# why 1/ infinity isn't indeterminate like other indeterminate?

$$1/\infty$$ tends to 0.

$$\mathbf {It\ doesn't \ satisfy\ the\ inverse \ process\ of\ multiplication \ and \\division\ i.e}$$

$$\infty * 0$$ is undefined or indeterminate.

So why $$1/\infty$$ is not indeterminate like other indeterminate $$0/0$$ , $$\infty/\infty,..$$

Thanks.

• Determinate or indeterminate forms doesn’t correspond to whether the form obeys traditional algebraic rules, just whether the limit is unambiguously defined. – eepperly16 Apr 25 at 5:14
• I’m not sure where but this question has multiple duplicates – gen-z ready to perish Apr 25 at 5:42

If $$a_n \to 0$$ and $$b_n \to 0$$ then $$\frac {a_n} {b_n}$$ may converge to any number or may not even converge. Same thing is true if $$a_n \to \infty$$ and $$b_n \to \infty$$. But if $$a_n \to 1$$ and $$b_n \to \infty$$ then $$\frac {a_n} {b_n} \to 0$$
• As an example of the latter, consider $a_n=n^2$ and $b_n=n$ vs. $a_n=n$ and $b_ n=n^2$ – J. W. Tanner Apr 25 at 5:09
1 over something very large ($$\infty$$) tends to 0. That much is clear. However, what does some very large number divided by another very large number even mean? Hence $$\infty/\infty$$ cannot be determined. Same with $$0/0$$ - it doesn't mean anything, hence indeterminate. You cannot determine what such a number is. Now, $$\infty*0$$ also is meaningless, since anything times infinity is infinity and anything times 0 is 0, so infinity times zero cannot be determined.
• It is not $0$. It tends to 0. – Paras Khosla Apr 25 at 7:52
• when we say 'anything times infinity is infinity', since 0 is different from rest numbers. So we can't say yet zero times infinity is infinity because that is the question. But with 'anything times zero is zero', here anything, include 0 or any numbers. Hence , $\infty$ * 0= 0, since no conflict as in the former. – Rajesh Marndi Apr 25 at 12:57
• "Indeterminate forms" refer to something that occurs while you are trying to take a limit. I think it is impossible to say anything useful about indeterminate forms unless it is understood how limits are involved. Contrary to what you wrote, the quotient of a very large number divided by a very large number is perfectly well defined. Moreover, $\lim_{x\to0^+}f(x)/g(x)=3$ when $f(x)=1+3/x$ and $g(x)=2+1/x$ even though $\lim_{x\to0^+}f(x)=\infty$ and $\lim_{x\to0^+}g(x)=\infty$. But if you write $f(0)/g(0)$, $f(0)$ is not defined, so you don't even get $\infty/\infty.$ – David K Nov 14 at 13:57