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Take a vanishing factor times a sum term by term in a one-sided limit - #2387
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From one side, a product whose factors tend to 0 and to an infinity, one of them a sum or a difference, is now asked as the sum of the products term by term, its whole limit once. Read as a product, l'Hopital's rule took x * (ln(-tanh(x)) + 1) at 0 from the left through nested limits for 4.5 s, where x ln(-tanh(x)) + x is 0 at once. The same gates as the quotient regrouping beside it: one-sided, and not while the sides of a two-sided limit are compared. x * (ln(-tanh(x)) + 1) at 0-: 4.5 s -> 0.44 s. e^(1/x) * (3 ln(-tanh(x)) + 1) at 0-: 7.2 s -> 0.46 s. (tanh(x)^3)^exp(1/x) at 0-: 7.5 s -> 0.47 s, the row the logarithm-of-a-power change (#2385) slowed. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
Codecov Report✅ All modified and coverable lines are covered by tests. Additional details and impacted files@@ Coverage Diff @@
## master #2387 +/- ##
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Coverage 86.80% 86.80%
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Files 312 312
Lines 69190 69204 +14
Branches 20177 20179 +2
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+ Hits 60060 60073 +13
- Misses 4916 4917 +1
Partials 4214 4214
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A factor tending to 0 times a sum tending to an infinity is asked term by term, from one side. Read as a product, l'Hôpital's rule took
x (ln(-tanh(x)) + 1)at 0 from the left through four and a half seconds of nested limits, wherex ln(-tanh(x)) + xis 0 at once.(tanh(x)^3)^exp(1/x)comes to the same product. #2385 slowed that row from 0.5 s to 7.5 s and accepted the slowdown, and this fixes it. The rewrite is exact, and it has the same gates as the quotient regroup in #2384: one side only, and not while the two sides of a limit are being compared. A term-by-term answer ofNaNis not taken. The product is read as before.x (ln(-tanh(x)) + 1)0, 3.2 s0, 4.5 s0, 0.44 se^(1/x) (3 ln(-tanh(x)) + 1)0, 1.0 s0, 7.2 s0, 0.46 s(tanh(x)^3)^exp(1/x)1, 0.5 s1, 7.5 s1, 0.47 ssin(x) (ln(-tanh(x)) + 1/x)1, 3.7 s1, 5.2 s1, 1.1 slimitcheck A/B against master (
DEPTH=3, the usual seeds): no answer lost. Five rows that timed out and five that were left unevaluated now finish. Two answers are new:log(pi, x)^sinh(x) sin(1)at 0, from each side, issin(1). The exponent tends to 0 faster thanln|log(pi, x)|grows, so the power tends to 1. Master leaves it unevaluated after 4 s, and 2.6.0 does not finish in 30 s.sin(e^x (x + 1/2))at -oo is 0. Master and 2.6.0 leave it unevaluated.log(2 pi, log(arctan(x), tanh(x)))at 0 moved from a timeout to a decline in 3–4 s.(x^2/sinh(x))^xat 0 from the left is still1, now in 5 s, where master took 9.5 s.Each changed answer is in BREAKING-CHANGES.md, measured on 2.6.0. The full suite (19,118), the wrapper suites, AOT and the ten harnesses pass.
Found by limitcheck, in the A/B of #2385. No issue was filed.
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