Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-01
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
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L'Hôpital's rule for the 0/0 form at finite or infinite, one-sided endpoints

Statement

Let c∈R and let f,g be differentiable on a deleted one-sided or two-sided neighbourhood of c, with g′≠0 there. Suppose f(x)→0, g(x)→0 as x→c in the chosen mode. If f′(x)/g′(x)→L∈R‾, then f(x)/g(x)→L in the same mode. The analogous statement at +∞ or −∞ follows after the substitution t=1/x, wherever the transformed functions are defined.

Facts & Assumptions

Given: The hypotheses and one fixed approach mode.

[L1]

Differentiability implies continuity, and the Cauchy quotient lemma gives a point between two arguments at which a secant quotient equals a derivative quotient (A function differentiable at c is continuous at c, Cauchy's mean-value theorem in quotient form when the denominator derivative is nonzero).

Proof

technique · direct
1.1

Extend f,g to c by f(c)=g(c)=0. Their continuity at c follows from the assumed zero limits, while differentiability gives continuity at every other point of the segment. For x≠c sufficiently close, the quotient lemma on the segment with endpoints c,x gives f(x)g(x)=f′(ξx)g′(ξx), where ξx lies strictly between c and x.

givenL1L2
2.1

As x→c in the chosen mode, ξx→c in that mode. Applying the defining finite or infinite limit inequality to the derivative quotient therefore gives f(x)/g(x)→L.

step 1.1L2
3.1

At infinity, put F(t)=f(1/t), G(t)=g(1/t). Then F′/G′=f′(1/t)/g′(1/t), since the common factor −1/t2 cancels. Apply steps 1.1 and 2.1 as t→0+ or 0−, and translate back.

L3step 2.1algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources