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I just wanted to know why some particular forms (seven of them) are called indeterminate forms. Why are there only seven indeterminate forms? Can anyone please prove why $0^\infty$ is not an indeterminate form?

Shyamal Jyoti Buragohain
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2 Answers2

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For these "indeterminate forms" it is important to note that numbers like $0$ and $1$ in these expressions are not necessarily identically $0$ or $1$ but are rather sequences of numbers "near" $0$ and $1$ who approach their respective values. As for $\infty$ it is in fact sequences of numbers who keep growing larger and larger without bound.

Expressions like "$1^\infty$" being indeterminate is referring to how things like $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n$ is in fact equal to $e$. Compare to $\lim\limits_{n\to\infty}(1-\frac{1}{n})^n$ which is $\frac{1}{e}$ and $\lim\limits_{n\to\infty}1^n$ which is identically $1$. These each gave different values despite all being of the form "a number close to $1$ raised to a large number." Just being told "$1^\infty$" without more context we don't know what the final result will be. It could be literally anything. That is to say if we know that $f(n)\to 1$ and $g(n)\to\infty$ as $n\to\infty$ we have no way of knowing what the value of $\lim\limits_{n\to\infty}f(n)^{g(n)}$ is equal to.

In the case of "$0^\infty$", a small number "near" $0$ raised to larger and larger exponents will always result in numbers near or equal to $0$. In every case, if you have $f(n)\to 0$ and $g(n)\to\infty$ as $n\to\infty$ you have $\lim\limits_{n\to\infty}f(n)^{g(n)}=0$. Tweaking the exact behavior of $f$ or $g$ does not change the overall result. In that way it is not "indeterminate"... it is very much determined. We know how $0^{\infty}$ acts regardless what form exactly "$0$" or "$\infty$" take.

As for a proof of why given that $f(n)\to 0,g(n)\to \infty$ as $n\to\infty$ implies that $\lim\limits_{n\to\infty}f(n)^{g(n)}=0$... note that there will necessarily be some $N$ such that for all $n>N$ you have $g(n)>1$ and $|f(n)|<1$. At that point you have $|f(n)^{g(n)}|\leq |f(n)|$ and the fact that $f(n)\to 0$ implies the rest.

JMoravitz
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  • Thanks for the answer So can I say indeterminate forms are the forms taken by sequence of numbers whose limit cannot be determined simply by knowing limits of its "constituent" sequences? But what about $\lim\limits_{n\to0}n^{1/n}$ which again is a $0^\infty$ but limit here DNE? – Shyamal Jyoti Buragohain Dec 07 '20 at 14:50
  • @ShyamalJyotiBuragohain that is not of the form $n\to\infty$. The only reason I can see you saying that limit doesn't exist is because of what happens if you approach from the left instead of the right... but then your expression isn't of the form $0^\infty$ but rather $0^{\pm\infty}$ or however you choose to notate it... from the left it was $0^{-\infty}$. The expression $+\infty$ is treated as different than $-\infty$ in this context. Approaching from the right, or rewording as $\lim\limits_{n\to\infty}\left(\frac{1}{n}\right)^n$ does in fact have a limit and approaches zero. – JMoravitz Dec 07 '20 at 15:34
  • As for understanding the phrase "indeterminate forms", yes that is exactly right. Each of the other indeterminate forms are equally ambiguous. If you haven't seen explicit examples of why, I encourage you to try to come up with some yourself. The trickiest for people is often $1^\infty$ which was already covered here with the example of limit definition of $e$. – JMoravitz Dec 07 '20 at 15:40
  • Thanks a lot – Shyamal Jyoti Buragohain Dec 07 '20 at 15:45
  • May I ask why $\vert f(n)^{g(n)}\vert < \vert f(n) \vert$ for all $n > N$? I don't really understand why that holds true, if $f(n)$ is negative. – Bman72 Nov 30 '23 at 16:45
  • @Bman72 this sure is an old post... rereading, I see a few typos I'll try to fix. In any event, if you want to avoid leaving the reals... it can be assumed that $f$ was nonnegative for every $n$ such that $g$ was not an integer... but that's no fun. If you were to allow complex numbers, you should be able to think of the pieces in polar form... $re^{i\theta}$, and the radius is going to behave like the ideal circumstance above as a strictly non-negative real being raised to an ever increasing power... thus it doesn't matter what angle it is pointing, $0$ radius pointed to any angle is $0$ – JMoravitz Nov 30 '23 at 17:15
  • Otherwise, if you did not want to allow for complex numbers and did not want to restrict $g$ like this, then the expression $f(n)^{g(n)}$ is not allowed in the first place as it has undefined values. – JMoravitz Nov 30 '23 at 17:17
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a op b is called an "indeterminate form" if you can't find the limit of f(x) op g(x) by knowing that the limit of f(x) is a and the limit of g(x) is b. And that is the case here: If you know that some limit of f(x) is 0, and the same limit of g(x) is infinity, that's not enough information to tell you what the limit of $f(x)^{g(x)}$ is, or whether it exists at all. You will find different functions f(x) and g(x) with different limits.

In your example $n^{1/n}$: When you try to find the limit you can go the wrong path and get to $0^{\inf}$ which means you took a wrong approach: Prove the limit some other way.

gnasher729
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