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. 
 A: 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 $y \in D$.
A: The 2D case. 
Given $y\in{\bf R}^2$, let $M=|y|$. By the triangle inequality $|x|-M\leq |x-y|\leq |x|+M$. Thus
$$
\ln\frac{|x|}{|x|+M}\leq\ln\frac{|x|}{|x-y|}\leq\ln\frac{|x|}{|x|-M}\tag{1}
$$
On the other hand, a simple application of L'Hopital rule implies that
$$
\lim_{|x|\to\infty}\bigg||x| \ln\frac{|x|}{|x|+M} \bigg|=1,\quad
\lim_{|x|\to\infty}|x| \ln\frac{|x|}{|x|-M} =1\tag{2}
$$
It follows from (2) that there exists some constant $C>0$ such that for all large enough $|x|$, 
$$
\ln\frac{|x|}{|x|+M}\geq -\frac{C}{|x|},\quad
\ln\frac{|x|}{|x|-M}\leq \frac{C}{|x|},\tag{3}
$$
Combining (1) and (3) together we get
$$
\ln\frac{|x|}{|x-y|}=O(\frac{1}{|x|}).
$$

The 3D case. 
Note that
$$
\lim_{|x|\to\infty}|x|\cdot\frac{1}{|x-y|}=1
$$
which can be proved by observing that for large $|x|$,
$$
\frac{|x|}{|x|+|y|}\leq\frac{|x|}{|x-y|}\leq\frac{|x|}{|x|-|y|}\tag{4}
$$
(Note: Be cautious that the expression $\frac{x}{x-y}$ is meaningless in ${\bf R}^3$.)
A: Below I assume $D\subset B(0,R)$ and $|x|>2R.$
$n=2:$ By the mean value theorem,
$$\ln|x-y|-\ln |x| \le \ln(|x|+|y|)-\ln |x| = \frac{1}{c}|y|,$$
where $c\in (|x|,|x|+|y|).$ Thus the last term is less than $\dfrac{R}{|x|}.$ We can argue similarly from below to get 
$$ \ln|x-y|-\ln |x| \ge \ln(|x|-|y|)-\ln |x|= \frac{1}{c}(-|y|) \ge \frac{-R}{|x|-|y|}\ge -\frac{R}{|x|/2}=-\frac{2R}{|x|}.$$
So we have
$$-\frac{2R}{|x|} \le \ln|x-y|-\ln |x| \le \frac{R}{|x|},$$
and this gives the desired $O\left (\dfrac{1}{|x|}\right) $ estimate.
$n=3:$ The result can be improved a bit to
$$\tag 1 \frac{1}{|x-y|} = \frac{1}{|x|} + O\left (\frac{1}{|x|^2}\right).$$
Using the MVT again, we obtain
$$\frac{1}{|x-y|} - \frac{1}{|x|} \ge \frac{1}{|x|+|y|} - \frac{1}{|x|} = \frac{-1}{c^2}|y| \ge -\frac{R}{|x|^2}.$$
We also have
$$\frac{1}{|x-y|} - \frac{1}{|x|} \le \frac{1}{|x|-|y|} - \frac{1}{|x|} \le \frac{-1}{c^2}(-|y|) \le \frac{R}{(|x|-|y|)^2}  < \frac{R}{(|x|/2)^2} = \frac{4R}{|x|^2}.$$
The two estimates give $(1).$
