# Why not make ε arbitrarily small in the (ε, δ)-definition of limit

The $(ε, δ)$-definition of limit:For every real $ε > 0$, there exists a real $δ > 0$ such that for all real $x$, $0 < | x − a | < δ$ implies $| f(x) − L | < ε$.

From my point of view, I think ,in fact we want to make $f(x)$ can be made arbitrarily close to $L$ here, so it would be better to change “every real $ε > 0$” in the definition to "arbitrarily small real $ε > 0$", what do you think about my opinion here ?

• They mean the same thing. Once you show it for one $\epsilon$, it also holds for all larger $\epsilon$. – Cheerful Parsnip Nov 6 '14 at 14:53
• Are you trying to rephrase the usual definition in a more "natural" way, or actually change the definition? How do you think the new definition would change the way we do math? – David K Nov 6 '14 at 14:55
• Then someone might ask: "doesn't this work for big $\varepsilon$?" and you would say: "yes, it works for both small and big epsilons." Also it is easier to write "$\forall \varepsilon >0$" – ThePortakal Nov 6 '14 at 15:10
• @ThePortakal So In math we want to be as general as possible? – iMath Nov 7 '14 at 4:53
• @DavidK I just think it is much more concrete to state the definition in my rephrased way, but I was inspired by your question and found some flaw in my changing, "arbitrarily small” is a feeling in our mind, it is vague, there seems no standard for "arbitrarily small” in math,is 0.001 arbitrarily small enough ? 0.0001? 0.00001? …all in all ,it is not precise. – iMath Nov 7 '14 at 5:02

That is exactly what we mean by "for every". We can take $\epsilon$ as small or large as we like. There is no need to change the statement, because it already hold for arbitrarily small $\epsilon$, since it holds for every $\epsilon \ge 0$. In math we want to be as general as possible, which here means saying that the statement must hold for every $\epsilon > 0$.