# Question on logic of epsilon and delta proof [duplicate]

To prove that $$\lim_{x\to a}f(x) = L$$, we need to show that for any $$\epsilon>0$$, there exists a $$\delta>0$$ such that if $$0<|x-a|<\delta$$, then $$|f(x)-L|<\epsilon$$.

What's the reasons for epsilon and delta have sing $$<$$ not a $$\leq$$ ? And Why do we assume for epsilon and some delta? Not either for all epsilon and all delta or some epsilon and all delta etc..

• The definition is also equivalent if you take $\leq \varepsilon$. Assume the original definition is satisfied. Then the alternate definition is also satisfied, since $|f(x) - L| < \varepsilon \implies |f(x) - L| \le \varepsilon$. And the alternate definition implies the original definition: pick any $\varepsilon > 0$, and apply the hypothesis of the old definition with $\frac{\varepsilon}{2}$. Commented Mar 14, 2023 at 3:58
• @K.Jiang your first comment has errors, since if the original definition (which means using $< \delta$) is satisfied then it doesn't let you use $\leq \delta$ for the same $\varepsilon$. Instead, if the definition using $\leq \delta$ (the "alternate definition") is satisfied then so is the one using $< \delta$ (the "original definition"), while if the definition using $< \delta$ is satisfied then so is the one using $\leq \delta/2$, so you could rename $\delta/2$ as $\delta$.
– KCd
Commented Mar 14, 2023 at 4:20
• @KCd Thanks, I overlooked that. Commented Mar 14, 2023 at 4:23
• You can use either, the important part is the $\epsilon>0$ and $\delta>0$. Commented Mar 14, 2023 at 4:43
• For the second question, the duplicate is math.stackexchange.com/q/1438766/139123. (Note that one of the guidelines here is to ask only one question per "question"), Commented Mar 14, 2023 at 6:09