From Wikipedia: In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.

  • The indeterminate forms include $0^{0},\frac{0}{0},(\infty - \infty),1^{\infty}, \ \text{etc}\cdots$

My question is can anyone give me a nice explanation of why $1^{\infty}$ is considered to be an indeterminate form? Because, i don't see any justification of this fact. I am still perplexed.

Elias Zamaria
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    Because the 1 may be approached from below (or, if you are working with complex numbers, from all sides!) – Mariano Suárez-Álvarez Nov 15 '10 at 22:44
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    @ Mariano: We are not approaching 1. $1$ is fixed, there is no limiting process to reach to one. We are letting only the power i.e. $x$ approach $\infty$. –  Nov 15 '10 at 23:39
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    @Sivaram: no, that is not the definition of an indeterminate form. If you fix 1 then clearly the limit is 1. – Qiaochu Yuan Nov 16 '10 at 00:42
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    Consider the *purpose* of the list of indeterminate forms. (Barring pathologies...) The first thing to try in an $x\to a$ limit is to "plug in" $a$ for $x$; if you get an expression that evaluates to $3$ or $\sqrt{\pi}$ or even $-\infty$, you're done. The "indeterminate forms" are labels (and/or warnings) for cases where there's more work to do. They capture the essence of the problem and guide you to appropriate follow-up strategies ... usually, "massage your limit into $\frac{0}{0}$ form". (See how I used "$\frac{0}{0}$" to *describe* a type of limit right there? *That's* the whole point!) – Blue Nov 16 '10 at 01:26
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    Ok. As Day Late Don pointed out my interpretation of the indeterminate form was wrong. Thanks for pointing it out. I have hence deleted my answer since it was wrong and could be misleading. –  Nov 16 '10 at 02:09
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    The reason why $1^\infty$ is indeterminate, is because what it really means intuitively is an approximation of the type $(\sim 1)^{\rm large \, number}$. And while $1$ to a large power is 1, a number very close to 1 to a large power can be anything..... – N. S. May 21 '11 at 18:47
  • 42nd upvote for this question. My life is made. – Ahaan S. Rungta Nov 22 '13 at 17:16
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    Because the Wikipedia entry is wrong. Seriously. –  Aug 12 '14 at 05:16

9 Answers9


Forms are indeterminate because, depending on the specific expressions involved, they can evaluate to different quantities. For example, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers.

$$\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n = 1$$

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$

$$\lim_{n \to \infty} \left(1 + \frac{1}{\ln n}\right)^n = \infty$$

To expand on this some (and this thought process can be applied to other indeterminate forms, too), one way to think about it is that there's a race going on between the expression that's trying to go to 1 and the expression that's trying to go to $\infty$. If the expression that's going to 1 is in some sense faster, then the limit will evaluate to 1. If the expression that's going to $\infty$ is in some sense faster, then the limit will evaluate to $\infty$. If the two expressions are headed toward their respective values at essentially the same rate, then the two effects sort of cancel each other out and you get something strictly between 1 and $\infty$.

There are some other cases, too, like $$\lim_{n \to \infty} \left(1 - \frac{1}{\ln n}\right)^n = 0,$$ but this still has the expression going to $\infty$ "winning." Since $1 - \frac{1}{\ln n}$ is less than 1 (once $n > 1$), the exponentiation forces the limit to 0 rather than $\infty$.

