I am reading complex analysis and searching the indeterminate form in the extended complex plane $\mathbb{C}_{\infty}.$ It is clear that $\infty+\infty$ is indeterminate form as $n+(-n)\rightarrow0$ and $n+(-2n)\rightarrow\infty.$ Now what about other Indeterminate form ? Is the following $$ \frac{0}{0},\frac{\infty}{\infty},0.\infty,\infty^{0},1^{\infty},0^{0}$$ are also Indeterminate form in $\mathbb{C}_{\infty}$? I am confused as in one of my book it is written that in $\mathbb{C}_{\infty}$ we define $a^{0}=1 $ for all $a\in \mathbb{C}_{\infty}. $ So can i say that $\infty^{0}$ and $0^{0}$ are not Indeterminate form in $\mathbb{C}_{\infty}?$ Please give me right way. Thanks in advance.
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When we say that $\infty ^0$ or $0^0 $ are indeterminate, we mean this in terms of limits. For example, $\lim_{x\rightarrow 0} x^0=1$ while $\lim_{x\rightarrow 0} 0^x=0$. These are both of the form $0^0$ when we plug in the value at the limit. Similarly for all of the other forms, they are still indeterminate in complex analysis.
This is slightly different if you are doing arithmetic with a point named infinity, but indeterminate refers to the taking of limits.
Konrad Wrobel
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ok its means all indeterminate forms are still valid? – neelkanth Oct 11 '15 at 16:29
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Yeah, most of the basic examples for why these are all indeterminate still hold in C. – Konrad Wrobel Oct 11 '15 at 16:31