## Why $1^\infty != 1$

A debate that came up recently in some of my work was over the right way to define $1^\infty$. Many people say that since $1^n = 1$ for any integer $n$, then $1^\infty = 1$ also. That's a very reasonable position, but when we start using more powerful mathematics, it's not good enough any more.

One very good way to approach this is to define a power function $f(x,y) = x^{1/y}$ when $y > 0$ and $x >0$. We are interested in the value of $f(1,0) = 1^{1/0} = 1^{\infty}.$ However, we need a formal framework, as $\infty$ is not formally a number. Acknowledging Berkeley's criticism of the ambiguity of infinitessials, we employ Weierstrass's idea of limits. $f(1,0) = \lim_{x\rightarrow 1, y\rightarrow +0} x^{1/y}$ We can not easily calculate a 2-d limit, but if a unique limit does exist, then for any parameterization $x(t), y(t)$ where $\lim_{t\rightarrow 0} x(t) = 1$ and $\lim_{t\rightarrow 0} y(t) = +0$ then we must have $f(1,0) = \lim_{t\rightarrow 0} x(t)^{1/y(t)}$ Well, if $x(t) = \exp(t)$ and $y(t) = t/log(k)$ for some positive constant $k$. These satisfy our requirements on $x(t) \rightarrow 1$ and $y(t) \rightarrow +0$ as long as we restrict $t$ to it's positive values. Then \begin{align*} f(1,0) &= \lim_{t\rightarrow 0} x(t)^{1/y(t)} \\ &= \lim_{t\rightarrow 0} ( \exp(t) )^{log(k)/t} \\ &= \lim_{t\rightarrow 0} \exp(log(k)) \\ &= k \end{align*} But k is any arbitrary positive constant, so the limit IS NOT uniquely defined. So, a reasonable person can make a sound argument that $1^\infty = 2$ or any other number, even though, at first pass, this seems ridiculus. This is one of the weird, counter-intuitive problems with $\infty$ that pops up when you haven't carefully defined what you mean and somebody else is working with a slightly different definition.

Formalism can be very hand-tying, but sometimes it's the only way to get out of some tight squeezes.