How to prove that a form like $0^\infty$ is not an indeterminate form? What makes a form indeterminate?

Question

I just wanted to know why some particular forms (seven of them) are called indeterminate forms. Why are there only seven indeterminate forms? Can anyone please prove why $0^\infty$ is not an indeterminate form?

Answer 1

For these "indeterminate forms" it is important to note that numbers like $0$ and $1$ in these expressions are not necessarily identically $0$ or $1$ but are rather sequences of numbers "near" $0$ and $1$ who approach their respective values. As for $\infty$ it is in fact sequences of numbers who keep growing larger and larger without bound.

Expressions like "$1^\infty$" being indeterminate is referring to how things like $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n$ is in fact equal to $e$. Compare to $\lim\limits_{n\to\infty}(1-\frac{1}{n})^n$ which is $\frac{1}{e}$ and $\lim\limits_{n\to\infty}1^n$ which is identically $1$. These each gave different values despite all being of the form "a number close to $1$ raised to a large number." Just being told "$1^\infty$" without more context we don't know what the final result will be. It could be literally anything. That is to say if we know that $f(n)\to 1$ and $g(n)\to\infty$ as $n\to\infty$ we have no way of knowing what the value of $\lim\limits_{n\to\infty}f(n)^{g(n)}$ is equal to.

In the case of "$0^\infty$", a small number "near" $0$ raised to larger and larger exponents will always result in numbers near or equal to $0$. In every case, if you have $f(n)\to 0$ and $g(n)\to\infty$ as $n\to\infty$ you have $\lim\limits_{n\to\infty}f(n)^{g(n)}=0$. Tweaking the exact behavior of $f$ or $g$ does not change the overall result. In that way it is not "indeterminate"... it is very much determined. We know how $0^{\infty}$ acts regardless what form exactly "$0$" or "$\infty$" take.

As for a proof of why given that $f(n)\to 0,g(n)\to \infty$ as $n\to\infty$ implies that $\lim\limits_{n\to\infty}f(n)^{g(n)}=0$... note that there will necessarily be some $N$ such that for all $n>N$ you have $g(n)>1$ and $|f(n)|<1$. At that point you have $|f(n)^{g(n)}|\leq |f(n)|$ and the fact that $f(n)\to 0$ implies the rest.