That depends on the numerator: lim_{x->1+} (x^2-1)/(x-1) == 2 Of course ideally evaluating this particular formula for x == 0 would give 0/0 which IEEE F

That depends on the numerator: lim_{x->1+} (x^2-1)/(x-1) == 2 Of course ideally evaluating this particular formula for x == 0 would give 0/0 which IEEE FP defines as NaN, but you could also get a non-zero finite number divided by zero due to rounding for x ≅ 1 and then the default result of +Infinity would be very far from the correct answer. Any situation where you might come close to a 0/0 or ∞/∞ term requires very careful consideration since there is very little separation between a NaN result and ±Infinity.