AngouriMath
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ApplyFirstRemarkable(AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity,AngouriMath.Core.ApproachFrom)
Method (no overloads)
Remarks
Each part is asked from the side of the limit. The equivalence of a sine and its argument holds on one side
of the point as much as on both, so a one-sided limit needs its parts to vanish on its side alone; asked
from both, (e^((x - 1)^ln(x)) - e^(-(x - 1)^ln(x))) / 2 at 1 from the right spent eight seconds on the left
of its dividend, where (x - 1)^ln(x) is not real. The simpler part is asked first, since a part that does
not vanish settles the question and a simple one is the cheaper to ask: the divisor of that quotient, a
constant, but the dividend of (-1 - x) / (ln(2) / ln(x))^(x - 1) at 1 from the left, whose divisor takes
seconds and is asked again by the descent.
of the point as much as on both, so a one-sided limit needs its parts to vanish on its side alone; asked
from both, (e^((x - 1)^ln(x)) - e^(-(x - 1)^ln(x))) / 2 at 1 from the right spent eight seconds on the left
of its dividend, where (x - 1)^ln(x) is not real. The simpler part is asked first, since a part that does
not vanish settles the question and a simple one is the cheaper to ask: the divisor of that quotient, a
constant, but the dividend of (-1 - x) / (ln(2) / ln(x))^(x - 1) at 1 from the left, whose divisor takes
seconds and is asked again by the descent.
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