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I've been taught that $1^\infty$ is undetermined case. Why is it so? Isn't $1*1*1...=1$ whatever times you would multiply it? So if you take a limit, say $\lim_{n\to\infty} 1^n$, doesn't it converge to 1? So why would the limit not exist?

Mihai Bujanca
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    See http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form – Meow Mar 03 '13 at 20:19
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    The notation is a bad one, $1^\infty$ stand for a limit like $$ \lim_{x\rightarrow x_0} f(x)^{g(x)}$$ where $\lim_{x\to x_0} f(x)=1$ and $\lim_{x\to x_0} g(x)=\infty$ – Dominic Michaelis Mar 03 '13 at 20:23
  • Oy. I, too, missed the duplicate. It is on the second page when you search for the word “indeterminate”. Oh, well. Voted to close. – Harald Hanche-Olsen Mar 03 '13 at 20:24
  • @HaraldHanche-Olsen Sorry, missed it, and already had answers when i reached the post. Otherwise I would have deleted it. Thanks anyway – Mihai Bujanca Mar 03 '13 at 20:31
  • What is amusing though is that at first it seems counter intuitive to many(including myself) that $$1^{\infty}=1×1×1×........\infty$$ is indeterminate but $$0×\infty=0+0+0+0+.......\infty$$ is relatively easier to accept as an indeterminate form. – Aurelius Feb 16 '24 at 17:16

4 Answers4

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It isn’t: $\lim_{n\to\infty}1^n=1$, exactly as you suggest. However, if $f$ and $g$ are functions such that $\lim_{n\to\infty}f(n)=1$ and $\lim_{n\to\infty}g(n)=\infty$, it is not necessarily true that

$$\lim_{n\to\infty}f(n)^{g(n)}=1\;.\tag{1}$$

For example, $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\approx2.718281828459045\;.$$

More generally,

$$\lim_{n\to\infty}\left(1+\frac1n\right)^{an}=e^a\;,$$

and as $a$ ranges over all real numbers, $e^a$ ranges over all positive real numbers. Finally,

$$\lim_{n\to\infty}\left(1+\frac1n\right)^{n^2}=\infty\;,$$

and

$$\lim_{n\to\infty}\left(1+\frac1n\right)^{\sqrt n}=0\;,$$

so a limit of the form $(1)$ always has to be evaluated on its own merits; the limits of $f$ and $g$ don’t by themselves determine its value.

Brian M. Scott
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The limit of $1^{\infty}$ exist:$$\lim_{n\to\infty}1^n$$ is not indeterminate. However$$\lim_{a\to 1^+,n\to\infty}a^n$$ is indeterminate..

KingLogic
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There are many reasons. For example, let $1^\infty=1$. Taking logarithm, you have $\infty\cdot 0=0$. Similarly for other operations you will obtain some absurd.

Boris Novikov
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The short answer is that it is because $x^y$ can tend to any nonnegative limit as $x\to1$ and $y\to\infty$. For one example, consider the classical limit $$\lim_{n\to\infty}\Bigl(1+\frac xn\Bigr)^n=e^x.$$

Paul Sinclair
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Harald Hanche-Olsen
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