[thaumasiotes]

You think it's possible for a bucket to contain a negative amount of water?

I should note that the product of zero and infinity being an indeterminate form is actually a result about the product of an infinitesimal (of small but indefinite magnitude) and an infinite value. If when you say "zero", you actually mean "zero", there is no ambiguity: zero is more infinitesimal than any infinite value is infinite, and the product of zero with anything, including an infinitely large value, is zero.

[Retric]

> You think it's possible for a bucket to contain a negative amount of water?

Irrelevant, the discontinuity occurs at 0 not a negative number.

The limit of f(X) = (X-2)/(X-2) as X approaches 2 is 1, that doesn’t mean the function has a defined value at 2. Limits seem easy because most students really don’t understand limits and thus misuse them.

[rbanffy]

Quantum physics tells us all particles are waves, so it’s possible that the amount of water will be negative in some point in time, as long as the average value is not negative. ;-)

[itishappy]

Zero times any number is zero, but infinity is not a number. In order for multiplication to be valid, your elements must share a field.

[thaumasiotes]

Are you trying to make a point? The indeterminate forms are statements about limits. Those limits are statements about the possible range of certain operations on infinitesimal and infinite values. It's perfectly valid to multiply those values. And when zero is one of the multiplicands, it's also the product.

You might want to think about why two times infinity is not an indeterminate form.

[itishappy]

I am, thanks for noticing.

If you extend your field to include infinity (e.g. the extended reals, or the extended positive reals), only then is it valid to multiply by infinity. One of the rules in such a system is that when infinity is one of the multiplicands, it's also the product. This gives us conflicting results for zero times infinity, therefore 0*∞ is an indeterminate form.

[thaumasiotes]

You do not appear to have the slightest idea what you're talking about. For example, the extended reals aren't a field.

But it's easy to extend the real numbers to a field that includes infinite and infinitesimal values. Limiting is then a projection from the hyperreals, which are a field, to the extended reals, which aren't. Instead of "lim", let's call this projection "f".

0·∞ is an indeterminate form because f(f⁻¹(0) · f⁻¹(∞)) is not well defined. f is a many-to-one function, and in this case the different possibilities that come up as we invert it interact differently. In contrast, 2·∞ is not an indeterminate form because, while f⁻¹(2) · f⁻¹(∞) might be any value that is greater than all real numbers, it must always be greater than all real numbers, and therefore f(f⁻¹(2) · f⁻¹(∞)) is always ∞.

> One of the rules in such a system is that when infinity is one of the multiplicands, it's also the product.

In the extended reals, this is a result, not a rule, and it doesn't always hold. Again, the extended reals aren't a field. But even ignoring the question we're actively discussing, you should have been able to think of e.g. -3 · ∞.

That's assuming that when you said "such a system", you meant the extended reals. If you meant a field that extends the reals to include infinite values, it's just meaningless noise - there is no value called "infinity" that would even let you evaluate the claim true or false. But in any multiplication of two values, any infinite value can only simultaneously be the product and one of the multiplicands if the other multiplicand is 1. When we say that 2·∞ = ∞, the ∞ on the left and the one on the right are both infinitely large, but they aren't the same value.

[itishappy]

> You do not appear to have the slightest idea what you're talking about. For example, the extended reals aren't a field.

Fair enough. You're right, of course, and you've learned me a thing. Appreciate your time.