---
title: Limits at Infinity and Monotone Functions
module: Limits and Continuity
moduleNumber: 4
lessonNumber: 6
order: 406
summary: >
  Treating infinity as a cluster point extends the epsilon–delta limit to
  x approaching plus or minus infinity, giving horizontal asymptotes and infinite
  limits. For monotone functions the one-sided limits always exist as suprema
  and infima, the discontinuities are jumps and at most countably many, the
  continuity is equivalent to the image being an interval, and a strictly
  monotone function always has a continuous inverse.
topics: [Limits and Continuity]
sources:
  - book: Lebl
    ref: "Ch. 3 — Continuous Functions; §3.5 Limits at infinity, §3.6 Monotone functions and continuity"
draft: false
---

Two features of the real line have no metric-space analogue: the line has two
ends, and it is ordered. The ends give the limit of $f(x)$ as $x \to +\infty$ or
$-\infty$, which makes asymptotes precise. The order gives the class of monotone
functions, whose structure alone forces their one-sided limits to exist and
restricts their discontinuities to countably many jumps. The two threads meet in
the continuity of inverse functions.

## Limits at infinity

Infinity is not a real number, but it can play the role of a
[cluster point](/real-analysis/continuity/limits-of-functions): a set clusters at
$\infty$ when it contains arbitrarily large elements.

> **Definition ($\infty$ as a cluster point; limit at infinity).** We say $\infty$
> is a cluster point of $S \subseteq \mathbb{R}$ if for every $M \in \mathbb{R}$
> there is an $x \in S$ with $x \ge M$. For such $S$ and $f : S \to \mathbb{R}$,
> we write $\lim_{x\to\infty} f(x) = L$ if for every $\epsilon > 0$ there is an
> $M \in \mathbb{R}$ such that
> $$
> x \in S \text{ and } x \ge M \implies |f(x) - L| < \epsilon.
> $$
> The limit as $x \to -\infty$ is defined symmetrically with $x \le M$.[^lebl-inf]

The threshold $M$ plays the role $\delta$ played at a finite point: it names how
far out you must go before $f$ stays within $\epsilon$ of $L$. When such an $L$
exists the line $y = L$ is a **horizontal asymptote**. As usual the limit, if it
exists, is unique.

$$
% caption: A horizontal asymptote: past the threshold M the graph stays inside the
% epsilon-band around L, and it never leaves again as x grows.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-0.3,0) -- (7.4,0) node[anchor=north] {$x$};
\draw[black, ->] (0,-0.3) -- (0,3.4) node[anchor=east] {$y$};
% band around L
\fill[acc!12] (0,2.05) rectangle (7.2,2.55);
\draw[acc!55] (0,2.3) -- (7.2,2.3);
\node[acc, anchor=east] at (-0.05,2.3) {$L$};
% threshold
\draw[black, dashed] (2.6,0) -- (2.6,2.9); \node[anchor=north] at (2.6,-0.02) {$M$};
% curve approaching asymptote
\draw[black, very thick] plot[domain=0.3:7.1, samples=80] (\x, {2.3 - 1.7/(\x+0.35)});
\node[black, anchor=north, font=\footnotesize] at (5.0,1.95) {inside the band beyond $M$};
\end{tikzpicture}
$$

> **Worked example (an explicit threshold).** $f(x) = \frac{1}{|x| + 1}$ has
> $\lim_{x\to\infty} f = 0$ and $\lim_{x\to-\infty} f = 0$. Given $\epsilon > 0$,
> take $M = 1/\epsilon$: for $x \ge M$,
> $$
> |f(x) - 0| = \frac{1}{|x| + 1} \le \frac{1}{x + 1} < \frac{1}{M} = \epsilon.
> $$
> The threshold $M(\epsilon) = 1/\epsilon$ is the analogue of the explicit
> $\delta(\epsilon)$ at a finite point.

> **Worked example (no limit at infinity).** $\sin(\pi x)$ has no limit as
> $x \to \infty$: it equals $1$ at $x = 2n + \tfrac12$ and $-1$ at $x = 2n +
> \tfrac32$, so past any threshold $M$ the function takes both values $1$ and
> $-1$, which cannot both lie in a band of half-height $\epsilon < 1$ around one
> $L$. Contrast the integer sequence $\sin(\pi n) = 0$, which converges: the
> continuous and sequential statements must not be confused, since the continuous
> limit quantifies over all large real inputs, not just integers.

