Definition. Let $a\in\mathbb{R}$, let $I$ be an open interval which contains $a$, and let $f$ be a real function defined everywhere on $I$ except possibly at $a$. Then $f(x)$ is said to be converge to $L$, as $x$ approaches $a$, if and only if for every $\varepsilon >0$ there is a $\delta > 0$ such that
$0 < \left| x-a\right| < \delta$ implies $\left| f\left( x\right) -L\right| < \varepsilon$.
Wade says (Wade's Introduction Analysis book, p.69, fourth edition)that Before continuing, we would like to draw your attention to two features of the definition: Assumption 1. The interval I is open; Assumption 2. $0 < \left| x-a\right|$. If $I=(c,d)$ is an open interval and $\delta _{0}:=\min \left\{ a-c,d-a\right\}$, then $\left| x-0\right| < \delta _{0}$ implies $x\in I$. Hence, Assumption 1 guarantees that for $\delta >0$ sufficiently small,$f(x)$ is defined for all $x\neq a$ satisfying $\left| x-a\right| < \delta$ (i.e., on BOTH sides of $a$). Since $\left| x-a\right| >0$ is equivalent to $x\neq a$, Assumption 2 guarantees that $f$ can have a limit at $a$ without being defined at $a$.
Can you explain last paragraph as geometrical?