Doubt related to the $\epsilon-\delta$ definition of limit $$\text{If}\quad\lim\limits_{x\to a}{f(x)}=L\quad\text{then:}$$
$$\text{If}\quad\forall (\epsilon > 0)\;\exists (\delta>0\;\text{and}\;\forall x\quad((x\neq a\;\text{and}\;|x-a|<\delta)\quad\Rightarrow\quad|f(x)-L|< \epsilon)).$$
I have understood the intiution 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 $a$.
So my doubt is that why definition is not defined by putting less than equal to in place of less than?
I mean what is wrong if we take equality sign, I mean why distance cannot be taken as $\delta$ or $\epsilon$
$$\text{If}\quad\lim\limits_{x\to a}{f(x)}=L\quad\text{then:}$$
$$\text{If}\quad\forall (\epsilon > 0)\;\exists (\delta > 0\;\text{and}\;\forall x\quad((x\neq a\;\text{and}\;|x-a|\leq\delta)\quad\Rightarrow\quad|f(x)-L|\leq\epsilon)).$$

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.
