One-sided limits of a monotone function always exist: for $f$ nondecreasing on an interval $I$ and $c\in I$, ${\lim }_{x\to c^{-}}f(x)=\sup \{f(x):x\in I,x<c\}$ whenever $I$ has points below $c$, ${\lim }_{x\to c^{+}}f(x)=\inf \{f(x):x\in I,x>c\}$ whenever it has points above $c$, and these satisfy ${\lim }_{x\to c^{-}}f(x)\le f(c)\le {\lim }_{x\to c^{+}}f(x)$

Let $I\subseteq \mathbb{R}$ be order-convex (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), let $f:I\to \mathbb{R}$ be nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences) and let $c\in I$. Write

$I^{-}:=I\cap (-\mathrm{\infty },c),I^{+}:=I\cap (c,\mathrm{\infty })$

(The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A\cap (-\mathrm{\infty },c)$ and $A\cap (c,\mathrm{\infty })$).

1. Left. If $I^{-}\mathrm{\ne }\mathrm{\varnothing }$ then $c$ is a limit point of $I^{-}$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), the set $\{f(x):x\in I^{-}\}$ is nonempty and bounded above by $f(c)$, and ${\lim }_{x\to c^{-}}f(x)=\sup \{f(x):x\in I,x<c\}\le f(c)\mathrm{.}$
2. Right. If $I^{+}\mathrm{\ne }\mathrm{\varnothing }$ then $c$ is a limit point of $I^{+}$, the set $\{f(x):x\in I^{+}\}$ is nonempty and bounded below by $f(c)$, and ${\lim }_{x\to c^{+}}f(x)=\inf \{f(x):x\in I,x>c\}\ge f(c)\mathrm{.}$
3. Together. If both $I^{-}$ and $I^{+}$ are nonempty then ${\lim }_{x\to c^{-}}f(x)\le f(c)\le {\lim }_{x\to c^{+}}f(x)\mathrm{.}$

In particular a nondecreasing function on an interval has, at every point of that interval, every one-sided limit that is well posed at all: no hypothesis of continuity, of boundedness, or of any other kind is needed.

The nonincreasing case is not a separate theorem. If $g:I\to \mathbb{R}$ is nonincreasing then $-g$ is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences), and a real $L$ is the left limit of $-g$ at $c$ exactly when $-L$ is the left limit of $g$ at $c$, since $\mathrm{\mid }(-g)(x)-L\mathrm{\mid }=\mathrm{\mid }g(x)-(-L)\mathrm{\mid }$; so claims 1 to 3 hold for $g$ with the suprema and infima exchanged and the inequalities reversed.

Order-convexity of $I$ is what makes the limits well posed. Without it the symbol ${\lim }_{x\to c^{-}}f(x)$ need not be defined even though $I^{-}$ is nonempty: for $I=\{0\}\cup [1,2]$ and $c=1$ the set $I^{-}=\{0\}$ is nonempty but $1$ is not a limit point of it, and The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A\cap (-\mathrm{\infty },c)$ and $A\cap (c,\mathrm{\infty })$ leaves the symbol undefined there for exactly that reason.

• Where completeness is spent. Exactly once on each side, in the existence of $\sup S^{-}$ and of $\inf S^{+}$; the rest of the proof is the definition of a one-sided limit and the monotonicity hypothesis. Over an ordered field that is not complete the statement fails, because the supremum need not exist.

• The two one-sided limits need not agree, and that is the point. When both are defined they satisfy ${\lim }_{x\to c^{-}}f(x)\le f(c)\le {\lim }_{x\to c^{+}}f(x)$, and a strict inequality between the outer two is exactly a jump discontinuity; A monotone function on an interval has no discontinuity of the second kind: at every point both relevant one-sided limits exist, and an interior point $c$ is a discontinuity exactly when ${\lim }_{x\to c^{-}}f(x)<{\lim }_{x\to c^{+}}f(x)$ turns that observation into the classification of the discontinuities of a monotone function, and Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used counts them.