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

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  • $\begingroup$ 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
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    $\begingroup$ A very small number raised to a large power is even smaller. $\endgroup$ – Simply Beautiful Art Sep 16 '17 at 23:36
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    $\begingroup$ @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
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    $\begingroup$ @stevengregory Yes... and why are you mentioning this? $\endgroup$ – Simply Beautiful Art Sep 16 '17 at 23:55
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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 $$ \left. \begin{align} & \lim_{x\to\infty} \left(1+\frac 1 x\right) = 1 \quad\text{and} \quad \lim_{x\to\infty} \left( 1 + \frac 1 x \right)^x = e \\[10pt] & \qquad \text{and} \\[10pt] & \lim_{x\to\infty} \left( 1 - \frac 1 x\right) = 1 \quad \text{and} \quad \lim_{x\to\infty} \left( 1 - \frac 1 x\right)^x = \frac 1 e. \end{align} \right\} \longleftarrow \text{two different numbers} $$

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    $\begingroup$ 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
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    $\begingroup$ @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
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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.)

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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