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.

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    $\begingroup$ Because the 1 may be approached from below (or, if you are working with complex numbers, from all sides!) $\endgroup$ Commented Nov 15, 2010 at 22:44
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    $\begingroup$ @ 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 $\infty$. $\endgroup$
    – user17762
    Commented Nov 15, 2010 at 23:39
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    $\begingroup$ @Sivaram: no, that is not the definition of an indeterminate form. If you fix 1 then clearly the limit is 1. $\endgroup$ Commented Nov 16, 2010 at 0:42
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    $\begingroup$ 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!) $\endgroup$
    – Blue
    Commented Nov 16, 2010 at 1:26
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    $\begingroup$ 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..... $\endgroup$
    – N. S.
    Commented May 21, 2011 at 18:47

9 Answers 9


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 $\infty$.

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    $\begingroup$ +1. Limits are about the journey (in this case, the race), not the destination. $\endgroup$
    – Blue
    Commented Nov 15, 2010 at 23:53
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    $\begingroup$ Did you mean $\ln n$ instead of $n$ in your very last sentence? As it stands, it is a little misleading, since $(1-1/n)^n \to e^{-1} \neq 0$. $\endgroup$ Commented Nov 16, 2010 at 15:35
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    $\begingroup$ 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 $\infty$?... $\endgroup$ Commented May 31, 2013 at 2:31
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    $\begingroup$ @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. $\endgroup$ Commented Jun 7, 2013 at 22:19
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    $\begingroup$ 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. $\endgroup$
    – Erika
    Commented Jul 22, 2013 at 18:07

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 the 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 $$\lim_{x \to \infty} f(x)^{g(x)} = \lim_{x \to \infty} \exp \Big(g(x) \log f(x)\Big) = \exp \Big(\lim_{x \to \infty} g(x) \log f(x)\Big)$$ and since the limit of a product is the product of the limits, that's $$ \exp \left[ \left(\lim_{x \to \infty} g(x)\right) \cdot \left(\lim_{x \to \infty} \log f(x)\right)\right] $$ or

$$ \exp \left[ \left(\lim_{x \to \infty} g(x)\right) \cdot \left(\lim_{x \to \infty} h(x)\right)\right] $$

But the first limit of $g(x)$ is infinity, and the second of $h(x)$ is zero.

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

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

  • $\begingroup$ TeX work not proper $\endgroup$
    – anonymous
    Commented Nov 16, 2010 at 1:40
  • $\begingroup$ Great idea Michael +1 $\endgroup$
    – anonymous
    Commented Nov 16, 2010 at 1:43
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    $\begingroup$ Thanks, Chandru. (Now maybe one of these days I'll get to teach calculus again.) $\endgroup$ Commented Nov 16, 2010 at 1:44
  • $\begingroup$ Nice answer... Michael Lugo........ $\endgroup$
    – juantheron
    Commented Nov 22, 2013 at 18:00

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.

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    $\begingroup$ very nice answer! +1 $\endgroup$ Commented Jul 13, 2015 at 12:40
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    $\begingroup$ You are wrong about that. $\lim_{x \to \infty}1^{x} = 1$ and there is no issue with this. Also this is equivalent to $\lim_{x \to 0^{+}}1^{1/x} = 1$. Further $\lim_{x \to 0^{-}}1^{1/x} = 1$ so that $\lim_{x \to 0}1^{1/x} = 1$. You seem to have put an argument which makes no sense. $\endgroup$
    – Paramanand Singh
    Commented Jul 26, 2015 at 7:33
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    $\begingroup$ Indeed, this makes no sense. $\endgroup$ Commented Jul 26, 2015 at 8:56
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    $\begingroup$ It is best to use comment boxes for "considerations", reserving answer boxes for, well, answers. $\endgroup$ Commented Jul 26, 2015 at 9:34
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    $\begingroup$ in a formal sense 1 / ∞ is an incorrect form, and it's also true that you can't chuck that around in an equation and start cancelling with other infinities etc - but I bet many mathematicians would either write or think 1 / ∞ as a valid route to 0 $\endgroup$
    – Cato
    Commented Aug 5, 2016 at 10:11

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.


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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    $\begingroup$ 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 $(\rightarrow 1)^{\rightarrow \infty}$. $\endgroup$
    – Mark
    Commented Feb 19, 2017 at 14:22
  • $\begingroup$ @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. $\endgroup$
    – Eric
    Commented Feb 19, 2017 at 15:01
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    $\begingroup$ 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. $\endgroup$
    – Mark
    Commented Feb 19, 2017 at 19:54

"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 $1^{\infty}$.


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 $e^{-1}$.

  • $\begingroup$ 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. $\endgroup$
    – Sam
    Commented Nov 16, 2010 at 17:16
  • $\begingroup$ 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. $\endgroup$
    – Shai Covo
    Commented Nov 16, 2010 at 17:36
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    $\begingroup$ 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. $\endgroup$ Commented May 4, 2012 at 0:28

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:

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

  • $\begingroup$ This argument (in the first part) worked best for me. $\endgroup$
    – trincot
    Commented Oct 30, 2022 at 19:34

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

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.


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