Elementary example for the indeterminate form $1^\infty$

Question

I am talking math with a bright middle schooler, who has not even seen logarithms in class yet. (We have successfully introduced logs as the inverse of exponentiation.) She is intrigued by this video and indeterminate forms. We have discussed how "$1^\infty$" is really the same as "$0/0$".

I would now like to discuss an example of "$1^\infty$". Unfortunately, every example I can come up with, and everything I find on the internet, uses that $\frac{\ln(1+t)}{t}\to 1$ as $t\to 0$, which is either "a well known fact", or an application of L'Hospital's rule - both of which I find unsatisfactory.

Is there a non-trivial example for the "$1^\infty$" indeterminate form (so, not just $1^t$ for $t\to\infty$) that can be analyzed using only the definition of the logarithm as the inverse function to exponentiation, without calculus or facts I would need to pull out of a hat?

Answer 1

Who can forget the classic example:

$\underset{n\to\infty}{\lim}\left(1+\dfrac{1}{n}\right)^{n}$?

If we expand $(1+\dfrac{1}{n})^{n}$ with the Binomial Theorem and compare terms with corresponding powers of $1/n$ for different values of $n$, we find that this function increases as $n$ increases without bound, but the function is bounded by the convergent series

$1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...$

So the limit is guaranteed to exis and is thus definable as $e$, from which the rule $[\ln(1+x)]/x\to1$ as $x\to 0$ follows.

Answer 2

We seek $f,\,g$ with $f\to1,\,g\to\infty$, say as $x\to0$, so that $f^g$ can have just any limit $L\in[0,\,\infty]$ or none. Examples:

• $f=e^{x^2},\,g=x^{-2}\ln L$ for $L>1$
• $f=e^{-x^2},\,g=-x^{-2}\ln L$ for $L\in(0,\,1)$
• $f=e^{x^4},\,g=x^{-2}$ for $L=1$
• $f=e^{-x^2},\,g=x^{-4}$ for $L=0$
• $f=e^{-x^4},\,g=x^{-2}$ for $L=\infty$
• $f=e^{x^2\sin(1/x)},\,g=x^{-2}$ for $\lim_{x\to0}f^g$ to be undefined.

The replacement $(f,\,g)\mapsto(1/f,\,-g)$ shows $1^{-\infty}$ works the same way, but nobody every lists that separately.

Answer 3

Why not just fix $k>0$ (e.g. $k=2$) and look at $(k^{1/n})^n$?

It is pretty clear intuitively that $k^{1/n}=\sqrt[n]{k}\to 1$ as $n\to\infty$; on the other hand, clearly $n\to\infty$ when $n\to\infty$. Thus, you have the case $1^\infty$ which actually converges to $k$ (and not just converges to $k$ but is constant), which you chose arbitrarily to start with.

Now this is easy to expand with $(k^{1/n})^{n^2}=k^n$ or $(k^{1/{n^2}})^n=k^{1/n}$, which converge to $0$ and $\infty$ (in some order, as long as $k\ne 1$).