Epsilon-delta definitions, inequality strict / non-strict? Reading some of the analysis related  posts, I have a question regarding the epsilon-delta language.
What we are taught is the inequality in the definition is strict. E.g
$$\forall\varepsilon >0\ \exists\delta >0: \forall x\in D\left ( |x-a|<\delta \Longrightarrow |f(x)-f(a)|<\varepsilon \right )$$
( definition of continuity at $a\in D$). If this is satisfied, we conclude for suitable choice of $x$ the difference between $f(x)$ and $f(a)$ is strictly less than any positive number hence it must be zero.
Intuitively, it also makes sense that it's sufficient if the inequality involving $\varepsilon$ is not strict. But how does one justify that?

Let $R(\varepsilon)$ represent a definition with strict $\varepsilon$ -inequality. Let $M(\varepsilon)$ be the same definition, but let $\varepsilon$-inequality be non-strict.  Then $R(\varepsilon)$ is satisfied iff $M(\varepsilon)$ is satisfied?

The question really is if $M(\varepsilon)$ is satisfied, is then $R(\varepsilon)$ satisfied? Does one simply say that since $M\left (\frac{\varepsilon}{2}\right )$, then $R(\varepsilon)$?
I can almost be sure that we can't allow the inequality involving $\delta$ to be non-strict, otherwise we could potentially permit points where $f$ tends to infinity? [On second thought, just make $\delta$ smaller]
 A: You're not correct, because we have the statement for all epsilon and delta, the strictness does not matter at all. The following are equivalent:
a) $\forall \varepsilon>0 \exists \delta>0:\forall x \in D: (|x-a|<\delta \Rightarrow |f(x)-f(a)| < \varepsilon)$
b) $\forall \varepsilon>0 \exists \delta>0:\forall x \in D: (|x-a|\leq \delta \Rightarrow |f(x)-f(a)| \leq \varepsilon)$
c) $\forall \varepsilon>0 \exists \delta>0:\forall x \in D: (|x-a|< \delta \Rightarrow |f(x)-f(a)| \leq \varepsilon)$
d) $\forall \varepsilon>0 \exists \delta>0:\forall x \in D: (|x-a|\leq \delta \Rightarrow |f(x)-f(a)| < \varepsilon)$
Let us prove a) $\Leftrightarrow$ b) and then you will understand the rest.
$a) \Rightarrow b)$: Given $\varepsilon >0$, by a) there is a $\tilde{\delta} > 0$ such that $\forall x \in D: (|x-a|<\tilde{\delta} \Rightarrow |f(x)-f(a)| < \varepsilon)$. Now choose $\delta = \frac{1}{2} \tilde{\delta}$, then $\forall x \in D: (|x-a|\leq\delta<2\delta \Rightarrow |f(x)-f(a)| < \varepsilon \Rightarrow |f(x)-f(a)| \leq \varepsilon)$.
$b) \Rightarrow a)$: Given $\varepsilon >0$, we define $\tilde{\varepsilon} = \frac{1}{2}\varepsilon$. By b) there is a $\delta > 0$ such that $\forall x \in D: (|x-a|\leq\delta \Rightarrow |f(x)-f(a)| \leq \tilde{\varepsilon})$. 
Thus we have:
$\forall x \in D: (|x-a|<\delta \Rightarrow |x-a|\leq\delta \Rightarrow |f(x)-f(a)| \leq \tilde{\varepsilon} < \epsilon)$. Q.E.D.
