Sunday, December 30, 2018

limits - "Proving" that $0^0 = 1$




I know that $0^0$ is one of the seven common indeterminate forms of limits, and I found on wikipedia two very simple examples in which one limit equates to 1, and the other to 0. I also saw here: Prove that $0^0 = 1$ using binomial theorem
that you can define $0^0$ as 1 if you'd like.



Even so, I was curious, so I did some work and seemingly demonstrated that $0^0$ always equals 1.



My Work:



$$y=\lim_{x\rightarrow0^+}{(x^x)}$$




$$\ln{y} = \lim_{x\rightarrow0^+}{(x\ln{x})} $$



$$\ln{y}= \lim_{x\rightarrow0^+}{\frac{\ln{x}}{x^{-1}}} = -\frac{∞}{∞} $$



$\implies$ Use L'Hôpital's Rule



$$\ln{y}=\lim_{x\rightarrow0^+}\frac{x^{-1}}{-x^{-2}} $$
$$\ln{y}=\lim_{x\rightarrow0^+} -x = 0$$
$$y = e^{0} = 1$$




What is wrong with this work? Does it have something to do with using $x^x$ rather than $f(x)^{g(x)}$? Or does it have something to do with using operations inside limits? If not, why is $0^0$ considered indeterminate at all?


Answer



Someone said that $0^0=1$ is correct, and got a flood of downvotes and a comment saying it was simply wrong. I think that someone, me for example, should point out that while saying $0^0=1$ is correct is an exaggeration, calling that "simply wrong" isn't quite right either. There are many contexts in which $0^0=1$ is the standard convention.



Two examples. First, power series. If we say $f(t)=\sum_{n=0}^\infty a_nt^n$ that's supposed to entail that $f(0)=a_0$. But $f(0)=a_0$ depends on the convention that $0^0=1$.



Second, elementary set theory: Say $|A|$ is the cardinality of $A$. The cardinality of the set off all functions from $A$ to $B$ should be $|B|^{|A|}$. Now what if $A=B=\emptyset$? There as well we want to say $0^0=1$; otherwise we could just say the cardinality of the set of all maps was $|B|^{|A|}$ unless $A$ and $B$ are both empty.



(Yes, there is exactly one function $f:\emptyset\to\emptyset$...)




Edit: Seems to be a popular answer, but I just realized that it really doesn't address what the OP said. For the record, of course the OP is nonetheless wrong in claiming to have proved that $0^0=1$. It's often left undefined, and in any case one does not prove definitions...


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