Why is $\infty^0$ an indeterminate form? 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.
You aren't trying hard enough. \begin{align}&amp;\lim_{x\to\infty} x^{1/\ln x}=\lim_{x\to\infty} e=e\\ &amp;\lim_{x\to\infty} x^{(\ln x)^{-1/2}}=\infty\end{align}
$\infty^0$ is directly related to $\infty\cdot0$ by taking the logarithm.