A sequential characterization holds here too: $\lim_{x\to\infty} f(x) = L$ iff
$f(x_n) \to L$ for every sequence $x_n \to \infty$ in $S$. Through it, the
[algebra of limits](/real-analysis/sequences-series/limit-laws-monotone) carries
over with the cluster point allowed to be $\pm\infty$.

### Infinite limits

It is convenient to name divergence that is orderly rather than oscillatory.

> **Definition (Divergence to infinity).** For $f : S \to \mathbb{R}$ with
> $\infty$ a cluster point of $S$, we write $\lim_{x\to\infty} f(x) = \infty$ if
> for every $N \in \mathbb{R}$ there is an $M$ with $f(x) > N$ whenever $x \ge M$.[^lebl-inf]

> **Worked example (divergence to infinity).** $\lim_{x\to\infty} \frac{1 +
> x^2}{1 + x} = \infty$: for $x \ge 1$,
> $$
> \frac{1 + x^2}{1 + x} \ge \frac{x^2}{x + x} = \frac{x}{2},
> $$
> so given $N$, taking $M = \max\{2N + 1, 1\}$ makes $\frac{1 + x^2}{1 + x} \ge
> \frac{x}{2} > N$ for all $x \ge M$.

Limits compose under a continuity hypothesis exactly as at finite points; for
instance $\lim_{x\to\infty} e^{-x^2 + x} = 0$, since $-x^2 + x \to -\infty$ and
$e^y \to 0$ as $y \to -\infty$.[^lebl-inf]

## Monotone functions

A function that only rises, or only falls, cannot oscillate, and this restriction
alone guarantees one-sided limits and restricts the discontinuities to jumps.

> **Definition (Monotone).** $f : S \to \mathbb{R}$ is **increasing** if $x < y$
> implies $f(x) \le f(y)$, and **strictly increasing** if $x < y$ implies $f(x) <
> f(y)$; **decreasing** and **strictly decreasing** reverse the inequality. A
> function that is increasing or decreasing is **monotone**.[^lebl-mono]

Since $-f$ reverses the direction, results for increasing functions convert
directly to decreasing ones. The key fact is that monotonicity forces one-sided
limits to exist, computed as suprema and infima of the values on one side.

> **Proposition (One-sided limits of monotone functions).** Let $f : S \to
> \mathbb{R}$ be increasing and $c \in \mathbb{R}$. If $c$ is a cluster point of
> $S \cap (-\infty, c)$, then
> $$
> \lim_{x\to c^-} f(x) = \sup\{f(x) : x \in S,\ x < c\},
> $$
> and if $c$ is a cluster point of $S \cap (c, \infty)$, then
> $\lim_{x\to c^+} f(x) = \inf\{f(x) : x \in S,\ x > c\}$.[^lebl-mono]

> **Proof.** Let $a = \sup\{f(x) : x \in S,\ x < c\}$, supplied by the
> [least-upper-bound property](/real-analysis/foundations/ordered-fields-completeness)
> when the set is bounded above (if it is not, the limit is $\infty$ in the
> extended sense by the same argument with a bound $M$ in place of
> $a - \epsilon$). Given $\epsilon > 0$, since $a - \epsilon$ is not an upper
> bound, some $x_0 \in S$ with $x_0 < c$ has $f(x_0) > a - \epsilon$. Take
> $\delta = c - x_0$: for $x \in S$ with $x_0 < x < c$, monotonicity gives
> $a - \epsilon < f(x_0) \le f(x) \le a$, so $|f(x) - a| < \epsilon$. The
> infimum case is symmetric.

**All** one-sided limits of a monotone function exist whenever they make sense.
Consequently, for $c$ interior to the domain,

$$
\lim_{x\to c^-} f(x) \le f(c) \le \lim_{x\to c^+} f(x),
$$

and $f$ is continuous at $c$ exactly when these two coincide. A discontinuity of a
monotone function is therefore always a **jump**: a gap between the left supremum
and the right infimum.

