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](https://math.stackexchange.com/questions/1032293/prove-that-00-1-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?