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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.