In a previous question, I learned that the equation $x^{\frac{1}{x}} = c$ have no solutions when $c = 0$. Below, I tried using Lambert's W function, and I found a solution at $x = 0$. Did I make a mistake?
If $x^{\frac{1}{x}} = c$ then ${\frac{1}{x}}\ln x = \ln c$ Let $x = e^{-y}$. So ${\frac{1}{x}}\ln x = -y e^{y} = \ln c$ The solution of $y e^y = -\ln c$ using Lambert's W is $ y = W(-\ln c).$ Hence $ x = e^{-W(-\ln c)} $ is a solution of $x^{\frac{1}{x}} = c$. Now, to solve $x^{\frac{1}{x}} = 0$, I tried $ x = \lim_{c\to 0} e^{-W(-\ln c)} = 0,$ since $\lim -\ln c \to \infty$ as $c\to 0$, and $\lim W(z) \to \infty$ as $z \to \infty$.
Hence, $x = 0$ is a solution to $x^{\frac{1}{x}} = 0$. But this contradicts the answers I got for my previous question. I must've made a mistake. Please point out where.