> Wolfram says that log(0) = -infinity. > > > Otherwise, limit(1/exp(log(-x), x, > > 0, "+") = +infinity ! > > Check this link: > http://www53.wolframalpha.com/input/?i=+limit+(1%2Fexp(log(-x))+as+x-%3E0%2B
I don't think Mathematica computes limits by substituting x with 0 in the expression, so it's not that relevant. My problem with using log(0) = -infinity is the following: In [19]: (1/f(log(-x))).subs(x,0).subs(f, exp) Out[19]: ∞ In [20]: (1/f(log(-x))).subs(f, exp).subs(x,0) Out[20]: -∞ Different results for the same expression depending on the order of the substitutions. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sympy?hl=en -~----------~----~----~----~------~----~------~--~---
