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A limit of an expression built on floor or sgn is answered by substitution at the jump #2266
Description
A limit of an expression built on floor or sgn is answered by substituting the destination, at the jump too, so a one-sided limit, and the two-sided one that does not exist, get the value at the point. A bare floor(x) or sgn(x) is declined, as it should be, but anything built on it is not. 2.6.0 and master:
| answered | is | |
|---|---|---|
limitleft(floor(x) - 1, x, 0) |
-1 |
-2 |
limit(floor(x) - 1, x, 0) |
-1 |
none: -2 from the left, -1 from the right |
limitleft(floor(x)^2, x, 0) |
0 |
1 |
limitleft(arcsin(floor(x)), x, 0) |
0 |
-pi/2 |
limitleft(-sgn(x), x, 0) |
0 |
1 |
limitright(-sgn(x), x, 0) |
0 |
-1 |
limit(floor(x) + x, x, 1) |
2 |
none: 1 from the left, 2 from the right |
The same substitution answers at a branch cut. limitright(ln(x^x), x, -1) is ln(-1), which is i pi, but x^x approaches -1 from below the cut there, so the limit is -i pi.
Found by a new workspace harness, work/limitcheck, which compares a limit with the values approaching it. Over 600 random expressions, 54 limits are of this kind.
Activity
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Rafael-SOWNet commentedon Oct 31, 2026
Done in three PRs, now on master (2c9dceb):
limitleft(floor(x) - 1, x, 0)-2limit(floor(x) - 1, x, 0)NaNlimitleft(floor(x)^2, x, 0)1limitleft(arcsin(floor(x)), x, 0)-pi/2limitleft(-sgn(x), x, 0)1limitright(-sgn(x), x, 0)-1limit(floor(x) + x, x, 1)NaNlimitright(ln(x^x), x, -1)One known cost: a limit that arrives on the cut from above is declined too, though its value was right. #2294 says why.
Leave a limit on the arctangent's cut unevaluated (#2334)