SignFromTheSlope​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​AngouriMath.​Entity,​AngouriMath.​Core.​ApproachFrom)

Where the slope tends to a non-zero L, infinities included, it has the sign of L near
a, and x - a is positive on the right and negative on the left, so g has the
sign of L on the right and the other sign on the left. This needs no derivative, and
answers where SignFromVanishingOrder(AngouriMath.Entity,AngouriMath.Entity.Variable,AngouriMath.Entity,AngouriMath.Core.ApproachFrom) cannot: sqrt(x) at 0, whose
derivative diverges there, has the slope 1 / sqrt(x), which tends to +oo, and
so floor(sqrt(x)) at 0 from the right is 0, where it was not answered.
A sign is a question about a real value, and the slope of a complex one can still tend to
a real infinity: x ^ tanh(pi) on the left of 0 is |x| ^ tanh(pi) times
e ^ (i pi tanh(pi)), and was read as negative. So the expression has to be real
throughout the side approached: real wherever it is defined, or real where the bases of its
roots are positive and those hold there, which sqrt(x)'s does on the right of 0 and
not on the left. Its real domain condition does not say this: that of x ^ tanh(pi) is true. Each base is a part of the expression, so asking about it does not come back here
with the same question.