Why is $1^{\infty}$ considered to be an indeterminate form

Solution 1:

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 $\infty$.

Solution 2:

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.)