Limits of Real Functions: Examples and Counterexamples

Every analysis course states the two examples of this page in the form

$\sin(1/x) \quad \text{has no limit at } 0, \qquad x \sin(1/x) \to 0 \text{ as } x \to 0 ,$

and a reader who has met them before will recognise $\psi(1/x)$ has no limit at $0$: two sequences tending to $0$ give values constantly $0$ and constantly $1/2$ and $x\,\psi(1/x) \to 0$ as $x \to 0$, by the squeeze theorem as those examples with $\psi$ in place of $\sin$. This remark records the correspondence, and it records that the correspondence is orientation only: the two displayed statements are reported as what the classical treatment proves, not asserted here, and nothing on this page uses or proves anything about $\sin$.

This library now constructs sine and cosine from their power series, proves their differential and addition laws, and defines pi from the first positive zero of cosine. Under this library's counterexample convention, sin(1/x) has no limit as x tends to zero ↗ displays the false proposition that the sine limit exists under Statement refuted, then proves it false; x sin(1/x) tends to zero despite its oscillation ↗ proves the squeezed limit for the product. Both occur later in the reading order, so the links are orientation-only forward references declared in this item's forward_refs; no proof on this earlier page depends on them.

The function $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ of The trigonometry-free oscillator $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ is well defined and attained at a nearest integer, takes values in $[0, 1/2]$, vanishes exactly on $\mathbb{Z}$, equals $1/2$ at half-integers, and is $1$-periodic is elementary — it needs only the integer part, the order and the absolute value — and it has the three properties that make the classical examples work:

• it is bounded, with values exactly in $[0, 1/2]$;
• it is periodic, with period $1$, so $\psi(1/x)$ oscillates without damping as $x \to 0$;
• it attains two distinct values on every punctured neighbourhood of $0$ after the substitution $x \mapsto 1/x$, namely $0$ at the reciprocals of the integers and $1/2$ at the reciprocals of the half-integers.

The third property is what $\psi(1/x)$ has no limit at $0$: two sequences tending to $0$ give values constantly $0$ and constantly $1/2$ uses. It is not sharper than what $\sin$ would give — the classical witnessing sequences hit the extreme values of $\sin$ exactly too — but it is available here: the two values $0$ and $1/2$ are read off from the integer part in one line (Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$), with no series and no $\pi$, whereas the corresponding facts about $\sin$ presuppose the whole construction described above.

Nothing on this page is weaker for using $\psi$. The two statements proved are exactly the statements usually proved with $\sin$, and their proofs are shorter.

What is lost is a connection to a different subject. The classical pair $\sin(1/x)$, $x\sin(1/x)$ also carries information about smoothness, about power series and about the topologist's sine curve, none of which $\psi$ can carry, since $\psi$ is assembled from the order, the absolute value and the integer part alone. Those notions occur only later in the reading order and are unavailable on this earlier page.

For a list $a_0, \dots, a_{n-1}$ of reals write

$p(x) \;:=\; \sum_{j < n} a_j x^j$

for the finite sum of Finite sums and finite products, by recursion applied to the list $j \mapsto a_j x^j$, with powers as in Integer powers $a^m$. So $\sum_{j < 0}$ is the empty sum $0$, and $p$ is a function $\mathbb{R} \to \mathbb{R}$; these are the polynomial functions.

Claim 1. For every polynomial function $p$ and every $c \in \mathbb{R}$, the limit of $p$ at $c$ exists and

$\lim_{x \to c} p(x) \;=\; p(c) .$

Claim 2. Let $p$ and $q$ be polynomial functions and let $c \in \mathbb{R}$ satisfy $q(c) \ne 0$. Put $A_q := \{\, x \in \mathbb{R} : q(x) \ne 0 \,\}$. Then $c \in A_q$, the point $c$ is a limit point of $A_q$, the quotient $p/q$ is defined on $A_q$, its limit at $c$ exists, and

$\lim_{x \to c} (p/q)|_{A_q}(x) \;=\; \frac{p(c)}{q(c)} .$

Everything is read off from Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero once two trivial limits are in hand: that of a constant function and that of the identity. Note that claim 1 is exactly the statement that $\lim_{x \to c} p(x) = p(c)$, the equality that FALSE: $\lim_{x \to c} f(x) = f(c)$ whenever both sides exist shows is not automatic; for polynomials it is a theorem, and the algebra of limits is what proves it.

