5 ms·

As long as lim(1/x)_x->0 = inf, 1/0 = 0 doesn't make a whole lot of sense, mathematically speaking. I might be wrong but I don't think it was addressed in the a

by rsp1984 2y ago

As long as lim(1/x)_x->0 = inf, 1/0 = 0 doesn't make a whole lot of sense, mathematically speaking. I might be wrong but I don't think it was addressed in the article either.

lupire 2y ago
It's fine. Infinity isn't a real number, so 1/x isn't continuous at 0, so it doesn't matter what the value of 1/0 is. All your open sets still behave the way you expect. Whether you choose "this function is undefined here" vs "it's impossible to ever reach the value of this function at this value, under any assumptions I'll ever care about" is purely a matter of convenience.
- rsp1984 2y ago
  Yea, but conceptually it's a bit smelly to have f(x) get larger and larger the closer x is to 0 and then suddenly have it be 0 once x reaches 0.
  - renewiltord 2y ago
    But once you go behind, it flips suddenly anyway so you could just as well have it be intuitively “halfway between the positive and negative infinities” which is at least fun and could spawn a few “Why is 1/x suddenly go to zero” articles on HN in 2053
  - _carbyau_ 2y ago
    As others have pointed out "larger and larger" is the same when it is negative too. So I think people are just going: positive infinity + negative infinity = 0. Intuitively nice in a sense but I honestly think '0' is misrepresenting what is going on here. I'm ok with it being ' "+ and/or -" infinity' as a new definition. Programmatically I think it should result in a NULL or VOID or similar. I mean, by definition it has no definition.
    - rsp1984 2y ago
      Good point.
- Turneyboy 2y ago
  Well if you consider 1/z as a function of a complex coordinate it definitely makes a lot of sense to set it to infty. That identifies +infty and -infty if you restrict yourself to the real numbers.
CharlieDigital 2y ago
There's a great Radiolab episode[0] that talks about divide by zero in perhaps more conceptual terms. KARIM ANI: If you take 10 and divide it by 10, you get one. 10 divided by five is two. 10 divided by half is 20. The smaller the number on the bottom, the number that you're dividing by, the larger the result. And so by that reasoning ... LULU: If you divide by zero, the smallest nothingness number we can conceive of, then your answer ... KARIM ANI: Would be infinity. LULU: Why isn't it infinity? Infinity feels like a great answer. KARIM ANI: Because infinity in mathematics isn't actually a number, it's a direction. It's a direction that we can move towards, but it isn't a destination that we can get to. And the reason is because if you allow for infinity then you get really weird results. For instance, infinity plus zero is ... LATIF: Infinity. KARIM ANI: Infinity plus two is infinity. Infinity plus three is infinity. And what that would suggest is zero is equal to one, is equal to two, is equal to three, is equal to four ... STEVE STROGATZ: And that would break math as we know it. Because then, as your friend says, all numbers would become the same number. [0] https://radiolab.org/podcast/zeroworld https://radiolab.org/podcast/zeroworld
- v4vvdq 2y ago
  Then take 10 and divide it by -10 = -1. 10 / -5 = -2. 10 / -0.5 = -20. So from the other side of the y-axis it behaves the exact opposite. It goes to minus infinity. So at x=0 we would have infinity and minus infinity at the same time. Imho that is why it is undefined.
  - WithinReason 2y ago
    on computers you can have negative zeros
    - recursive 2y ago
      Negative zero is equal to zero, so it's not really a distinct number, just another representation of the same value.
      - 2y ago
v4vvdq 2y ago
I was also looking for this. And would like to add: lim(-1/x)_x -> 0 = -inf That is (in my opinion) the whole point why it is actually undefined. On one side of the y-axis it goes to infinity, on the other to minus infinity. I don't see a solution to this and therefore always have accepted that it is undefined.
- lupire 2y ago
  No. 1/x^2 is undefined at 0 but has the same limit behavior, because limit behavior is not a function from "pairs of (functions from R to R, R)" to R Infinity is not a real number.
  - ganzuul 2y ago
    Can limit behavior be defined as Cauchy sequences?
deleted 2y ago
[deleted]
judofyr 2y ago
The arguments around limits are addressed towards the end (under "Update 8/12/2018"): > > If 0/0 = 0 then lim_(x -> 0) sin(x) / x = sin(0) / 0 = 0, but by L’Hospitals’ Rule lim_(x -> 0) sin(x) / x = lim_(x -> 0) cos(x) / 1 = 1. So we have 0 = 1. > This was a really clever one. The issue is that the counterargument assumes that if the limit exists and f(0) is defined, then lim_(x -> 0) f(x) = f(0). This isn’t always true: take a continuous function and add a point discontinuity. The limit of sin(x) / x is not sin(0) / 0, because sin(x) / x is discontinuous at 0. For the unextended division it’s because sin(0) / 0 is undefined, while for our extended division it’s a point discontinuity. Funnily enough if we instead picked x/0 = 1 then sin(x) / x would be continuous everywhere. Similar examples can be constructed for any regular function which is discontinuous (e.g. Heaviside step function).

