What is the importance of the difference between these two?
$$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x)([0 < \lvert x-a \rvert < \delta ] \Rightarrow [\lvert f(x)-L \rvert < \varepsilon ])$$ $$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x)([0 < \lvert x-a \rvert \leq \delta ] \Rightarrow [\lvert f(x)-L \rvert \leq \varepsilon ])$$
Thanks
Assume (1). Consider an arbitrary $\epsilon>0$. Then because $\epsilon>0$, there exists a $k> 0$ such that $$0 < \lvert x-a \rvert < k \Rightarrow \lvert f(x)-L \rvert < \varepsilon $$ In particular, when $\delta=k/2$, $|x-a| \le \delta$ implies $|x-a| < k$ implying $\lvert f(x)-L \rvert < \varepsilon$ implying $\lvert f(x)-L \rvert \le \varepsilon$. Thus (2) holds.
Conversely, assume (2). Consider an arbitrary $\epsilon>0$. Then because $\epsilon/2>0$, there exists a $k> 0$ such that $$0 < \lvert x-a \rvert \le k \Rightarrow \lvert f(x)-L \rvert \le \varepsilon/2 $$ In particular, when $\delta=k$, $|x-a| < \delta$ implies $|x-a| \le \delta$ implying $\lvert f(x)-L \rvert \le \varepsilon/2$ implying $\lvert f(x)-L \rvert < \varepsilon$. Thus (1) holds.