$$
% caption: At a jump of an increasing function the left limit is the supremum of
% earlier values and the right limit the infimum of later ones; the gap between
% them is the height of the jump, with f(c) sitting inside it.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-0.3,0) -- (6.6,0) node[anchor=north] {$x$};
\draw[black, ->] (0,-0.3) -- (0,4.0) node[anchor=east] {$y$};
\node[anchor=north] at (3.0,-0.02) {$c$};
\draw[black, dashed] (3.0,0) -- (3.0,3.5);
% left branch rising to a
\draw[black, very thick] (0.4,0.7) .. controls (1.6,1.2) and (2.4,1.5) .. (2.98,1.7);
\fill[white] (3.0,1.7) circle (2pt); \draw[acc] (3.0,1.7) circle (2pt);
\draw[black, dashed] (0,1.7) -- (3.0,1.7); \node[acc, anchor=east] at (-0.05,1.7) {$a$};
% right branch from b
\draw[black, very thick] (3.02,2.9) .. controls (3.8,3.1) and (5.0,3.4) .. (6.2,3.7);
\fill[acc] (3.0,2.9) circle (1.7pt);
\draw[black, dashed] (0,2.9) -- (3.0,2.9); \node[acc, anchor=east] at (-0.05,2.9) {$b$};
% f(c) value
\fill[acc] (3.0,2.3) circle (1.5pt); \node[acc, anchor=west, font=\footnotesize] at (3.08,2.3) {$f(c)$};
\node[black, anchor=west, font=\footnotesize] at (3.15,1.7) {left limit $a$};
\node[black, anchor=west, font=\footnotesize] at (3.15,3.25) {right limit $b$};
\end{tikzpicture}
$$

> **Worked example (one-sided limits of the floor function).** Let $f(x) =
> \lfloor x \rfloor$, the largest integer $\le x$, an increasing step function.
> At an integer $n$:
> $$
> \lim_{x\to n^-} \lfloor x \rfloor = \sup\{\lfloor x \rfloor : x < n\} = n - 1,
> \qquad
> \lim_{x\to n^+} \lfloor x \rfloor = \inf\{\lfloor x \rfloor : x > n\} = n,
> $$
> since $\lfloor x \rfloor = n - 1$ on $[n-1, n)$ and $\lfloor x \rfloor = n$ on
> $[n, n+1)$. The two one-sided limits differ by $1$, so each integer is a jump
> discontinuity of height $1$, with $f(n) = n$ equal to the right limit; at
> every non-integer both one-sided limits equal the value, and $f$ is
> continuous there.

$$
% caption: The floor function: constant on each half-open interval between
% integers, jumping by one at each integer, where the value joins the upper step.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-0.4,0) -- (5.4,0) node[anchor=north] {$x$};
\draw[black, ->] (0,-0.4) -- (0,3.6) node[anchor=east] {$y$};
\foreach \n in {1,2,3,4} \draw[black] (\n,0.06) -- (\n,-0.06);
\node[anchor=north, font=\footnotesize] at (1,-0.1) {1};
\node[anchor=north, font=\footnotesize] at (2,-0.1) {2};
\node[anchor=north, font=\footnotesize] at (3,-0.1) {3};
\node[anchor=north, font=\footnotesize] at (4,-0.1) {4};
\foreach \n/\yy in {1/0.75, 2/1.5, 3/2.25, 4/3.0} {
  \draw[acc, very thick] (\n,\yy) -- (\n+0.96,\yy);
  \fill[acc] (\n,\yy) circle (1.7pt);
  \fill[white] (\n+1,\yy) circle (2pt); \draw[acc] (\n+1,\yy) circle (2pt);
}
\draw[acc, very thick] (0,0) -- (0.96,0);
\fill[white] (1,0) circle (2pt); \draw[acc] (1,0) circle (2pt);
\node[acc, anchor=west, font=\footnotesize] at (4.4,2.55) {each jump has height 1};
\end{tikzpicture}
$$

### Countably many discontinuities

The jumps of a monotone function cannot pile up too densely, because each jump
encloses an open gap and the gaps are disjoint.

> **Corollary (Countable discontinuities).** A monotone function on an interval
> has at most countably many discontinuities.[^lebl-mono]

> **Proof.** At each interior discontinuity $c$ the gap $(a, b)$ with
> $a = \lim_{x\to c^-} f$ and $b = \lim_{x\to c^+} f$ is a nonempty open
> interval, and monotonicity makes gaps at distinct discontinuities disjoint.
> Pick a rational $q_c \in (a, b)$ for each; the assignment $c \mapsto q_c$ is
> injective into
> [$\mathbb{Q}$](/real-analysis/foundations/intervals-uncountability), so the set
> of discontinuities is countable.

