# In the $\epsilon-\delta$ definition of limit, why aren't the strict inequalities weak?

$$\text{If}\lim_{x\to a}{f(x)}=L\text{ then:}$$
$$\text{If} \forall (\epsilon > 0)\;\exists (\delta>0\;\text{ and }\forall x((x\neq a\;\text{and}\;|x-a|<\delta)\to|f(x)-L|< \epsilon)).$$

I have understood the intuition behind this definition as: it says that for every $$x$$ closer and closer to $$a$$, if we have $$f(x)$$ closer and closer to $$l$$, here closeness is in terms of $$\delta$$ and $$\epsilon$$ then we say that limit of function equals $$l$$ as $$x$$ approaches $$a$$.

But why doesn't this definition use ≤, in place of <? What is wrong if we take ≤? Why can't distance be taken as follows?

$$\text{If}\lim\limits_{x\to a}{f(x)}=L\text{ then, }$$
$$\text{If} \forall (\epsilon > 0)\exists (\delta > 0\text{ and }\forall x((x\neq a \text{ and }|x-a|\leq\delta)\to|f(x)-L|\leq\epsilon)).$$

There is nothing wrong if you put $\leq$ instead of $<$, its just that conventionally a topology is described by its open sets, even if it can also be defined in terms of the closed sets. Here, behind the idea of taking less than, the topological aspect plays a big role so that the definition can easily be extended in a more general setting.
They are equivalent. If there exists a $\delta$ for the first definition, taking $\delta' = \delta/2$ for the second works. If there exists a $\delta$ for the second definition, then using that same $\delta$ works in the first.