Why is $1^\infty$ an indeterminate form while $0^\infty = 0$? If $0\cdot0\cdot0\cdots = 0$ shouldn't $1\cdot1\cdot1\cdots = 1$?

  • $\begingroup$ Check out these two links: math.stackexchange.com/questions/520795/… math.stackexchange.com/questions/10490/… $\endgroup$ – Brenton Sep 16 '17 at 23:35
  • $\begingroup$ $0^{+\infty}=0$, but $|0^{-\infty}|=+\infty$. Here on Wikipedia it is mentioned that $0^{\infty}$ is not an indeterminate form because $0^{+\infty}$ is $0$ and $0^{-\infty}$ is $1/0$, which is not $0$, but still we know that $(0^+)^{-\infty}=+\infty$ and $(0^{-})^{-\infty}=-\infty$, or $|0^{-\infty}|=+\infty$. $\endgroup$ – user236182 Sep 16 '17 at 23:36
  • 3
    $\begingroup$ A very small number raised to a large power is even smaller. $\endgroup$ – Simply Beautiful Art Sep 16 '17 at 23:36
  • 1
    $\begingroup$ @stevengregory $$\left(\frac1n\right)^n=\frac1{n^n}\stackrel{n\to\infty}\longrightarrow\frac1\infty=0$$ $\endgroup$ – Simply Beautiful Art Sep 16 '17 at 23:51
  • 1
    $\begingroup$ @stevengregory Yes... and why are you mentioning this? $\endgroup$ – Simply Beautiful Art Sep 16 '17 at 23:55

To say that $1^\infty$ is an indeterminate form means that there is more than one object that can be $\lim\limits_{x\,\to\,\text{something}} f(x)^{g(x)}$ where $f(x)\to1$ and $g(x)\to\infty,$ so that the limit depends on which functions $f$ and $g$ are.

Thus $$ \left. \begin{align} & \lim_{x\to\infty} \left(1+\frac 1 x\right) = 1 \quad\text{and} \quad \lim_{x\to\infty} \left( 1 + \frac 1 x \right)^x = e \\[10pt] & \qquad \text{and} \\[10pt] & \lim_{x\to\infty} \left( 1 - \frac 1 x\right) = 1 \quad \text{and} \quad \lim_{x\to\infty} \left( 1 - \frac 1 x\right)^x = \frac 1 e. \end{align} \right\} \longleftarrow \text{two different numbers} $$

  • 1
    $\begingroup$ Sir, I'm not sure if you can count, but that's three different numbers ;-) $\endgroup$ – Simply Beautiful Art Sep 17 '17 at 1:21
  • 1
    $\begingroup$ @SimplyBeautifulArt : That depends on which numbers are referred to. However, there are of course three kinds of people in the world: those who can count and those who can't. $\endgroup$ – Michael Hardy Oct 5 '17 at 0:37
  • $\begingroup$ LMAO $\vphantom{....}$ $\endgroup$ – Simply Beautiful Art Oct 5 '17 at 1:01

I want to address two questions here:

1. What does we mean when we say $1^\infty$ is indeterminate?

First of all, we should understand what $1^\infty$ means. It is not the product $ 1 \cdot 1 \cdot 1 \cdots$. Instead, what it represents is that if you have two limits $\lim_{n \to \infty} x_n = 1$ and $\lim_{n \to \infty} y_n \to \infty$ then we cannot determine the value of $\lim_{n \to \infty} (x_n)^{y_n}$.

What does it mean not to be able to determine the value of a limit? It means that the value of the limit depends on the sequences we choose. That is, we could have two pairs of sequences $(x_n, y_n)$ and $(x_n', y_n')$ where $x_n, x_n' \to 1$ and $y_n, y_n' \to \infty$ but

$$ \lim_{n \to \infty} (x_n)^{y_n} \ne \lim_{n \to \infty} (x_n')^{y_n'}. $$

This is different than a determined form. For instance if $x_n \to 1$ and $y_n \to 2$ then we will always have

$$ \lim_{n \to \infty} x_n + y_n = 3, \qquad \lim_{n \to \infty} x_n y_n = 2 \quad\text{ and }\quad \lim_{n \to \infty} (x_n)^{y_n} = 1 $$

regardless of what the sequences $x_n$ and $y_n$ are.

2. What makes $1^\infty$ different than $0^\infty$?

For the form $0^\infty$ we are trying to find the limit of $(x_n)^{y_n}$ where $x_n \to 0$ and $y_n \to \infty$. By definition, $x_n \to 0$ means that for every $\varepsilon > 0$, $|x_n| < \varepsilon$ eventually (for all $n \ge$ some $N = N(\varepsilon)$). In particular, we can take $\varepsilon = 1/2$. Then if $n$ is large enough,

$$ 0 \le \left| (x_n)^{y_n} \right| \le \left| (1/2)^{y_n} \right| \to 0 $$

as $n \to \infty$. It follows that $\lim_{n \to \infty} (x_n)^{y_n} = 0$.

What you'll notice is that $1/2$ could have been any number between $0$ and $1$. That is, if $x_n$ is "close to $0$" in the sense that $|x_n| < r$ and $r < 1$ then we can conclude that $\lim_{n \to \infty} (x_n)^{y_n} = 0$.

