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

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

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

1. Left. If $I^{-} \ne \varnothing$ 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 $\lim_{x \to c^{-}} f(x) \;=\; \sup\{\, f(x) : x \in I,\ x < c \,\} \;\le\; f(c) .$
2. Right. If $I^{+} \ne \varnothing$ then $c$ is a limit point of $I^{+}$, the set $\{\, f(x) : x \in I^{+} \,\}$ is nonempty and bounded below by $f(c)$, and $\lim_{x \to c^{+}} f(x) \;=\; \inf\{\, f(x) : x \in I,\ x > c \,\} \;\ge\; f(c) .$
3. Together. If both $I^{-}$ and $I^{+}$ are nonempty then $\lim_{x \to c^{-}} f(x) \;\le\; f(c) \;\le\; \lim_{x \to c^{+}} f(x) .$

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 $|(-g)(x) - L| = |g(x) - (-L)|$; 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 $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ 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.