How to prove the asymptotic behavior of the fundamental solutions to the 2D and 3D Laplace's equations? Consider the fundamental solution of Laplace's equation in 2d and 3d: 
$$\Phi(x,y):=
\begin{cases}
\frac{1}{2\pi}\ln\frac{1}{|x-y|},\quad x,y\in{\mathbb R}^2\\
\frac{1}{4\pi}\frac{1}{|x-y|},\quad x,y\in{\mathbb R}^3.
\end{cases}$$
Let $D\subset{\mathbb R}^m$ $(m=2,3)$ be a bounded domain of class $C^1$. 
Question: How can I show the following asymptotic behavior of $\Phi$?

$$\Phi(x,y)=
\begin{cases}
\frac{1}{2\pi}\ln\frac{1}{|x|}+O(\frac{1}{|x|}),\quad x,y\in{\mathbb R}^2\\
O(\frac{1}{|x|}),\quad x,y\in{\mathbb R}^3
\end{cases}$$
  for $|x|\to\infty$, which holds uniformly for all directions $x/|x|$ and all $y\in\partial D$.

Note: The statement above appears without a proof in Rainer Kress's Linear Integral Equations (2nd edition), Chapter 6.

[My thoughts.]
According to the definition of the big O notation, for the case in ${\mathbb R}^2$, it suffices to show that there exist constants $C,M>0$ such that
$$
\frac{1}{2\pi}\bigg(\ln\frac{1}{|x-y|}-\ln\frac{1}{|x|}\bigg)\leq C\frac{1}{|x|},\quad \forall |x|>M.
$$
This boils down to doing the estimate
$$
\ln\frac{|x|}{|x-y|}\leq \tilde{C}\frac{1}{|x|}
$$ 
which I have no idea how to go on. 
Similarly for the case in ${\mathbb R}^3$, one needs to show for some $M$ that
$$
\frac{1}{|x-y|}\leq C\frac{1}{|x|},\quad \forall |x|>M.
$$
This one is much simpler: one can simply rewrite $|x-y|$ and $x$ in terms of Cartesian coordinates. 

The first estimate is actually best done without combining the two logarithms.  Let $L_y$ be the segment between $x-y$ and $x$.  Then, a standard estimate tells us that
$$\left|\ln \frac{1}{|x-y|} - \ln\frac{1}{|x|}\right| \le |(x-y) - x| \sup_{z \in L_y} \left|\nabla_z \ln\frac{1}{|z|}\right| = \sup_{z \in L_y} \frac{|y|}{|z|}$$
For $z \in L_y$, $|x| - |y| \le |z|$.  If we let $A = \sup_{y \in D} |y|$, then
$|x| - A \le |z|$.  In particular, for $|x| > 2A$, this shows that 
$$\frac{1}{|z|} \le \frac{2}{|x|}.$$
From this, we can conclude
$$\left|\ln \frac{1}{|x-y|} - \ln\frac{1}{|x|}\right| \le \frac{2A}{|x|}$$
independent of the choice of $y \in D$.