Mike Spivey
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    +1. Limits are about the journey (in this case, the race), not the destination. – Blue Nov 15 '10 at 23:53
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    Did you mean $\ln n$ instead of $n$ in your very last sentence? As it stands, it is a little misleading, since $(1-1/n)^n \to e^{-1} \neq 0$. – Hans Lundmark Nov 16 '10 at 15:35
  • Yes, I did. Thank you, Hans; I will correct that. – Mike Spivey Nov 16 '10 at 16:25
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    Forgive me for adding to this discussion so late; though I was looking for some decent explanation of the same claim and don't find this as satisfactory as I know it should be. Indeed, the limits you have defined in your answer evaluate to different quantities; however, just looking at $\lim_{n \to \infty} 1^\infty$, what reason do we have to go out of our way to look for such alternate expressions? Surely, we must take the same approach in evaluating limits such as $\lim_{n \to \infty} 2^\infty$? Though, we find it acceptable to just as easily conclude the latter evaluates to $\infty$?... – ThisIsNotAnId May 31 '13 at 02:31
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    In other words, what makes 1 special? And if someone was looking at this for the first time, what reason would they have to take this approach and not simply conclude the former limit is after all just 1? – ThisIsNotAnId May 31 '13 at 02:32
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    @ThisIsNotAnId: I'm not sure that the comment box is an adequate place to address your question, but I'll give it a shot. The number 1 is special in the $1^{\infty}$ form because it is a boundary case. Take any fixed number slightly smaller than 1 and raise it to an arbitrarily large power, and you're heading toward 0. Fix 1, and take it to an arbitrarily large power, and you still have 1. Take any fixed number slightly larger than 1 to an arbitrarily large power, and you're heading toward $\infty$. The number 2 as the base does not represent such a boundary case. – Mike Spivey Jun 07 '13 at 21:31
  • @ThisIsNotAnId: Now, suppose you don't fix the base on your expression and you allow it to vary instead. What happens to the value of $\lim_{n \to \infty} f(n)^{g(n)}$, where $f(n) \to 1$ and $g(n) \to \infty$? Does it go to $\infty$, or to 0, or to something, like 1, in between 0 and $\infty$? Because 1 is a boundary case for the base of the exponential expression the answer turns out to depend on the *relative rates* at which $f(n)$ goes to 1 and $g(n)$ goes to $\infty$. My original post expands on that idea. – Mike Spivey Jun 07 '13 at 21:32
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    @MikeSpivey Thank you for expanding; but my main concern still remains. When dealing with $1$, why should we *have* to consider something like $\lim_{n \to \infty} f(n)^{g(n)}$, where $f(n) \to 1$ and $g(n) \to \infty$? In the expression, $1^\infty$, $1$ is already fixed, and there is nothing to suggest, or require, that it should vary. – ThisIsNotAnId Jun 07 '13 at 22:19
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    I think the thing is that if 1 is fixed, then it's no longer an indeterminate form. If you look at the other indeterminate forms, they do not contain "0" or "$\infty$", but objects that tend towards 0 or $\infty$. Likewise in the indeterminate form $1^\infty$, 1 is understood as an object tending towards 1. See comments on the original question. – Erika Jul 22 '13 at 18:07
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    @MikeSpivey: Do you know where I can find the proof for the last expression? – user541686 Nov 19 '13 at 22:20
  • @Mehrdad: You should ask that as a question on the main site. – Mike Spivey Nov 19 '13 at 23:18
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    I think it's preferable the term undefined than undeterminated, because limits aren't equations with multiple solutions. Limits are expressions that must be evaluated as a whole (not part by part), at least if you haven't theorems or axioms that allow you to evaluate them part by part. – castarco Mar 19 '15 at 09:58
  • @Blue, how will you implement your "philosophical" approach to L'Hopital's rule?. – Aditya Agarwal Aug 02 '15 at 13:49
  • @AdityaAgarwal: It's a *slogan*, not an *approach*. :) I'm just observing that the limit (or lack thereof) of $f(x)$ as $x$ approaches $a$ has nothing to do with the value (or lack thereof) of $f(a)$. With regard to indeterminant forms, the up-shot is that "plugging in $a$" is absolutely no help at all: $0/0$, $1^\infty$, $0^0$, etc, are nonsense algebraic expressions. Any associated limiting value requires (as described nicely in Mike's answer above) comparing the *relative* journeys of two components as they race against each other. L'Hopital's Rule does exactly this kind of comparing. – Blue Aug 02 '15 at 14:33

Look at the logarithm.

More specifically, consider $f(x)^{g(x)}$ as $x \to \infty$, where $\lim_{x \to \infty} g(x) = \infty$ and $\lim_{x \to \infty} f(x) = 1$. (This is something of form $1^\infty$.)

Now say $f(x) = e^{h(x)}$, so $h(x) = \log f(x)$. Then $\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \log f(x) = \log \lim_{x \to \infty} f(x) = \log 1 = 0$.

