3 ms·
Well, it’s incorrect when dividing by 0, yes. But when you do the limit as the bottom factor approaches 0, it makes sense. So, below, the first one is defined,
by colejohnson66 6y ago
Well, it’s incorrect when dividing by 0, yes. But when you do the limit as the bottom factor approaches 0, it makes sense.
So, below, the first one is defined, while the second one isn’t.
lim_{x->0+} 1/x == +Infinity
1/0 != (anything)
- corty 6y agoYes, but thats only a small part of the story, your limit only works if the numerator is constant, positive and > 0. Actually, the correct answer for lim{x->0} a/x is really "anything" and "strongly depends on a and da/dx"... So in general, assuming anything like \infty for a division by zero is wrong wrong wrong.
- BeetleB 6y agoIEEE 754 handles all the cases you mentioned - it doesn't always prescribe Inf.
- colejohnson66 6y agoSorry if I wasn’t clear. I was not implying that any a/b where b approaches 0 is always +infinity. I was simply saying that a/0 isn’t defined, but that in the case of 1/x (positive 0), it is somewhat defined as positive infinity.
- nybble41 6y agoThat 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.