1. I don't agree that this discussion is tedious. I think your reply was rather tedious. Personally I don't beleive that $\displaystyle 0^{0} := 1$, I think that this definition only holds for convenience in special cases.

It's not hard to think of examples where $\displaystyle f(x), g(x)$ approach zero continuously, yet $\displaystyle f(x)^{g(x)}$ does not approach $\displaystyle 1$

I'm still interested in your rigorous proof though.

-pomp.

2. Originally Posted by Moo
... and the limit as x comes from the left : $\displaystyle x\to 0^-$ ?...
As for the function $\displaystyle \sqrt{x}$ the point x=0 for the function $\displaystyle f(x)=x^{x}$ is a branch point. The value of the function $\displaystyle f(x)=x^{x}$ for $\displaystyle x \le 0$ are illustrated in the figure...

Of course is...

$\displaystyle lim_{ x\rightarrow 0^{+}} = lim_{ x \rightarrow 0^{-}} = 1$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ‘Lady’ Moo : … I like the way you refer to yourself... it would be more interesting to give an outline of the proof... (I've seen some, I'm thinking about anyone who reads this…)

pomp : … I don't agree that this discussion is tedious. I think your reply was rather tedious. Personally I don't believe that $\displaystyle 0^{0}=1$, I think that this definition only holds for convenience in special cases… I'm still interested in your rigorous proof though…

Mi past experience about the question $\displaystyle 0^{0}= ?$ suggests that for some people that is a sort of ‘tabu’ which ‘hits a nerve’… I wonder why but that is the true

I likely accept the proposal of ‘Lady’ Moo and plomp to explain a rigorous proof of the fact that is $\displaystyle 0^{0}=1$ at only one condition: a moderator [or more than one moderator..] will assume the role of ‘impartial judge’ and will decide if from the ‘mathematical point of view’ the ‘proof’ is correct or not… is someone disposable to that?…

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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