As long as lim(1/x)_x->0 = inf, 1/0 = 0 doesn't make a whole lot of sense, mathematically speaking. I might be wrong but I don't think it was addressed in the article either.

lupire 2y ago

It's fine. Infinity isn't a real number, so 1/x isn't continuous at 0, so it doesn't matter what the value of 1/0 is. All your open sets still behave the way you expect. Whether you choose "this function is undefined here" vs "it's impossible to ever reach the value of this function at this value, under any assumptions I'll ever care about" is purely a matter of convenience.

rsp1984 2y ago
Yea, but conceptually it's a bit smelly to have f(x) get larger and larger the closer x is to 0 and then suddenly have it be 0 once x reaches 0.
- renewiltord 2y ago
  But once you go behind, it flips suddenly anyway so you could just as well have it be intuitively “halfway between the positive and negative infinities” which is at least fun and could spawn a few “Why is 1/x suddenly go to zero” articles on HN in 2053
- _carbyau_ 2y ago
  As others have pointed out "larger and larger" is the same when it is negative too. So I think people are just going: positive infinity + negative infinity = 0. Intuitively nice in a sense but I honestly think '0' is misrepresenting what is going on here. I'm ok with it being ' "+ and/or -" infinity' as a new definition. Programmatically I think it should result in a NULL or VOID or similar. I mean, by definition it has no definition.
  - rsp1984 2y ago
    Good point.
Turneyboy 2y ago
Well if you consider 1/z as a function of a complex coordinate it definitely makes a lot of sense to set it to infty. That identifies +infty and -infty if you restrict yourself to the real numbers.

CharlieDigital 2y ago

There's a great Radiolab episode[0] that talks about divide by zero in perhaps more conceptual terms. KARIM ANI: If you take 10 and divide it by 10, you get one. 10 divided by five is two. 10 divided by half is 20. The smaller the number on the bottom, the number that you're dividing by, the larger the result. And so by that reasoning ... LULU: If you divide by zero, the smallest nothingness number we can conceive of, then your answer ... KARIM ANI: Would be infinity. LULU: Why isn't it infinity? Infinity feels like a great answer. KARIM ANI: Because infinity in mathematics isn't actually a number, it's a direction. It's a direction that we can move towards, but it isn't a destination that we can get to. And the reason is because if you allow for infinity then you get really weird results. For instance, infinity plus zero is ... LATIF: Infinity. KARIM ANI: Infinity plus two is infinity. Infinity plus three is infinity. And what that would suggest is zero is equal to one, is equal to two, is equal to three, is equal to four ... STEVE STROGATZ: And that would break math as we know it. Because then, as your friend says, all numbers would become the same number. [0] https://radiolab.org/podcast/zeroworld https://radiolab.org/podcast/zeroworld

v4vvdq 2y ago
Then take 10 and divide it by -10 = -1. 10 / -5 = -2. 10 / -0.5 = -20. So from the other side of the y-axis it behaves the exact opposite. It goes to minus infinity. So at x=0 we would have infinity and minus infinity at the same time. Imho that is why it is undefined.
- WithinReason 2y ago
  on computers you can have negative zeros
  - recursive 2y ago
    Negative zero is equal to zero, so it's not really a distinct number, just another representation of the same value.
    - 2y ago

v4vvdq 2y ago

I was also looking for this. And would like to add: lim(-1/x)_x -> 0 = -inf That is (in my opinion) the whole point why it is actually undefined. On one side of the y-axis it goes to infinity, on the other to minus infinity. I don't see a solution to this and therefore always have accepted that it is undefined.

lupire 2y ago
No. 1/x^2 is undefined at 0 but has the same limit behavior, because limit behavior is not a function from "pairs of (functions from R to R, R)" to R Infinity is not a real number.
- ganzuul 2y ago
  Can limit behavior be defined as Cauchy sequences?

deleted 2y ago

[deleted]

judofyr 2y ago

The arguments around limits are addressed towards the end (under "Update 8/12/2018"): > > If 0/0 = 0 then lim_(x -> 0) sin(x) / x = sin(0) / 0 = 0, but by L’Hospitals’ Rule lim_(x -> 0) sin(x) / x = lim_(x -> 0) cos(x) / 1 = 1. So we have 0 = 1. > This was a really clever one. The issue is that the counterargument assumes that if the limit exists and f(0) is defined, then lim_(x -> 0) f(x) = f(0). This isn’t always true: take a continuous function and add a point discontinuity. The limit of sin(x) / x is not sin(0) / 0, because sin(x) / x is discontinuous at 0. For the unextended division it’s because sin(0) / 0 is undefined, while for our extended division it’s a point discontinuity. Funnily enough if we instead picked x/0 = 1 then sin(x) / x would be continuous everywhere. Similar examples can be constructed for any regular function which is discontinuous (e.g. Heaviside step function).