Note that we cannot conclude this for $1^\infty$. That is, when we try to approximate $x_n$ by $1 +$ some error then it matters what the error is. For $0^\infty$ as long as the error is $< 1$ then we can conclude that the limit is $0$. But for $1^\infty$ if the error is

  • positive, then $(1 + \text{error})^{y_n} \to \infty$
  • zero, then $(1 + \text{error})^{y_n} \to 1$
  • negative, then $(1 + \text{error})^{y_n} \to 0$

This gives us no information about the value of $\lim_{n \to \infty} (x_n)^{y_n}$ and indeed, we can find sequences $x_n$ and $y_n$ where $(x_n)^{y_n}$ tends to $\infty$ or $1$ or $0$. In fact, we can make $(x_n)^{y_n}$ converges to any given real number $\ge 0$ or infinity.

Appendix: a sketch of a construction of sequences $x_n$ and $y_n$ such that $x_n \to 1$, $y_n \to \infty$ and $(x_n)^{y_n} \to r$ where $r \in (0, \infty)$. (Try finding sequences for $r = 0, \infty$ as an exercise.)

Let $p_n/q_n$ be a sequence of rational numbers converging to $\log(r)$ where the denominators, $q_n \to \infty$. For instance if $\log(r) = 2$ we could take the sequence $2/1, 20/10, 200/100, 2000/1000$ and so on.

Consider the limit

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

which we can take logarithms and use a Taylor series for $\log(1 + x)$to get

\begin{align} \log\left( \lim_{n \to \infty} \left( 1 + \frac{1}{q_n} \right)^{p_n} \right) &= \lim_{n \to \infty} p_n \log\left( 1 + \frac{1}{q_n} \right) \\ &= \lim_{n \to \infty} p_n \left( \frac{1}{q_n} - \frac{1}{2q_n^2} + \frac1{3q_n^3} - \frac{1}{4q_n^4} + \cdots \right) \\ &= \lim_{n \to \infty} \frac{p_n}{q_n}\left( 1 - \frac{1}{2q_n} + \frac1{3q_n^2} - \frac{1}{4q_n^3} + \cdots \right) \\ &= \lim_{n \to \infty} \frac{p_n}{q_n} \left( \lim_{n \to \infty} 1 - \lim_{n \to \infty} \frac{1}{2q_n} + \lim_{n \to \infty}\frac1{3q_n^2} - \cdots \right) \\ &= \log(r)(1 - 0 + 0 - \cdots) = r. \end{align}

Therefore $\left( 1 + \frac{1}{q_n} \right)^{p_n} \to r$. (I am sweeping some details about swapping limits with summations under the rug.)

  • $\begingroup$ Very nice answer. +1 the idea is represented via sequences and can be easily transformed into the limit of functions of a real variable. So the use of sequences makes no essential difference. $\endgroup$ – Paramanand Singh Sep 17 '17 at 3:31
  • 1
    $\begingroup$ I like this answer because it makes a point at the very beginning that calling something an "indeterminate form" has a very particular meaning. The phrasing of the question suggests that OP has forgotten (or has never received) this meaning, so this seems to go at the heart of the matter. $\endgroup$ – David K Sep 17 '17 at 9:31

It is far easier to think about these indeterminate forms by taking logarithms. $$ \ln\left( f(x)^{g(x)} \right) = g(x) \ln f(x) $$ If $f \rightarrow 1$ and $g \rightarrow \infty$, you have an $\infty \cdot 0$ indeterminate form.

If $f \rightarrow 0$ and $g \rightarrow \infty$ you have an $\infty \cdot - \infty$ form, which is not indeterminate at all. In fact $$ \exp \ln(f(x)^{g(x)} ) \xrightarrow{f(x) \rightarrow 0, g(x) \rightarrow \infty} \exp(- \infty) \rightarrow 0 \text{.} $$

Without logarithms...

Here are three different $1^\infty$s:

  • $\lim_{n \rightarrow \infty} (2^{1/\ln n})^n = \infty$
  • $\lim_{n \rightarrow \infty} (2^{1/n})^n = 2$
  • $\lim_{n \rightarrow \infty} (2^{1/n^2})^n = 1$

Only the last one is doing what you seem to expect. This is because $1/n^2$ is going to $0$ faster than $n$ can overpower. When that doesn't happen we can arrange for other limits.

Now any indeterminate form $0^\infty$ is $f(x)^{g(x)}$ with $\lim_{x \rightarrow L} f(x) = 0$ and $\lim_{x \rightarrow L} g(x) = \infty$, where $L$ is either a real number or, if it is either of $\pm \infty$, adjust the following as appropriate. Since $\lim_{x \rightarrow L} f(x) = 0$, there is a neighborhood of $L$, $(L-d, L+d)$, on which $|f(x)| < 1/2$. So let's look at what happens to $(1/2)^{g(x)}$ as $x$ gets close to $L$. For $\lim_{x \rightarrow L} g(x) = \infty$, there is an $e_1$ such that $g(x) > 1$ on $(L-e_1, L+e_1)$. Similarly, $g(x) > 2$ on some $(L-e_2, L+e_2)$, and so on for some sequence of nested open sets collapsing toward $L$. Eventually, there is some $N$ such that $(L-e_N, L+e_N) \subseteq (L-d, L+d)$, so talk about choices of $n > N$. On $(L-e_n, L+e_n)$, $(1/2)^{g(x)} < (1/2)^n = 2^{-n}$. That is, $(1/2)^{g(x)}$ shrinks to $0$ as $x$ approaches $L$. If, instead of $1/2$, we use the actual, smaller magnitude values of $f$, $f(x)^{g(x)}$ approaches $0$ also. There's no indeterminacy here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.