Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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FALSE: lim⁡x→cf(g(x))=M whenever lim⁡x→cg=L and lim⁡y→Lf=M

Statement

False claim: let A,B⊆R, let g:A→R with g(A)⊆B and f:B→R, let c be a limit point of A and L a limit point of B. If

lim⁡x→cg(x)=Landlim⁡y→Lf(y)=M,

then the limit of f∘g at c exists and lim⁡x→cf(g(x))=M (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

This is the statement of Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c with both of its extra hypotheses removed, and it is false. It is refuted below by a pair in which g is constant and f has a removable defect at the value of that constant.

Where the naive argument breaks. The inner limit gives ∣g(x)−L∣<ρ for x near c; the outer limit gives ∣f(y)−M∣<ε for y∈B with 0<∣y−L∣<ρ. To combine them at y=g(x) one needs ∣g(x)−L∣>0, and nothing in the hypotheses supplies that. Where g(x)=L, the only information available about f is its value f(L), and The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A says nothing whatever about that value (FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist). The two hypotheses of Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c are exactly the two ways of closing that gap.

Facts & Assumptions

Given: The sets A:=R and B:=R; the point c:=0; the function f:R→R of FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist, namely f(y):=0 for y≠0 and f(0):=1; and the constant function g:R→R, g(x):=0 for every x.

[L1]

The limit condition (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A): lim⁡x→ch(x)=P means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−P∣<ε.

[L3]

Absolute value: ∣0∣=0 (Basic properties of the absolute value).

[L4]

Order in R: trichotomy, and 0<1, so 1≠0 (The multiplicative identity is positive, Ordered field).

[L5]

The function f above satisfies f(0)=1 and has limit 0 at 0: for every real ε>0 the radius δ=1 works, since 0<∣y−0∣<1 forces y≠0 and then ∣f(y)−0∣=0<ε; this is the computation carried out in FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist.

Refutation

technique · direct
1.1

The point 0 is a limit point of R, and g(R)={0}⊆R=B, so f∘g is a function on R.

L2
1.2

By [L5], lim⁡y→0f(y)=0; so the outer hypothesis holds with L=0 and M=0.

L5
1.3

The reals 0 and 1 are distinct.

L4
2.1

The inner hypothesis holds with L=0: for the constant function g and any real ε>0, every δ>0 works, since ∣g(x)−0∣=∣0∣=0<ε for every x. So lim⁡x→0g(x)=0.

step 1.1L1L3
3.1

But f∘g is the constant function 1: for every x∈R, g(x)=0 and hence f(g(x))=f(0)=1. Therefore, by the same computation as in step 2.1, the limit of f∘g at 0 exists and equals 1.

step 2.1L1L3L5
3.2

Both extra hypotheses of Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c fail for this pair: hypothesis (i) fails because L=0 lies in B=R while f(L)=f(0)=1≠0=M; and hypothesis (ii) fails because g(x)=0=L for every x, so no punctured neighbourhood of 0 avoids the value L.

step 2.1L5L6
4.1

So lim⁡x→0g(x)=0=L and lim⁡y→0f(y)=0=M, while lim⁡x→0f(g(x))=1≠0=M: the claim is false, and step 3.2 identifies exactly which hypotheses of the true theorem are missing.

step 1.2step 1.3step 3.1step 3.2∎

Remarks

Depends on

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Dependency tree · two levels

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Sources