$0^\infty$ and indeterminate forms

Question

I have actually two related questions:
1. Should indeterminate forms be able to attain an infinite number of values to be considered indeterminate?
I’m asking because Wikipedia says:

The expression 1/0 is not commonly regarded as an indeterminate form, because there is not an infinite range of values that f/g could approach.

2. I know that $0^\infty$ is not indeterminate.
But i want to know why this is wrong:
$\lim_{(x,y)\to(0,\infty)}x^y=e^{y\cdot \ln(x)}$ and if we take the path $y=\frac{a}{\ln(x)}$ it becomes $e^a$ and could attain infinite number of values and therefore is indeterminate.

Answer 1

The issue is that as $x\to 0$ then $y:=a/\ln(x)\to 0$, so what you have written is a justification of $0^0$ being indeterminate.

Whatever path we take, $0^\infty$ will give $0$: this is because if $y\geq 1$ and $0<x<1$ then $0<x^y\leq x$.

Answer 2

Short answer:$$0^\infty=(\exp-\infty)^\infty=\exp(-\infty\times\infty)=\exp(-\infty)=0.$$Long answer: suppose $f(x) \to0^+,\,g(x)\to\infty$ as $x\to c$. Then$$\ln f\to-\infty\implies g\ln f\to-\infty\implies f^g\to0.$$