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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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L'Hôpital's rule for the /\infty/\infty form at finite or infinite, one-sided endpoints

Statement

Let f,gf,g be differentiable on a one-sided neighbourhood of cc, or on a tail at ++\infty or -\infty, with g0g'\ne0. Suppose f(x)|f(x)|\to\infty and g(x)|g(x)|\to\infty in the selected mode, with each numerator and denominator eventually of fixed sign. If f(x)/g(x)LRf'(x)/g'(x)\to L\in\overline{\mathbb R}, then f(x)/g(x)Lf(x)/g(x)\to L.

Facts & Assumptions

Proof

technique · direct
1.1

Fix a base point aa inside the domain. For variable xx farther toward the limiting end, [L1] gives f(x)f(a)g(x)g(a)=f(ξx)g(ξx)\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(\xi_x)}{g'(\xi_x)}, where ξx\xi_x lies between aa and xx.

givenL1
2.1

First choose aa sufficiently far toward the end that the derivative quotient is as close to LL as required throughout the remaining tail. Then the quotient of increments has the same bound for every later xx.

step 1.1L2choose
3.1

Since g(x)|g(x)|\to\infty, g(a)/g(x)0g(a)/g(x)\to0; since the increment quotient is bounded in the finite-LL case, the identity f(x)g(x)=f(x)f(a)g(x)g(a)(1g(a)g(x))+f(a)g(x)\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}\left(1-\frac{g(a)}{g(x)}\right)+\frac{f(a)}{g(x)} gives the finite conclusion. For L=±L=\pm\infty, choose the derivative-quotient lower or upper bound first and then make the two fixed-base terms negligible, obtaining the defining arbitrary bound.

step 2.1L2algebra
4.1

Thus the quotient has limit LL in every stated mode.

step 3.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 63 results over 20 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