Green's function for Laplace Equation and the unit ball In order to find a solution to

$\Delta u(x) =0$ in $B(0,1)$ and $u(x)=g(x)$ on $\partial B(0,1)$

the book I am reading ((Graduate Studies in Mathematics) Lawrence C. Evans - Partial Differential Equations_ Second Edition -AMS (2010) Page 39) uses the Greens function to solve this problem. Finding a function that satifies

$\Delta \Phi^x(y) =0$ in $B(0,1)$ and $\Phi^x(y)=\Phi(y-x)$ on $\partial B(0,1)$

($\Phi $ being the fundamental solution to the laplace equation) will give us the Greens function $G(x,y)=\Phi(y-x)-\Phi^x(y)$, which we can solve the problem with.
For the unit ball $B(0,1)$ we used $\Phi^x(y)=\Phi(|x|(y-\tilde x))$ with $\tilde x =\frac{x}{|x|^2}$. And showed that this is harmonic and also satisfies the boundary condition - but only for $x \neq 0$. And that's where my question comes from:
Why isn't it a problem, that $G(x,y)=\Phi(y-x)-\Phi(|x|(y-\tilde x))$ is not well defined at $x = 0$ since the inversion used previously is not possible?

Although the inversion is not possible, $G$ can be extended to $x = 0$, provided $y \neq 0$. This follows by observing that $|x|\tilde{x}$ has norm one for every $x \neq 0$ and upon recalling that the function $\Phi$ is radial. To be precise, for every sequence $x_n \to 0$, $x_n \neq 0$, we have that $$\lim_n \left[\Phi(x_n - y) - \Phi(|x_n|y - |x_n|x_n)\right]$$exists and is independent of the choice of $\{x_n\}$. We then set $G(0,y)$ equal to this value.
