LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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If $\lim_{x \to c} f (x) = L \neq 0$ then $∣ f ∣ > ∣ L ∣ / 2$ on a punctured neighbourhood of $c$ ; in particular if $L > 0$ then $f > L / 2 > 0$ there

Statement

Let $A \subseteq R$ , let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ), let $f : A \to R$ and suppose the limit of $f$ at $c$ exists with $\lim_{x \to c} f (x) = L$ and $L \neq 0$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ). Then there is a real $δ > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ$ satisfies

$∣ f (x) ∣ > \frac{∣ L ∣}{2} > 0;$

in particular $f (x) \neq 0$ for every such $x$ . Moreover:

if $L > 0$ then $f (x) > L / 2 > 0$ for every such $x$ ;
if $L < 0$ then $f (x) < L / 2 < 0$ for every such $x$ .

Consequently, writing

$A_{0} : = {x \in A : f (x) \neq 0},$

the point $c$ is a limit point of $A_{0}$ .

The bound $∣ L ∣ / 2$ , and not merely " $f \neq 0$ ", is what later proofs need. The quotient case of Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero estimates $1 / ∣ f ∣$ near $c$ and therefore needs a positive lower bound on $∣ f ∣$ there, and the last claim is what lets a limit be taken on the smaller domain $A_{0}$ at all.

Facts & Assumptions

Given: A set $A \subseteq R$ , a limit point $c$ of $A$ , a function $f : A \to R$ and a real $L \neq 0$ with $\lim_{x \to c} f (x) = L$ ; and $A_{0} : = {x \in A : f (x) \neq 0}$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L1]

The limit condition: for every real $ε > 0$ there is a real $δ > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ$ satisfies $∣ f (x) - L ∣ < ε$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

[L2]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ if and only if $u = 0$ ; $∣ u ∣ = u$ for $u \geq 0$ and $∣ u ∣ = - u$ for $u \leq 0$ ; and for $t > 0$ , $∣ u ∣ < t$ is equivalent to $- t < u < t$ (Basic properties of the absolute value).

[L3]

Reverse triangle inequality: $∣ ∣ u ∣ - ∣ v ∣ ∣ \leq ∣ u - v ∣$ (The reverse triangle inequality).

[L4]

Limit point: for every real $ρ > 0$ there is $x \in A$ with $0 < ∣ x - c ∣ < ρ$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L5]

Order arithmetic in $R$ : trichotomy, so $u \neq 0$ with $∣ u ∣ \geq 0$ and $∣ u ∣ \neq 0$ forces $∣ u ∣ > 0$ ; $0 < 1$ (The multiplicative identity is positive), hence $2 > 0$ and $2^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order), so $t / 2 > 0$ and $t - t / 2 = t / 2$ for $t > 0$ (Sign rules for products and monotonicity of multiplication); adding a constant to an inequality (Order is preserved by adding a constant and by adding inequalities); and of two positive reals the smaller is positive, the order being total (Ordered field).

Proof

technique · direct

1.1

Since $L \neq 0$ we have $∣ L ∣ \neq 0$ while $∣ L ∣ \geq 0$ , so trichotomy gives $∣ L ∣ > 0$ , and $ε : = ∣ L ∣ / 2 > 0$ with $∣ L ∣ - ∣ L ∣ / 2 = ∣ L ∣ / 2$ .

givenL2L5

2.1

Apply [L1] with this $ε$ : fix a real $δ > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ$ satisfies $∣ f (x) - L ∣ < ∣ L ∣ / 2$ .

step 1.1L1choose

3.1

For every such $x$ the reverse triangle inequality gives $∣ ∣ f (x) ∣ - ∣ L ∣ ∣ \leq ∣ f (x) - L ∣ < ∣ L ∣ / 2$ , hence $∣ f (x) ∣ - ∣ L ∣ > - ∣ L ∣ / 2$ and so $∣ f (x) ∣ > ∣ L ∣ - ∣ L ∣ / 2 = ∣ L ∣ / 2 > 0$ ; in particular $∣ f (x) ∣ \neq 0$ and therefore $f (x) \neq 0$ .

step 2.1L2L3L5

3.2

If $L > 0$ then $∣ L ∣ = L$ , and for every such $x$ the estimate $∣ f (x) - L ∣ < L / 2$ gives $- L / 2 < f (x) - L$ , that is $f (x) > L - L / 2 = L / 2 > 0$ .

step 2.1L2L5

3.3

If $L < 0$ then $∣ L ∣ = - L$ , and for every such $x$ the estimate $∣ f (x) - L ∣ < - L / 2$ gives $f (x) - L < - L / 2$ , that is $f (x) < L - L / 2 = L / 2 < 0$ .

step 2.1L2L5

4.1

Let $η > 0$ be an arbitrary real and let $ρ$ be the smaller of $δ$ and $η$ , so $ρ > 0$ . Since $c$ is a limit point of $A$ there is $x \in A$ with $0 < ∣ x - c ∣ < ρ$ ; that $x$ satisfies $0 < ∣ x - c ∣ < δ$ , hence $f (x) \neq 0$ by step 3.1, so $x \in A_{0}$ and $0 < ∣ x - c ∣ < η$ . As $η$ was arbitrary, $c$ is a limit point of $A_{0}$ .

step 3.1L4L5

5.1

So on $A \cap N_{δ}^{*} (c)$ the function is bounded away from $0$ by $∣ L ∣ / 2$ and carries the sign of $L$ , and $c$ remains a limit point of the set $A_{0}$ where $f$ does not vanish.

step 3.1step 3.2step 3.3step 4.1∎

Remarks

Why $∣ L ∣ / 2$ and not some other fraction. Any $ε$ strictly between $0$ and $∣ L ∣$ gives a positive lower bound $∣ L ∣ - ε$ ; the choice $ε = ∣ L ∣ / 2$ makes the bound $∣ L ∣ / 2$ , which is the form used downstream and needs only that $2$ is invertible and positive (The multiplicative identity is positive, Inverses of positives are positive, and reciprocation reverses order).
The last claim is the one that is easy to forget. Restricting a quotient to the set where the denominator does not vanish is useless unless $c$ is still a limit point of that set, since The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ is stated only at a limit point. Step 4.1 is exactly that check, and it is what Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero cites when it forms $A_{0}$ .
Nothing here says anything about $f (c)$ . As always the point $c$ itself is excluded by $0 < ∣ x - c ∣$ , so $f (c)$ may be $0$ even when $L \neq 0$ ; the function of FALSE: $\lim_{x \to c} f (x) = f (c)$ whenever both sides exist, read with the roles of $0$ and $1$ exchanged, is such an example.

Depends on

Used by

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, §3.1: Limits of functions (standard reference, not scraped)
Limit of a function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.1 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)

If lim⁡x→cf(x)=L≠0limx→c​f(x)=L=0 then ∣f∣>∣L∣/2∣f∣>∣L∣/2 on a punctured neighbourhood of cc; in particular if L>0L>0 then f>L/2>0f>L/2>0 there