ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The sign function has both one-sided limits at $0$ and no two-sided limit

Example

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 $A$ ).

Facts & Assumptions

Given: The function $\operatorname{sgn} : \mathbb{R} \to \mathbb{R}$ above, with $A := \mathbb{R}$ , $c := 0$ , $A^{+} = \mathbb{R} \cap (0,\infty) = (0,\infty)$ and $A^{-} = \mathbb{R} \cap (-\infty,0) = (-\infty,0)$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L1]

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 $|f(x) - P| < \varepsilon$ .

[L2]

One-sided limits are the limits at $c$ of the restrictions of $f$ to $A^{-}$ and $A^{+}$ , and are well posed exactly when $c$ is a limit point of the set in question (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ ).

[L3]

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 $\mathbb{R}$ ).

[L4]

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

[L5]

Order in $\mathbb{R}$ : trichotomy, so every real satisfies exactly one of $x > 0$ , $x = 0$ , $x < 0$ ; $0 < 1$ and hence $2 > 0$ , $\varepsilon/2 > 0$ and $\varepsilon/2 < \varepsilon$ for $\varepsilon > 0$ ; and $1 - (-1) = 1 + 1 > 0$ , so $-1 < 1$ and in particular $1 \ne -1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Ordered field).

[L6]

Two-sided versus one-sided: if $c$ is a limit point of both $A^{-}$ and $A^{+}$ and $\lim_{x \to c} f(x) = L$ , then $\lim_{x \to c^{-}} f(x) = L$ and $\lim_{x \to c^{+}} f(x) = L$ (If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree).

[L7]

At a limit point of its domain a function has at most one limit (At a limit point of the domain a function has at most one limit); applied to the restrictions, each one-sided limit is a single real.

Verification

technique · direct

1.1

$\operatorname{sgn}$ is a well-defined function on $\mathbb{R}$ : by trichotomy every real satisfies exactly one of the three defining conditions.

1.2

$0$ is a limit point of $A^{+} = (0,\infty)$ and of $A^{-} = (-\infty,0)$ : given a real $\varepsilon > 0$ , the real $\varepsilon/2$ is positive, hence lies in $A^{+}$ , and satisfies $0 < |\varepsilon/2 - 0| = \varepsilon/2 < \varepsilon$ ; and $-\varepsilon/2$ is negative, hence lies in $A^{-}$ , and satisfies $0 < |-\varepsilon/2 - 0| = \varepsilon/2 < \varepsilon$ .

L3L4L5

1.3

The reals $1$ and $-1$ are distinct, since $1 - (-1) = 1 + 1 > 0$ .

2.1

$\lim_{x \to 0^{+}} \operatorname{sgn}(x) = 1$ : by [L2] this is the limit at $0$ of the restriction of $\operatorname{sgn}$ to $A^{+}$ , which is well posed by step 1.2. Given a real $\varepsilon > 0$ , any $\delta > 0$ serves, since every $x \in A^{+}$ has $x > 0$ , hence $\operatorname{sgn}(x) = 1$ and $|\operatorname{sgn}(x) - 1| = |0| = 0 < \varepsilon$ .

step 1.1step 1.2L1L2L4

2.2

$\lim_{x \to 0^{-}} \operatorname{sgn}(x) = -1$ : identically, every $x \in A^{-}$ has $x < 0$ , hence $\operatorname{sgn}(x) = -1$ and $|\operatorname{sgn}(x) - (-1)| = |0| = 0 < \varepsilon$ for every $\varepsilon > 0$ and every $\delta > 0$ .

step 1.1step 1.2L1L2L4

3.1

Suppose $\operatorname{sgn}$ had a limit at $0$ , say $\lim_{x \to 0} \operatorname{sgn}(x) = L$ . Since $0$ is a limit point of both $A^{-}$ and $A^{+}$ by step 1.2, [L6] gives $\lim_{x \to 0^{-}} \operatorname{sgn}(x) = L$ and $\lim_{x \to 0^{+}} \operatorname{sgn}(x) = L$ ; each one-sided limit is single valued by [L7], so steps 2.1 and 2.2 force $L = 1$ and $L = -1$ , contradicting step 1.3. Hence $\operatorname{sgn}$ has no limit at $0$ .

step 1.3step 2.1step 2.2L6L7∎

Remarks

The failure is not about the value at $0$ . Redefining $\operatorname{sgn}(0)$ to be $1$ , or $-1$ , or anything else changes nothing: The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ never evaluates the function at the point, and the two one-sided limits are computed on sets that exclude $0$ (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ ). This is a genuine jump, not a removable defect of the kind FALSE: $\lim_{x \to c} f(x) = f(c)$ whenever both sides exist exhibits.
Away from $0$ the function is locally constant, so it has a limit at every other point of $\mathbb{R}$ , equal to its value there: for $c > 0$ take $\delta$ to be $c$ itself, and every $x$ with $0 < |x - c| < c$ has $x > 0$ and $\operatorname{sgn}(x) = 1 = \operatorname{sgn}(c)$ ; symmetrically for $c < 0$ . So the single point $0$ carries the whole phenomenon.
Contrast with the two other failures on this page. Here both one-sided limits exist and differ; for $\psi(1/x)$ at $0$ ( $\psi(1/x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1/2$ ) the failure is already one-sided, both witnessing sequences there having positive terms; and for the indicator of $\mathbb{Q}$ (The indicator of $\mathbb{Q}$ has a limit at no point of $\mathbb{R}$ ) the failure occurs at every point at once.

Depends on

Used by

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Sources

Sign function (Wikipedia) (standard reference, not scraped)
One-sided limit (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.1 (standard reference, not scraped)

The sign function has both one-sided limits at 000 and no two-sided limit