A monotone function can still be discontinuous on a dense set such as the
rationals, but never on more than a countable set.

$$
% caption: An increasing function's jumps enclose disjoint open gaps on the
% y-axis; choosing one rational inside each gap injects the jumps into the
% rationals, so there are at most countably many.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{wrn}{HTML}{B0672F}
\draw[black, ->] (-0.3,0) -- (6.6,0) node[anchor=north] {$x$};
\draw[black, ->] (0,-0.3) -- (0,4.0) node[anchor=east] {$y$};
% rising staircase with jumps
\draw[acc, very thick] (0.3,0.4) -- (1.4,0.9);
\draw[acc, very thick] (1.4,1.5) -- (2.8,2.0);
\draw[acc, very thick] (2.8,2.6) -- (4.3,3.0);
\draw[acc, very thick] (4.3,3.4) -- (5.9,3.7);
\foreach \x/\yl/\yh in {1.4/0.9/1.5, 2.8/2.0/2.6, 4.3/3.0/3.4} {
  \fill[white] (\x,\yl) circle (2pt); \draw[acc] (\x,\yl) circle (2pt);
  \fill[acc] (\x,\yh) circle (1.6pt);
}
% gaps on y-axis with rationals
\foreach \yl/\yh in {0.9/1.5, 2.0/2.6, 3.0/3.4} {
  \draw[wrn, thick] (0.06,\yl) -- (0.06,\yh);
  \pgfmathsetmacro\ym{(\yl+\yh)/2}
  \fill[wrn] (0.06,\ym) circle (1.5pt);
}
\node[wrn, anchor=west, font=\footnotesize] at (0.15,1.2) {one rational per gap};
\end{tikzpicture}
$$

## Continuity, images, and inverses

For monotone functions on an interval, continuity is equivalent to a purely
set-theoretic condition on the image, via the
[intermediate value theorem](/real-analysis/continuity/evt-ivt).

> **Corollary (Image characterization).** If $I$ is an interval and $f : I \to
> \mathbb{R}$ is monotone and not constant, then $f(I)$ is an interval if and only
> if $f$ is continuous.[^lebl-mono]

> **Proof.** If $f$ is continuous, IVT makes $f(I)$ contain every value between
> any two of its values, so $f(I)$ is an interval. If $f$ is discontinuous at an
> interior $c$, the gap $(a,b)$ around the jump misses $f(I)$ apart from $f(c)$,
> so $f(I)$ has a hole and is not an interval.

The order structure also yields continuity of inverses. A strictly monotone
function is injective, so it has an inverse on its range.

> **Proposition (Continuity of the inverse).** If $I$ is an interval and $f : I
> \to \mathbb{R}$ is strictly monotone, then $f^{-1} : f(I) \to I$ is continuous —
> even if $f$ itself is not.[^lebl-mono]

> **Proof.** The inverse of a strictly increasing function is strictly
> increasing, so its one-sided limits exist by the monotone-limit proposition; a
> short supremum–infimum comparison shows they agree, so $f^{-1}$ is continuous
> at every point of its domain.

When $f(I)$ happens to be an interval, the image characterization applies to
both $f$ and $f^{-1}$: an onto strictly monotone map between two intervals is a
continuous bijection with continuous inverse, a
[homeomorphism](/real-analysis/continuity/continuity-metric-spaces) of intervals.

> **Worked example ($x^k$ and its root).** For $k \in \mathbb{N}$ the map
> $f(x) = x^k$ is strictly increasing on $[0, \infty)$: for $0 \le x < y$,
> $y^k - x^k = (y - x)(y^{k-1} + y^{k-2}x + \cdots + x^{k-1}) > 0$. Its image is
> $[0, \infty)$ (an interval, by the
> [$k$-th root existence argument](/real-analysis/continuity/evt-ivt)), so $f$
> is continuous by the image characterization, and the proposition makes the
> inverse $f^{-1}(y) = y^{1/k}$ continuous on $[0, \infty)$ with no
> $\epsilon$–$\delta$ estimate for the root function ever written down.

This continuity of the inverse is the topological half of the
[inverse function theorem](/real-analysis/differentiation/inverse-function-1d):
once differentiability is added, the inverse of a differentiable strictly
monotone function is differentiable wherever $f' \neq 0$, with derivative the
reciprocal of $f'$.

[^lebl-inf]: **Lebl**, _Basic Analysis I_, §3.5 — Definitions 3.5.1 and 3.5.6, Lemma 3.5.5, Proposition 3.5.8, and Examples 3.5.3–3.5.9 (limits at infinity, infinite limits, compositions).
[^lebl-mono]: **Lebl**, _Basic Analysis I_, §3.6 — Definition 3.6.1, Proposition 3.6.2, Corollaries 3.6.3–3.6.4, and Proposition 3.6.6 (monotone functions, one-sided limits, countable discontinuities, continuity of inverses).