Is $(0,0)$ a solution to $x^y-y^x=0$? I am trying to determine if $(0,0)$ a solution to $x^y-y^x=0$.  My hunch is that it is undefined since $0^0$ is an indeterminate form.  To attempt to prove this, I have tried the usual "different paths give different limits" trick with
$(x,ax^n)\rightarrow(0,0)$
$(x,\sin(x))\rightarrow(0,0)$
$(x,e^x-1)\rightarrow(0,0)$
None of the above accomplished my goal.  
I did find this older post ($x^y = y^x$ for integers $x$ and $y$) which included an answer by "Yuval Filmus" which states $y=x=0$ is $\it{trivially}$ a solution.  He accomplished this by defining $0^0$ to some value and moves on.  I would like to see something more rigorous, if it exists.
Any hints on how to proceed?
Edited for wording.
Edit:  Stefan Smith confirms my suspicion that the comment of Yuval Filmus cannot be made rigorous.  

First of all, this function may not be well-defined if $x<0$ or $y<0$ (for example, if $x=-1$ and $y=1/2$, then $x^y=i$, which is not real). Therefore, I suggest you do the following:


*

*Define the function only on $\mathbb R_{++}^2\equiv\{(x,y)\in\mathbb R^2\,|\,x>0,\, y>0\}$

*Show that for any sequence $(x_n,y_n)_{n\in\mathbb Z_+}\subset\mathbb R_{++}^2$ that converges to $(0,0)$, you have that $f(x_n,y_n)$ converges to $0$.


In this way, you can continuously extend the function to $(0,0)$ by defining $f(0,0)\equiv0$.

Too bad this cannot be done! To see this, let
\begin{align*}
(x_n,y_n)\equiv\left(\frac 1 n,\frac{1}{\ln(n+1)}\right)\quad\forall n\in\mathbb Z_+
\end{align*}
Clearly, $(x_n,y_n)\to 0$ as $n\to\infty$. However,
\begin{align*}
f(x_n,y_n)=\left(\frac{1}{n}\right)^{1/\ln(n+1)}-\left(\frac{1}{\ln (n+1)}\right)^{1/n}
\end{align*}
converges to $1/e-1$ as $n\to\infty$. Because of this, $0$ is not a good definition of $f(0,0)$, since $0$ may not be approached by the function $f$ when evaluated close to $(0,0)$.

In terms of how you put it originally, try
$$\left(x,\ln\left(\dfrac{x+1}{x}\right)^{-1}\right)$$
This converges to $(0,0)$ as $x\searrow0$, but the function value converges to $1/e-1$.
