $0 < q < r$, and use the $\delta_q$ for $q$ to say that $\left|f(x) - A\right| < q < r$ whenever $\left|x - x_0\right| < \delta_q$. So, if you show the inequality holds for positive rational $\epsilon$, you have actually shown it for all positive real $\epsilon$. Here we are using the fact that the rationals are dense in the reals, and you can find many proofs of this online or in a basic real analysis text (also refer to Ian Coley's comment).
As for "only having access to" rational $\delta$ (and I cannot see why this would be the case), this is also not a problem: $\delta$ only needs to be found - it doesn't matter what it is!
