Question #d30b9

1 Answer
Aug 24, 2016

Answer:

Most versions on an epsilon-delta definition require #delta > 0#.

Explanation:

Perhaps the definition you're working with is something like:

#lim_(xrarra^-)f(x) = -oo# if and only if, for every #B#, there is a #delta > 0# such that for every #x#, if #a- delta < x < a#, then #f(x) < B#.

This allows for the possibility that #B# is a positive number, in which case #-1/B# would be negative..

Without knowing exactly what definition you're working with, I can only suggest that that's the problem.