> If x slightly exceeds 1, then log x is a small positive value, i.e. a denormal, and therefore log() will incorrectly report failure. No it won't, for the nei

> If x slightly exceeds 1, then log x is a small positive value, i.e. a denormal, and therefore log() will incorrectly report failure. No it won't, for the neighboring values of 1, which are 1-2^-53 and 1+2^-52, log will return approximately -2^-53 and 2^-52, which are not denormal.