I stumbled across a list of indeterminate forms on Wolfram Alpha. On top of the common ones, plus a few ones like $(-1)^\infty$ that usually aren't mentioned in textbooks but I can make sense of, one captured my attention.
$$\sqrt[0]{x}\hskip1.0cm\forall x \in \mathbb{R}$$
I'm trying to figure out why this would be an indeterminate form.
Let $f(x)$ be a function and $g(x) = \frac{1}{x}$.
Then, we have
$$\lim_{x\to0^+}f(x)^{g(x)} = \sqrt[0]{f(x)}$$
Such a limit evaluates to $\infty$ for every $f(x)$ greater than $1$, to $1$ for every $f(x)$ constantly equal to $1$, to $0$ for every $f(x)$ smaller than $1$, and is an indeterminate form if $f(x)$ tends to $1$ or $0$.
In other words, I would say that $\sqrt[0]{x} = x^\infty$, the latter not being an indeterminate form for every $x$ but just for some $x$'s.
So is it really correct to say that $\sqrt[0]{x}$ is an indeterminate form $\forall x \in \mathbb{R}$?
Best Answer
$$\lim_{t\to0}\sqrt[t^2]2=\infty,$$
$$\lim_{t\to0}\sqrt[-t^2]2=0,$$
$$\lim_{t\to0}\sqrt[t]2\ \ \not\exists.$$