Limiting value of functions - From Tao Analysis II Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces,$E \subseteq X$ and $f:E \rightarrow Y$ a function. If $x_0 \in \overline E$ we define
$$
 \lim_{x \rightarrow x_0;x \in E} f(x) = L 
$$ iff
$$
\exists L \in Y \forall \epsilon > 0 \exists \delta > 0 \forall x\in E: d_X(x,x_0) < \delta \rightarrow d_Y(f(x),L) < \epsilon \qquad (1)
$$
He then gives the following exercise: Let $x_0 \in E$. Then $\lim_{x \rightarrow x_0; x \in E} f(x)$ exists $\iff$ $\lim_{x \rightarrow x_0; x \in E \setminus \{x_0\}} f(x)$ exists. Further we have that if $\lim_{x \rightarrow x_0; x \in E} f(x)$ exists, then the limit equals $f(x_0)$.
First the implications: From left to right is trivial. But what if $x \in E \setminus \{x_0\}$ ? Why does then (1) also hold for $x_0$ itself ?

I got some edits: The implication $\Leftarrow )$ also requires that the limit equals $f(x_0)$. Sorry for that. Then the equivalence is true.
Further I see that Tao doesn't intend that for example the function $f(x)= 1$ if $x = x_0$ and $f(x) = 0$ otherwise should have a limit where $E = \mathbb R$. First, if we consider $E \setminus \{x_0\}$ we have that the limit equals $0$. In fact if the limit exists and $x_0 \in E$ then the function must be continous.

As indicated in the comments above, I don't think the definition of the limit that you've given above is the right one. Assuming that we're using the definition of the limit suggested by @user49684 in the comments (which I think is the appropriate definition), then I think it would be obvious to you why the two limits are equal.
