Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The logarithm grows more slowly than every positive real power

Statement

For every α>0\alpha>0, limx+logxxα=0.\lim_{x\to+\infty}\frac{\log x}{x^\alpha}=0.

Facts & Assumptions

Given: A real α>0\alpha>0.

[L1]

exp(t)\exp(t) tends to ++\infty as t+t\to+\infty, and its range is (0,)(0,\infty) (The exponential tends to ++\infty at ++\infty and to 00 at -\infty, The natural logarithm as the inverse of the exponential function).

[L2]

For a>0a>0, t/exp(at)0t/\exp(at)\to0 as t+t\to+\infty (The exponential dominates every fixed nonnegative integer power at ++\infty).

[L3]

xα=exp(αlogx)x^\alpha=\exp(\alpha\log x) for x>0x>0, and algebra of limits permits substitution through the displayed identities (Real powers for positive bases, with the zero-base positive-exponent convention, Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · direct
1.1

Put t=logxt=\log x. As x+x\to+\infty, the inverse relation and [L1] give t+t\to+\infty.

L1
1.2

By [L3], logx/xα=t/exp(αt)\log x/x^\alpha=t/\exp(\alpha t).

L3
2.1

The right-hand side tends to 00 by [L2], which proves the claim.

step 1.1step 1.2L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 85 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources