Let $(X, d_X) $ and $(Y, d_Y)$ be two metric spaces, let $E$ be a subset of $X$, and let $f: X \to Y $ be a function. If $x_0 \in X$ is an adherent point of $E$, and $L \in Y$.
we say that $f(x)$ converges to $L$ in $Y$ as $x$ converges to $x_0$ in $E$, and we write $\limd{ x }{ x_0 }{ x \in E \setminus \set{ x_0 }} f(x) = L$ iff
$$\forall \epsilon > 0, \exists \delta > 0 \text{ such that } \forall x \in E, d_X(x, x_0) < \delta \implies d_Y(f(x), L) < \epsilon $$
According to the definition of function continuity at a point, we see that
$f$ is continuous at $x_0$ iff
$$\limd{ x }{ x_0 }{ x \in X } f(x) = f(x_0) $$
Thus $f$ is continuous on $X$ iff we have
$$\limd{ x }{ x_0 }{ x \in X } f(x) = f(x_0) \forall x_0 \in X $$
Often, we omit the condition $x \in X $ and abbreviate $\limd{ x }{ x_0 }{ x \in X } f(x) $ as simply $\limu{ x }{ x_0 } f(x) $ when it is clear what space $x $ will range in.
Let $(X, d_X) $ and $(Y, d_y) $ be two metric spaces. $E \subseteq X$, and let $f: X \to Y$ be a function. Let $x_0 \in X$ be an adherent point of $E$ and $L \in Y$. Then the following four statements are logically equivalent:
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$\limd{ x }{ x_0 }{ x \in E } f(x) = L$.
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$\forall (x^{(n)})^\infty_{ n =1} \in E, \limu{ n }{ \infty }(x^{(n)}) = x_0$ with respect to the metric $d_X$, $\limu{ n }{ \infty }(f(x^{(n)}))^\infty_{ n =1} = L $ with respect to the metric $d_Y $.
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For every open set $V \subset Y$ which contains $L$, there exists an open set $U \subset X$ containing $x_0$ such that $f(U \cap E) \subseteq V$.
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If one defines the function $g: E \cup \set{ x_0 } \to Y $ by defining $g(x_0) := L$, and $g(x) := f(x) $ for $x \in E \setminus \set{ x_0 } $, then $g $ is continuous at $x_0 $. Furthermore, if $x_0 \in E $, then $f(x_0) = L $.
Observe that a function $f(x)$ can converge to at most one limit $L$ as $x$ converges to $x_0$. In other words, if the limit
$$\limd{ x }{ x_0 }{ x \in E } f(x)$$
exists at all, then it can only take at most one value.
The requirement that $x_0 $ be an adherent point of $E$ is necessary, because it ensures that for $\delta $ sufficiently small, there are points $x \in E $, so that $d(x, x_0) < \delta $.
$(X, d_X), (Y, d_Y)$. $E \subseteq X$. $f: E \to Y$. $x_0 \in E$.
Show that $\limd{ x }{ x_0 }{ x \in E } f(x)$ exists if and only if $\limd{ x }{ x_0 }{ x \in E \setminus \set{ x_0 }} f(x) $ exists and is equal to $f(x_0)$.
Show that $\limd{ x }{ x_0 }{ x \in E } f(x)$ exists $\implies \limd{ x }{ x_0 }{ x \in E } f(x) = f(x_0)$.