Why would $1^{-\infty}$ not be 1?

Question

It would seem to me that $1^{-∞}=\lim_\limits{x→∞}1^{-x}=\lim_\limits{x→∞}\frac1{1^x}=\frac11=1$ no matter how we approach it. However, Wolfram Alpha answers with a mysteriously unqualified “$\text{(undefined)}$”. Similarly, JavaScript also thinks that the result isn't a number. On the other hand, the very mathematically inclined APL languages NARS2000 and J both have it give $1$.

What reasons are there to reject $1^{-∞}=1$?

Answer 1

Since for any $x$ we have $1^x=1$ as you noticed

$$\lim_{x\to -\infty} 1^x =\lim_{x\to -\infty} 1=1$$

and you are completely right on that, but as $f(x)\to 1$

$$\lim_{x\to -\infty} \left(f(x)\right)^x $$

is an indeterminate form, that is we can obtain any result depending on the nature of $f(x)$.

Probably by this input wolfram refer symbolically to this latter case.

Answer 2

The value $1^\infty$ is indeterminate. It can be interpreted as $$\lim_{x\to \infty} 1^x=1,$$ as $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n =e,$$ among others. This ambiguity makes it indeterminate. BPRP has some good videos delving deeper into this concept.

Answer 3

$$\begin{array}{} \left( 1+\frac{1}{n} \right)^{-n} \to e^{-1} \\ \left( 1+\frac{1}{n^2} \right)^{-n} \to 1 \\ \left( 1+\frac{1}{n} \right)^{-n^2} \to 0 \end{array}$$