I know there are 7 indeterminate forms as follows- $$0^0$$ $$1^{\infty}$$ $${\infty}^0$$ $$\frac{0}{0}$$ $$\frac{\infty}{\infty}$$ $$0\cdot\infty$$ $${\infty}-{\infty}$$
I cant help but wonder if these are also indeterminate- $$(-1)^{\infty}$$ $$1^{-\infty}$$ $$({-\infty})^0$$
If these are not indeterminate forms can someone give an explanation regarding this dilemma ?
On
As for $1^{-\infty}$, you simply treat it as $$\frac{1}{1^{\infty}}$$ and proceed as usual.
Treating $(-1)^{\infty}$ and $(- \infty)^0$ is delicate, though. Look at definitions of exponential: have you ever defined $a^b$ for negative $a$? The answer is: yes, but only when $b$ is an integer. In particular, if $b$ is a quantity apporaching to $0$ (but $b \neq 0$), the symbol $$a^b$$ is not defined. Let's make an example: $$\left( 1+ x \right)^{1/x}$$ is not defined for $x < -1$, so it does not make sense to consider its limit as $x \to - \infty$. In general, the following forms actually do not make sense: $$(- \infty)^0 , (-1)^0 , (-2)^0, (-53)^{\pi}, (-3)^{1/4}$$ and so on. I'm not saying that these are indeterminate forms, but simply that they do not make sense: you will never find them, as you will never find someone asking you "what is the volume of $4$".
Some of them are easily seen to reduce to the old ones
$$\begin{cases}1^{-\infty} = (1^{-1})^\infty=1^\infty \\ (-\infty)^0=(-1)^0(\infty^0) = \infty^0 \end{cases}$$
Based on your comments on the original post I chose the interpretation you said in the comments. $(-1)^\infty$ does not exist since if you're taking a limit on continuous things you cannot pass through non-integer values, and even if you could the way you show $1^\infty$ is an indeterminate form is because you do $\log$ to it, and you cannot do $\infty\cdot\log(-1)$ since log is not defined on negatives.