• Two inductions, and why they are separate. The first builds the monomials from the identity by repeated multiplication; the second builds the polynomial from the monomials by repeated addition. Each is an induction on the recursion clause of the object it builds (Integer powers $a^m$ and Finite sums and finite products, by recursion respectively), and neither can be replaced by dots.

• Index hygiene. The sum is written $\sum_{j<n}$, whose first index is $j = 0$ and whose empty case $n = 0$ is the zero function; the base case of step 4.1 is that empty case, and $x^0 = 1$ holds for every real $x$ including $0$ (Integer powers $a^m$), so no index or value is left undefined.

• What claim 2 does not say. It says nothing at a zero of $q$. There the quotient is undefined, and whether it has a limit depends on $p$ as well. It may have one: on $\{\, x : x \ne 0 \,\}$ the quotient $x^2/x$ equals $x$, whose limit at $0$ is $0$ by step 1.3 and The limit at $c$ depends only on the restriction of $f$ to a punctured neighbourhood of $c$, and passes to any subset of the domain having $c$ as a limit point. It may also fail to have one. Nothing on this page decides such cases in general.

Identify $\mathbb{Z}$ with its canonical copy in $\mathbb{R}$ (The integers as equivalence classes of pairs of naturals, The integers embed in the rationals, The rationals embed densely in the reals) and for $x \in \mathbb{R}$ put

$D(x) \;:=\; \{\, |x - n| \ : \ n \in \mathbb{Z} \,\}, \qquad \psi(x) \;:=\; \inf D(x)$

(Greatest lower bound (infimum)). Write $m := \lfloor x \rfloor$ for the integer part of $x$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$) and $t := x - m$, so $0 \le t < 1$. Then:

1. Existence and attainment. $\psi(x)$ exists and is attained: $\psi(x) \;=\; \min\{\, t,\ 1 - t \,\} \;=\; \min\bigl\{\, |x - m|,\ |x - (m+1)| \,\bigr\} ,$ so $\psi(x) = |x - n|$ for $n = m$ or $n = m + 1$, and $\psi(x) = \min D(x)$ (Maximum and minimum of a set).
2. Range. $0 \le \psi(x) \le 1/2$ for every real $x$, and every value in $[0, 1/2]$ occurs: the range of $\psi$ is exactly the interval $[0, 1/2]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).
3. Zero set. $\psi(x) = 0$ if and only if $x \in \mathbb{Z}$.
4. Half-integers. $\psi(m + 1/2) = 1/2$ for every $m \in \mathbb{Z}$.
5. Periodicity. $\psi(x + 1) = \psi(x)$ for every real $x$.

What this function is for. It is the elementary, trigonometry-free substitute for $\sin$: it is bounded, it oscillates, and on every punctured neighbourhood of $0$ the composite $\psi(1/x)$ attains both the value $0$ and the value $1/2$. Claims 3 and 4 are exactly what the companion counterexample $\psi(1/x)$ has no limit at $0$: two sequences tending to $0$ give values constantly $0$ and constantly $1/2$ evaluates, and claim 2 is what the squeeze argument of $x\,\psi(1/x) \to 0$ as $x \to 0$, by the squeeze theorem uses.

• No completeness of $\mathbb{R}$ is needed for the infimum here. The general existence theorem Every nonempty set bounded below has an infimum would supply $\inf D(x)$ from the least-upper-bound property, but step 3.1 does not use it: the infimum is produced by exhibiting an element of $D(x)$ that is also a lower bound, which is Greatest lower bound (infimum) read directly. Completeness does enter, once, through Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$, whose existence half is the Archimedean property.

