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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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Because the 1 may be approached from below (or, if you are working with complex numbers, from all sides!) –  Mariano Suárez-Alvarez 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 $\infty$. –  user17762 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 0:42
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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 1:26
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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

4 Answers 4

up vote 87 down vote accepted

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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+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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@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 7 '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

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 $e^{-1}$.

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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 4 '12 at 0:28

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.)

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TeX work not proper –  anonymous Nov 16 '10 at 1:40
    
Great idea Michael +1 –  anonymous Nov 16 '10 at 1:43
    
Thanks, Chandru. (Now maybe one of these days I'll get to teach calculus again.) –  Michael Lugo Nov 16 '10 at 1:44
    
Nice answer... Michael Lugo........ –  juantheron Nov 22 '13 at 18:00

Adding up to the existing answers look:

$$ 1^\infty = 1^{\frac{1}{0}} = 0\sqrt{1}$$ so in limits $1^\infty$ isn't necessarily $1\times 1 \times1 ....$

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protected by Marvis Jun 28 '12 at 1:27

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