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What makes $0^0$ indeterminate. Here is a video by numberphile that claims that $z^z$ does not exist as $z \to 0$ where $z \in \mathbb C $. I tried tried $\lim_{x \to 0}(x+ix)^{(x+ix)}$ and replaced $x$ by $-x$ and couple of tricks Limit[(x + I x^2)^(x + I x^2), x -> 0] in Mathematica but I am getting $1$. Did I misunderstand what is tried to be shown on that video?

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0^0 = 1 is fine. – user58512 Jan 28 '13 at 9:26
This seems to be based on the same old pitfall as in this previous question of yours. Did you get something from the answers there? – Did Jan 28 '13 at 10:08
there was an answer by user58512 but it seems he/she deleted it. I got the same. $z^z = e^{z \ln z} = e^{z(\ln |z| + i \arg (z))}$ which goes to $1$ as $z \to 0$ – hasExams Jan 28 '13 at 10:14
which goes to $1$ as $z\to0$... is pure fantasy from your part. – Did Jan 30 '13 at 22:26
there was an answer by user58512 but it seems he/she deleted it... What are you talking about? No answer was deleted. – Did Jan 30 '13 at 22:27
up vote 2 down vote accepted

Answering, instead, your first question: What makes $0^0$ indeterminate?

This is not just a matter of $$ \lim_{x \to 0} x^x = ? \tag{1} $$ but more generally of $$ \lim_{t \to a} f(t)^{g(t)} = ? \qquad\text{where } \lim_{t \to a} f(t) = 0\text{ and } \lim_{t\to a}g(t) = 0. \tag{2}$$ We say "$0^0$ is indeterminate" to mean that the answer in (2) can vary, depending on $f$ and $g$.

For example, if $$ f(t) := e^{-(\log 2)/|t|},\qquad g(t) := |t|, $$ then $$ \lim_{t \to 0} f(t) = 0,\qquad \lim_{t\to 0}g(t) = 0,\qquad \lim_{t\to 0} f(t)^{g(t)} = \frac{1}{2} . $$ You can probably make minor changes in this to get other results instead of $1/2$.

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I disagree.

The subtlety is $z^{z}$ is a multi-valued complex function. If it is $x^{x}$ then it is easy to find the limit by $e^{x\log[x]}$, and we know $\lim_{x\rightarrow 0}x\log[x]=\frac{1/x}{-1/x^{2}}=-x=0$(repeat use of L'Hospital rule at here). So you can conclude $x^{x}\rightarrow 1$.

In the complex case the same strategy would give you complications, because $\log[z]=\log[|z|]+irad(z)+2\pi ni$. This is because if we let $z=re^{i\theta}$, then we have $\log[z]=\log[r]+i\theta+2\pi n i$. So we have extra complication aring from $z* rad(z)$, or $re^{i\theta}*i\theta$. Clearly when $r\rightarrow 0$ this goes to $0$ as well. So we can conclude $e^{z\log[z]}$ goes to 1 as we did earlier.

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