In floating point terms infinity tends to indicate overflow (any number that is too big to represent) and the 0's indicate underflows. So in more verbose terms,
1/(positive overflow) = (positive underflow)
While
-1/(positive overflow) = (negative underflow)
In this case, since the positive overflow isn't really infinity and the underflows aren't really 0, they are not equal. In practice, -0 and the 0 can both also arise from situations that produce 0 exactly, too, but this is not that case.
You may be thinking about how lim{x->inf}(1/x) = 0 = lim{x->inf}(-1/x), which is true. Infinity in floating point does not necessarily represent that limit, though, just any number that is too big.
You may also notice that the limit is not in the range of the functions inside the limit. For all real x, 1/x != 0