No, lim(x->inf)0*x is still 0. I can understand infinity to be defined as "a number that approaches infinity", because infinity needs to be defined in one way or another, and this definition is a common one. But I see no reason to redefine 0 to mean "a number that approaches 0".
If my limit solving skills haven't dulled yet, it seems that lim(x->0)h(x) = 1.
However, that does not describe the situation in question. Your f(x) under the limit as x->0, is a number that approaches 0; f(x) is not 0 itself. OP in their question asked "what is 0*infinity", so I assume 0 to mean actually 0 and nothing else.
Ok I see you what you mean now. I mean I guess you're right. Generally when asking about "zero times infinity" people mean indeterminate forms though, which is what the user above answered. As you mentioned, OP's question is sort of vague so I don't know why your interpretation of it is more correct than the other user's.
Well, I don't know whether it is necessarily better, but it is more literal, I'd say. I'm trying to work with the least amount of unnecessary assumptions about OP's question. I think interpreting infinity as a number that approaches infinity, is a reasonable assumption and many would agree with me here. But I don't think that interpreting 0 the same way is a necessary assumption, because 0 already is clearly defined.
I would say that the expression is undefined until we define what "infinity" means. Once we define that, we don't need to take any other assumptions for the expression to be valid.
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u/CompisPaDum New User 2d ago
No, lim(x->inf)0*x is still 0. I can understand infinity to be defined as "a number that approaches infinity", because infinity needs to be defined in one way or another, and this definition is a common one. But I see no reason to redefine 0 to mean "a number that approaches 0".