• Why $\min\{t, 1-t\}$ and not "the distance to the nearest integer". The phrase presupposes that a nearest integer exists, which is exactly what step 2.2 establishes and what the picture cannot. When $t = 1/2$ there are two nearest integers, $m$ and $m+1$, and the formula is indifferent to which is chosen, so nothing has to be selected.

• $\psi$ is the triangle wave of amplitude $1/2$ and period $1$ — not the sawtooth $x - \lfloor x \rfloor$, which drops discontinuously at every integer: on $[0, 1/2]$ it is $\psi(s) = s$ by step 5.1, and periodicity and the reflection $\psi(-x) = \psi(x)$ — immediate from $D(-x) = \{\, |-x - n| : n \in \mathbb{Z} \,\} = \{\, |x + n| : n \in \mathbb{Z} \,\} = D(x)$, using $|-u| = |u|$ and the bijection $n \mapsto -n$ of $\mathbb{Z}$ — determine it everywhere.

Let $A := \mathbb{R} \setminus \{0\}$ and define $h : A \to \mathbb{R}$ by

$h(x) \;:=\; x \cdot \psi(1/x),$

with $\psi$ as in The trigonometry-free oscillator $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ is well defined and attained at a nearest integer, takes values in $[0, 1/2]$, vanishes exactly on $\mathbb{Z}$, equals $1/2$ at half-integers, and is $1$-periodic. Then $0$ is a limit point of $A$, the limit of $h$ at $0$ exists, and

$\lim_{x \to 0} h(x) \;=\; 0 .$

The point of the example. The factor $\psi(1/x)$ has no limit at $0$ at all ($\psi(1/x)$ has no limit at $0$: two sequences tending to $0$ give values constantly $0$ and constantly $1/2$), so Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero cannot be applied to the product: its product rule requires both factors to have limits. What is available is that $\psi(1/x)$ stays inside $[0,1/2]$, and a bounded factor multiplied by one tending to $0$ is killed. That is exactly what If $f \le g \le h$ near $c$ and $f$ and $h$ have the same limit at $c$, then so does $g$ delivers, and it delivers the existence of the limit, not merely its value.

• Where the hypotheses of the squeeze theorem are met. The order hypothesis holds on all of $A$, so any $\eta > 0$ serves and $\eta = 1$ is taken; the two outer limits are computed by hand in step 1.3; and $0$ is a limit point of $A$ by step 1.1, which is what makes every limit here well posed (The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$).

• Nothing about $\psi$ beyond its range is used. Replacing $\psi$ by any function with values in a fixed bounded set would give the same conclusion by the same three steps. What makes the example worth stating is the contrast with $\psi(1/x)$ has no limit at $0$: two sequences tending to $0$ give values constantly $0$ and constantly $1/2$: the same oscillating factor, multiplied by $x$ or not, is the difference between a limit existing and not.

• The classical version of this example is $x \sin(1/x) \to 0$ as $x \to 0$; see The classical form of the oscillator above is $\sin(1/x)$, which this library can only construct much later for why this library writes $\psi$ and not $\sin$.

Define $\operatorname{sgn} : \mathbb{R} \to \mathbb{R}$ by

$\operatorname{sgn}(x) := \begin{cases} 1 & x > 0, \\ 0 & x = 0, \\ -1 & x < 0. \end{cases}$

Then $0$ is a limit point of both $(0,\infty)$ and $(-\infty,0)$, both one-sided limits at $0$ exist (The left and right limits of $f$ at $c$, as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$),

$\lim_{x \to 0^{+}} \operatorname{sgn}(x) = 1, \qquad \lim_{x \to 0^{-}} \operatorname{sgn}(x) = -1 ,$

and $\operatorname{sgn}$ has no limit at $0$.

This is the standard illustration of If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree: the two one-sided limits both exist, so nothing is missing on either side, yet they disagree, and disagreement is exactly what the theorem converts into the failure of the two-sided limit. Note also that the value $\operatorname{sgn}(0) = 0$ is equal to neither one-sided limit, and is irrelevant to all three assertions (The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$).