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?

The question has been answered, but for sorting out the $(2^5 - 1)$ different ways of replacing strict inequalities by weak ones in the definition, the following might help.
The condition to be met is more stringent for smaller $\epsilon$. If you allow $\epsilon \geq 0$ there is no need for the $\forall \epsilon > 0$ quantifier, one can just replace $\epsilon$ by $0$ everywhere in  the definition.  The logical formula will then either express the condition that a function be equal to $L$ on a neighborhood of $a$, or be so strict that no function meets the condition.   Assume, then, that the formula begins $\forall \epsilon > 0 \dots \quad$.  In that case it makes no difference whether in the final inequality $|f(x)-L|$ is $ < \epsilon$ or $\leq \epsilon$.
The condition to be met is less stringent for smaller $\delta$. If $\delta =0$ is allowed then the $\exists \delta \dots$ can be satisfied if and only it is satisfied by $\delta=0$, and one can replace $\delta$ by zero everywhere instead of quantifying over $\delta$.  In that case one gets either a condition that is true for every function, or the condition that $f(a)=L$, according to whether $x=a$ is allowed.
The requirement that $0 < |x-a| < \delta$ is the one that is most natural to modify.  It defines the type of neighborhood of $a$ on which the convergence to $L$ occurs.  Here it is a punctured two-sided neighborhood (usually to allow discussion of derivatives where ratios of type 0/0 appear, like $\sin(x)/x$ near $x=a=0$) but allowing $x=a$ gives a definition of continuity, or one might want one-sided limits with $ 0 < x-a < \delta$ or $0 < a - x < \delta$.  If $\delta=0$ is permitted then the natural neighborhood to use would be $0 \leq |x-a| \leq \delta$ but this would only lead to a complicated restatement of "$f(a)=L$".  Finally, changing the upper bound to $|x-a| \leq \delta$ would not affect anything (except in the useless case where $\delta=0$ is allowed).
To summarize, allowing $0 \leq |x-a|$ gives a definition of continuity, but changes to any of the other inequalities $\epsilon > 0$, $\delta > 0$, $|x-a| < \delta$ or $|f(x)-L| < \epsilon$ either do not affect the definition, or trivialize it.
