Let $S \subseteq \R, a, L \in \R \cup \set{ \infty, -\infty }$.
$\limu{ x }{ a } f(x) = L $ if
$f $ is a function defined on $S $.
$\forall (x_n) \in S, (x_n) \in S, \limu{ n }{ \infty } x_n = a$. We have $\limu{ n }{ \infty }f(x_n) = L $.
$\limu{ x }{ a } f(x) $ is the limit of $f(x)$ as $x$ tends to $a$ along $S$.
$f $ does not need to be defined at $a $. i.e. $a $ is not necessarily in $S $.
The definition of continuity can now be expressed as $f $ is continuous at $x_0 $ if and only if $\limu{ x }{ a } f(x) = f(a)$.
The limits of functions, when they exist, are unique.
Let $f_1 $ and $f_2 $ be functions for which the limits $L_1 = \limu{ x }{ a } f_1(x) $ and $L_2 = \limu{ x }{ a } f_2 (x) $ exist and are finite. Then
- $\limu{ x }{ a }(f_1 + f_2)(x) $ exists and equals $L_1 + L_2 $.
- $\limu{ x }{ a }(f_1f_2)(x) $ exists and equals $L_1L_2 $.
- $\limu{ x }{ a }(f_1 / f_2)(x) $ exists and equals $L_1 / L_2 $ provided $L_2 \neq 0 $ and $f_2(x) \neq 0 $ for $x \in S $.
Let $f $ be a function for which the limit $L = \limu{ x }{ a }f(x) $ exists and is finite. If $g $ is a function defined on $\set{ f(x) \mid x \in S } \cup \set{ L }$ that is continuous at $L $, then $\limu{ x }{ a } g \circ f(x) = g(L)$.
Let $f $ be a function, $S = \dom{ f } \subseteq \R $, $\forall (x_n) \in S, \limu{ x }{ \infty }x_n = a$. Let $L $ be a real number, then $\limu{ x }{ a }f(x) = L $ if and only if
$$\forall \epsilon > 0, \exists \delta > 0, x \in S \text{ and } \abs{ x - a } < \delta \implies \abs{ f(x) - L } < \epsilon $$
Let $f $ be a function defined on $J \setminus \set{ a } $ for some open interval $J $ containing $a $, and let $L $ be a real number. Then $\limu{ x }{ a } f(x) =L$ if and only if
$$\forall \epsilon > 0, \exists \delta > 0, x \in S \text{ and } 0 < \abs{ x - a } < \delta \implies \abs{ f(x) - L } < \epsilon $$
Let $f $ be a function defined on some interval $(a, b) $, and let $L $ be a real number. Then $\limu{ x }{ a^+ } f(x) = L$ if and only if
$$\forall \epsilon > 0, \exists \delta > 0, x \in S \text{ and } < a + \delta \implies \abs{ f(x) - L } < \epsilon $$
Let $f $ be a function defined on $J \setminus \set{ a } $ for some open interval $J $ containing $a $. $\limu{ x }{ a } f(x) $ exists if and only if the limits $\limu{ x }{ a^+ } f(x) $ and $\limu{ x }{ a^- } f(x) $ both exist and are equal, in which case all three limits are equal.