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$x\,\psi(1/x) \to 0$ as $x \to 0$ , by the squeeze theorem

Example

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.

Facts & Assumptions

Given: The set $A = \mathbb{R} \setminus \{0\}$ and the function $h : A \to \mathbb{R}$ , $h(x) = x \cdot \psi(1/x)$ , with $\psi$ the function 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.

[L1]

Range of $\psi$ : $0 \le \psi(u) \le 1/2$ for every real $u$ (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, claim 2).

[L2]

The limit condition (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ): $\lim_{x \to c} f(x) = P$ means that for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that every $x$ in the domain with $0 < |x - c| < \delta$ satisfies $|f(x) - P| < \varepsilon$ .

[L3]

Squeeze theorem: if $f \le g \le k$ on $A \cap N^{*}_{\eta}(c)$ for some real $\eta > 0$ , and the limits of $f$ and of $k$ at $c$ exist and are equal to $L$ , then the limit of $g$ at $c$ exists and equals $L$ (If $f \le g \le h$ near $c$ and $f$ and $h$ have the same limit at $c$ , then so does $g$ ).

[L4]

Absolute value: $|u| \ge 0$ ; $|u| = 0$ exactly when $u = 0$ ; $|uv| = |u|\,|v|$ ; $|-u| = |u|$ ; $|u| = u$ for $u \ge 0$ ; and $-|u| \le u \le |u|$ (Basic properties of the absolute value).

[L5]

Order and field arithmetic: $x \ne 0$ has an inverse $1/x$ (Field); $0 < 1$ , so $2 > 0$ and $1/2 > 0$ with $t/2 < t$ for $t > 0$ (The multiplicative identity is positive, Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication); multiplying an inequality by a non-negative factor, and adding inequalities (Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities); the order is total (Ordered field). Those two sources state their moves in their STRICT forms only; the non-strict forms used below follow by adjoining the equality case, in which the two sides coincide (Ordered field).

[L6]

Limit point and neighbourhoods (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

Verification

technique · direct

1.1

$0$ is a limit point of $A = \mathbb{R} \setminus \{0\}$ : given a real $\varepsilon > 0$ , the real $\varepsilon/2$ satisfies $\varepsilon/2 > 0$ , so it lies in $A$ , and $0 < |\varepsilon/2 - 0| = \varepsilon/2 < \varepsilon$ .

L4L5L6

1.2

$h$ is defined on all of $A$ , and $|h(x)| \le |x|/2$ there: for $x \in A$ we have $x \ne 0$ , so $1/x$ exists, and $|h(x)| = |x| \cdot \psi(1/x)$ by [L4], while $0 \le \psi(1/x) \le 1/2$ by [L1] and $|x| \ge 0$ , so multiplying the inequality $\psi(1/x) \le 1/2$ by the non-negative factor $|x|$ gives $|h(x)| \le |x|/2$ .

L1L4L5

1.3

The two functions $x \mapsto -|x|/2$ and $x \mapsto |x|/2$ on $A$ each have limit $0$ at $0$ : given a real $\varepsilon > 0$ , take $\delta := \varepsilon$ ; every $x \in A$ with $0 < |x - 0| < \delta$ satisfies $\bigl| |x|/2 - 0 \bigr| = |x|/2 < \varepsilon/2 < \varepsilon$ , and likewise $\bigl| -|x|/2 - 0 \bigr| = |x|/2 < \varepsilon$ .

L2L4L5

2.1

Hence $-|x|/2 \le h(x) \le |x|/2$ for every $x \in A$ , by [L4] applied to $|h(x)| \le |x|/2$ .

step 1.2L4

3.1

The three functions satisfy $-|x|/2 \le h(x) \le |x|/2$ on all of $A$ , in particular on $A \cap N^{*}_{1}(0)$ , and the two outer ones have limit $0$ at $0$ ; since $0$ is a limit point of $A$ , the squeeze theorem [L3] gives that the limit of $h$ at $0$ exists and equals $0$ .

step 1.1step 1.3step 2.1L3∎

Remarks

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 $\sin$ .

Depends on

Used by

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Sources

Squeeze theorem (Wikipedia) (standard reference, not scraped)
Limit of a function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.1 (standard reference, not scraped)

x ψ(1/x)→0x\,\psi(1/x) \to 0xψ(1/x)→0 as x→0x \to 0x→0, by the squeeze theorem