Then $$\lim_{x \to \infty} f(x)^{g(x)} = \lim_{x \to \infty} \exp (g(x) \log f(x)) = \exp \lim_{x \to \infty} (g(x) \log f(x)) $$ and since the limit of a product is the product of the limits, that's $$ \exp [ (\lim_{x \to \infty} g(x)) \cdot (\lim_{x \to \infty} \log f(x))] $$ or $$ \exp [ (\lim_{x \to \infty} g(x)) \cdot (\lim_{x \to \infty} h(x)) ]. $$ But the first limit is infinity, and the second is zero.

So the indeterminacy of $1^\infty$ follows directly from the indeterminancy of $\infty \cdot 0$.

(The indeterminacy of $\infty^0$ actually follows in the same way, by taking the factors in the other order.)

Michael Lugo
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This is just one more consideration $1^\infty$ can be roughly rewritten as:

$1^{\frac 10}=\sqrt[0]{1}$

Now just think to the zeroth root of 1: every number raised to 0 is one so the zeroth root of 1 could be every number! This is why $1^\infty$ is an indeterminate form.

Renato Faraone
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If you're confused about this part let me try to clarify:


$$\lim_{x\to\infty} 1^x=1$$

$$\lim_{x\to\infty} (1-\frac{1}{x} )^x=???$$

Only the last one is indeterminate. We can't be sure if the expression in parentheses goes to 1 "faster" than the exponent takes the entire expression to infinity. The indeterminate forms are often abbreviated with stuff like "$1^\infty$" but that's not what they mean. This "$1^\infty$" (in regards to indeterminate forms) actually means: when there is an expression that approaches 1 and then it is raised to the power of an expression that approaches infinity we can't determine what happens in that form. Hence, indeterminate form.

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Here is some intuitive explanation, suitable also for non-mathematicians. Suppose that an imaginary basketball player has a probability $p = 0.999$ of making a free throw. The probability that he makes $10000$ free throws in a row is very small, the probability that he makes $100$ is high, and the probability that he makes $1000$ is approximately $e^{-1}$.

Shai Covo
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  • I think this misses the point somewhat, because his probability definitely will go to 0 (even monotonically), so it does not portray an indeterminate form. A better example would perhaps be a basketball player that gets better and better with the number of throws. – Sam Nov 16 '10 at 17:16
  • The purpose of my answer was to give a simple intuitive explanation for why something close to $1$ raised to a power of large $N$ may result in an arbitrary value. – Shai Covo Nov 16 '10 at 17:36
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    I think this is a nice answer. It makes the point that it's not how close to $1$ you are in an absolute sense that matters, but rather how this compares to the size of the exponent. – Pete L. Clark May 04 '12 at 00:28

"Indeterminate forms" are a vague concept and it is better to keep them "vague" rather than define them properly.

Limit evaluations are done on the basis of certain limit theorems which include the "algebra of limits" in particular. Theorems dealing with "algebra of limits" suffice to calculate limit of expressions which are composed of sub-expressions combined with $+, -, \times, /$ and the hope is that each sub-expression has a limit (perhaps calculated by expressing is as a combination of sub-sub-expressions) and then we use the algebra of limits to calculate limit of the expression by combining limits of sub-expressions via operations of $+, -, \times, /$.

These rules of "algebra of limits" however have two main limitations:

1) Limits of sub-expressions must exist (meaning they are finite, sorry I had to be explicit here to use the word "finite" as some textbooks treat limit $\infty$ also as "existing").

2) Rule dealing with division says that the limit of sub-expression in denominator should not be $0$.

"Indeterminate forms" were conceived to enumerate the cases where "algebra of limits" fails because of the above two limitations and for each of these cases certain other tactics / methodologies were developed. A classic case is expression of type $f(x)/g(x)$ where both $f(x), g(x)$ tend to $0$. Now to classify such cases the indeterminate form $0/0$ was invented. Similarly to deal with expressions of type $f(x)g(x)$ where $f(x) \to \infty$ and $g(x) \to 0$ the form $\infty\times 0$ was used. Also in each case where an indeterminate form was invented all the following options were possible: 1) limit exists, limit is $\pm \infty$ or there is oscillation. So classifying certain cases into "forms" did not guarantee the eventual limit, it only allowed us to use tactics and tools suitable to that form. Hence the word "indeterminate" was also added (we could not determine the limit by the form).

