Is $1^{1/0}= 1?$

Question

If: $1^0 = 1$
then: $1 = \sqrt[0] 1$
but that implies that: $1 = 1^{1/0}$
But since 1/0 doesn't exists, how is this possible?

Answer 1

Your assumption that $a^0=1$ means that $a=\sqrt[0]1$ is nonsense.

Since $a^0=1$ is true for all $a,$ you’d have, by the same reasoning, that $1=\sqrt[0]1=2$ by comparing $a=1$ and $a=2.$

We just don’t define $\sqrt[n]{b}$ for $n=0,$ because it has no value for $b\neq 1$ and all values when $b=1.$ It is undefined because there is no useful definition.

On a related but different note, if $f,g>0$ are functions, and $f(x)\to 1$ and $g(x)\to 0^,$ then we can’t say anything about the convergence of $f(x)^{1/g(x)}.$ This can be stated as “$1^\infty$ is an indeterminate form.” Without knowing more about $f$ and $g$, $f^{1/g}$ could converge to anything or not converge at all.