TITLE: Indeterminate form $1^\infty$ vs. $0^\infty$
QUESTION [6 upvotes]: Why is $1^\infty$ an indeterminate form while $0^\infty = 0$? If $0\cdot0\cdot0\cdots = 0$ shouldn't $1\cdot1\cdot1\cdots = 1$?

REPLY [3 votes]: 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.)