Cases like $f(x)g(x)$ where $f(x) \to \infty$ and $g(x) \to 1$ can't be handled by "algebra of limits" but these are not classified into indeterminate forms because there are theorems in this case which say that the resulting limit is $\infty$ so that the form is no longer "indeterminate" and we could perhaps classify such cases into "determinate forms" if we wanted.

Coming to the form $1^{\infty}$ it is obvious that it is designed to handle expressions of type $\{f(x)\}^{g(x)}$ where $f(x) \to 1$ and $g(x) \to \infty$. In such case we can write the expression as $\exp\{g(x)\cdot\log f(x)\}$. By properties of $\log$ function if $f(x) \to 1$ then $\log f(x) \to 0$ and hence $\{g(x)\cdot\log f(x)\}$ is already an indeterminate form of the type $\infty\times 0$. Therefore $\exp\{g(x)\log f(x)\} = \{f(x)\}^{g(x)}$ also has to be considered as an indeterminate form and it is usually written in the notation $1^{\infty}$.

Paramanand Singh
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In fact, a better notation of this type of indeterminate form should be $(\rightarrow 1)^\infty$, where the right arrow means the number $1$ is the limit of the base function, not all of the value of the base function is literally $1$ (i.e. not the case like $\displaystyle\lim_{n\to\infty}1^{n}$).

This is the same when we write other indeterminate form, for example like $\frac{\to 0}{\to 0}$. If the nominator function is TRULY $0$, such like $\displaystyle\lim_{n\to\infty}\frac{0}{\frac{1}{n}}$, then it is not the indeterminate form $\frac{\to 0}{\to 0}$ that the calculus books are talking; it is definite form, $\displaystyle\lim_{n\to\infty}\frac{0}{\frac{1}{n}}=0$. To repeat, it seems to be, but NOT the indeterminate form, and we had better denoted it as $\frac{0}{\to 0}$.

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    I strongly agree with this. The current notation for indeterminate forms, such as $\frac00$ is ambiguous; your notation $\frac{\rightarrow 0}{\rightarrow 0}$ is much better. The question "why is $1^{\infty}$ an indeterminate form" is caused by our ambiguous notation, it would be asked much less frequently if we adopted the much clearer notation $(\rightarrow 1)^{\rightarrow \infty}$. – Mark Feb 19 '17 at 14:22
  • @Mark Yes. So many people say, "the reason why we don't define $0^0$ to be a exact number is, from calculus we know that $0^0$ is indeterminate form, so it shouldn't be assigned a particular value." I don't consider it reasonable. There may be other reasons that we don't want to define $0^0$, but not due to the so-called indeterminate issue. Well, since, it is not a process that a variable in either base or the power approaches zero. – Eric Feb 19 '17 at 15:01
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    Indeed. There are no arguments against defining $0^0=1$ that follow the rules of mathematics. Confusion is what keeps that debate going. Better notation would definitely help. As long as for one group of people $0$ simply means $0$, while for others it means "something that converges to $0$" (yes: that uses circular logic, but arguments against defining $0^0$ do not follow the rules of math), then it is hard to see how the debate would end. – Mark Feb 19 '17 at 19:54

$$ \lim_{n\to\infty} \left( 1 + \frac a n\right)^n = e^a. $$

This limit depends on $a$. In other words, if the base approaches $1$ and the exponent approaches $\infty$, that's not enough to tell you what the limit is.

Michael Hardy
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By continuity of the logarithm,

$$\log(\lim a^b)=\lim\log(a^b)=\lim(\log(a)\cdot b)$$

so that an indeterminate form $1^\infty$ is equivalent to a form $0\cdot\infty$.

A concrete example:

$$\log\left(1.1^{10}\right)={0.9531017980\cdots}=\log(2.5937424601\cdots)$$ $$\log\left(1.01^{100}\right)={0.9950330853\cdots}=\log(2.7048138294\cdots)$$ $$\log\left(1.001^{1000}\right)={0.9995003330\cdots}=\log(2.7169239322\cdots)$$ $$\log\left(1.0001^{10000}\right)={0.9999500033\cdots}=\log(2.7181459268\cdots)$$ $$\cdots$$ $$\log\left(1^\infty\right)=1=\log(2.7182818284\cdots)$$