What is the significance of the three nonzero requirements in the $\varepsilon-\delta$ definition of the limit? What are the consequences of the three nonzero requriments in the definition of the limit: 

$\lim_{x \to a} f(x) = L \Leftrightarrow \forall$ $\varepsilon>0$, $\exists$ $\delta>0 :\forall$ $x$, $0 < \lvert x-a\rvert <\delta \implies \lvert f(x)-L \rvert < \varepsilon$

I believe I understand that: 


*

*if $0 = \lvert x-a\rvert$ were allowed the definition would require that $f(x) \approx L$ at $a$ ($\lvert f(a)-L \rvert < \varepsilon$); 

*if $\varepsilon=0$ and $\lvert f(a)-L \rvert \le \varepsilon$ were allowed the theorem would require that $f(x) = L$ near $a$ (for $0 < \lvert x-a\rvert <\delta$); and

*if $\delta=0$ were allowed (and eliminating the tautology by allowing $0 \le \lvert x-a\rvert \le \delta$) the definition would simply apply to any function where $f(a) = L$, regardless of what happened in the neighborhood of $f(a)$.
Of course if (2'.) $\varepsilon=0$ were allowed on its own, the theorem would never apply ($\lvert f(a)-L \rvert \nless 0$). 
What I'm not clear about is [A] the logical consequences of (3'.) allowing $\delta=0$ its own, so that:

$\lim_{x \to a} f(x) = L \Leftrightarrow \forall$ $\varepsilon>0$, $\exists$ $\delta≥0 :\forall$ $x$, $0 < \lvert x-a\rvert <\delta \implies \lvert f(x)-L \rvert < \varepsilon$

and [B] whether allowing both 1. and 2. would be equivalent to requiring continuity?

For (3), if $\delta = 0$ was allowed the definition would apply to everything: since $|x-a| < 0$ is impossible, it implies whatever you like.
