What if epsilon, delta are rational? The precise formulation of the (ε, δ)-definition of limit as written here in wikipedia https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit doesn't specify which kind of number the epsilon, delta are. 
I wonder if it matters. 
 A: If $\lim_{x \to a} f(x) = \ell$, then for all (real) $\epsilon > 0$, there exists a (real) $\delta > 0$, such that $|x - a| < \delta$ gives $|f(x) - \ell| < \epsilon$. Hence, for $\epsilon \in \Bbb Q \cap(0,\infty)$, there exists $\delta > 0$ such that $|x - a| < \delta$ gives $|f(x) - \ell| < \epsilon$. Now choose $\delta '$ to be a rational number such that $0 < \delta ' < \delta$. Then, $|x - a| < \delta '$ gives $|f(x) - \ell| < \epsilon$.
Conversely, suppose that for all $\epsilon \in \Bbb Q \cap (0,\infty)$, there exists $\delta \in \Bbb Q \cap (0,\infty)$ such that $|x - a| < \delta$ gives $|f(x) - \ell| < \epsilon$. Let $\epsilon > 0$ be any real number. Let $\epsilon '$ be a rational number such that $0 < \epsilon ' < \epsilon$. Then, there exists a $\delta > 0$ (rational, but this is immaterial) such that $|x - a| < \delta$ gives $|f(x) - \ell| < \epsilon ' $, from which we get $|f(x) - \ell| < \epsilon$.
Conclusion: the two are equivalent.
A: If you restrict the pair ($\epsilon, \delta$) to some subset of allowed (pairs of) positive values, nothing changes as long as 


*

*arbitrarily small values of $\epsilon$ are allowed

*for every $\epsilon$, arbitrarily small values of $\delta$ are allowed
This allows for some dependence between epsilon and delta, although the question was about independently restricting each one to be rational. 
