Why is $\infty^0$ an indeterminate form?

Question

I can't seem to find a situation where $\lim_{x\to a}f(x)=\infty$ and $\lim_{x\to a}g(x)=0$, but $\lim_{x\to a} f(x)^{g(x)}\neq 1$. This makes me think that, even though this form ($\infty^0$) is often stated as an indeterminate form, it is not actually indeterminate (since the limit is always 1).

Moreover, I cant find a question asking this on stackexchange, and the calculus book I use neglects to provide an explanation for why it is considered indeterminate.

I have tried to prove that the limit is always 1, but it has been a few years since my real analysis days, and I am getting stuck in the proof. I'll continue to look for a proof, but in the mean time, I figured I would ask the question, and see if anyone can provide an example I've missed.

Answer 1

You aren't trying hard enough. \begin{align}&\lim_{x\to\infty} x^{1/\ln x}=\lim_{x\to\infty} e=e\\ &\lim_{x\to\infty} x^{(\ln x)^{-1/2}}=\infty\end{align}

Answer 2

Another example $$\lim_{x \rightarrow \infty}(e^x)^{1/x}=e$$ Before finding the limit $f(x)=(e^x)^{1/x}$ gives $f(\infty)= \infty^0$ which is only a `warning sign' that our calculational intuition may fail us when we deal with very small and very big numbers. But evaluation of $f(x)$ for large values of $x$ e.g. 10,100,1000, ... $f(x)$ may go down to a good number which would be 2.71.... But normally we do not do numerics, so we should follow and appropriate method for this warning sign, the method is $\lim_{x\rightarrow a} f(x)^{g(x)}=\exp[\lim_{x \rightarrow a} g(x) \log f(x)]$