How can I show that $\lim_{\epsilon\to 0+}\int_{B_1\setminus B_\epsilon}\frac{x\cdot\nabla f}{\|x\|^2}dx_1dx_2=Cf(0)$? 
Let $B_r:=\{x\in{\Bbb R}^2:\|x\|<r\}$ and $f:C^1_c(B_1\to{\Bbb R})$, where $C^1_c$ means $f$ is continuously differentiable and has compact support. Show that
  $$\lim_{\epsilon\to 0+}\int_{B_1\setminus B_\epsilon}\frac{x\cdot\nabla f}{\|x\|^2}dx_1dx_2=Cf(0)$$
  for some constant $C$. 


I don't know how to begin with this problem. Any idea?
 A: If $C$ is a smooth positively oriented Jordan curve in $\mathbb{R}^2$ and $F:U\to \mathbb{R}^2$ is a smooth vector-field so that $C$ is contained in $U$ then, 
$$ \int_C F\cdot N ~ ds = \iint_D \nabla \cdot F  $$
Where $N$ is the outward unit-normal and $D$ is the interior of $C$. 
Now if $C$ is the unit-circle then $N(x,y) = \left( \frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}} \right)$ for any point $(x,y)$ on the circle. 
This is beginning to look like what you want, see if you carry this further. 
A: You don't tell us your background. What's going on here is that if you take $\vec G = \dfrac1{x^2+y^2}(x,y)$, then $\text{div}\,\vec G = 2\pi\delta_0$, if you speak physicist's (or distribution) language. @I Love Mr. Paul has indeed gotten you on the right track. You need to apply Green's Theorem to the region $B_1\backslash B_\epsilon$ and use compact support, on one hand, and a limiting argument, on the other, to show that the boundary integral gives you $Cf(0)$ in the limit. (Note that $f$ is continuous at $0$.) 
As a further hint, you need a vector field $\vec F$ whose divergence is $\displaystyle\frac{x\cdot\nabla f(x)}{x^2+y^2}$. The key point in your search will be the fact that $\text{div}\,\vec G = 0$ on the region in question. (It's interesting to figure out why. Start with $\ln(\sqrt{x^2+y^2})$ and see that it is harmonic. Polar coordinates are useful here.)
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
{\vec{r}\cdot\nabla\fermi\pars{\vec{r}} \over r^{2}}
&=
\nabla\bracks{\ln\pars{r}}\cdot\nabla\fermi\pars{\vec{r}}
=
\nabla\cdot\bracks{\fermi\pars{\vec{r}}\cdot\nabla\ln\pars{r}}
-
\fermi\pars{\vec{r}}\nabla\cdot\nabla\ln\pars{r}
\\[3mm]&=
\nabla\cdot\bracks{\fermi\pars{\vec{r}}\cdot\nabla\ln\pars{r}}
-
\fermi\pars{\vec{r}}\ \overbrace{\quad\nabla^{2}\ln\pars{r}\quad}^{\ds{2\pi\,\delta\pars{\vec{r}}}}
\end{align}

$$
{\vec{r}\cdot\nabla\fermi\pars{\vec{r}} \over r^{2}}
=
\nabla\cdot\bracks{\fermi\pars{\vec{r}}\cdot\nabla\ln\pars{r}}
-
2\pi\,\delta\pars{\vec{r}}\fermi\pars{\vec{0}}
$$
  Integration of the RHS first term vanishes out and we are left with:
  $$\color{#0000ff}{\large%
\int{\vec{r}\cdot\nabla\fermi\pars{\vec{r}} \over r^{2}}\,\dd S
=
\overbrace{\quad -2\pi\int\delta\pars{\vec{r}}\,\dd S\quad}^{\ds{\equiv\ {\rm C}}}\
\fermi\pars{\vec{0}} = {\rm C}\fermi\pars{\vec{0}}}
$$

$$
{\rm C}
=
-2\pi\int_{0}^{2\pi}\int_{0}^{\epsilon}
{\delta\pars{r}\delta\pars{\theta} \over r}\,r\,\dd r\,\dd\theta = -2\pi
$$
$\large\tt ADDENDUM:$
\begin{align}
\nabla\ln\pars{r}&=\totald{\ln\pars{r}}{r}\,{\vec{r} \over r}
=
{x \over r^{2}}\,\hat{x} + {y \over r^{2}}\,\hat{y}
\quad\imp\quad
\nabla^{2}\ln\pars{r}
=
\nabla\cdot\bracks{\nabla\ln\pars{r}} = \partiald{\pars{x/r^{2}}}{x} + \partiald{\pars{y/r^{2}}}{y}
\end{align}
$$
\nabla^{2}\ln\pars{r}=
\hat{z}\cdot\nabla\times\pars{-\,{y \over r^{2}}\,\hat{x} + {x \over r^{2}}\,\hat{y}}
$$

\begin{align}
\color{#0000ff}{\large\int_{S}\nabla^{2}\ln\pars{r}\,\dd S}&=
\int\nabla\times
\pars{-\,{y \over r^{2}}\,\hat{x} + {x \over r^{2}}\,\hat{y}}\cdot
\overbrace{\hat{z}\,\dd S}^{\ds{\equiv \dd\vec{S}}}
=
\oint
\pars{-\,{y \over r^{2}}\,\hat{x} + {x \over r^{2}}\,\hat{y}}\cdot\dd\vec{r}
\\[3mm]&=
\oint{-y\,\dd x + x\,\dd y \over r^{2}}
\\[3mm]&=
\int_{0}^{2\pi}{-r\sin\pars{\theta}\bracks{-r\sin\pars{\theta}\,\dd\theta}
+
r\cos\pars{\theta}\bracks{r\cos\pars{\theta}\,\dd\theta} \over r^{2}}
\\[3mm]&=
\int_{0}^{2\pi}\overbrace{\bracks{\sin^{2}\pars{\theta} + \cos^{2}\pars{\theta}}}^{\ds{=\ 1}}\,\dd\theta
=
2\pi = 2\pi\int_{0}^{2\pi}\dd\theta\int_{0}^{\infty}\dd r\,r\,{\delta\pars{r}\delta\pars{\theta} \over r}
\\[3mm]&=
2\pi\int\delta\pars{\vec{r}}\,\dd S
=
\color{#0000ff}{\large\int\bracks{2\pi\delta\pars{\vec{r}}}\,\dd S}
\end{align}

A: Let $r=(x_1^2+x_2^2)^{1/2}$, $\,x=(x_1,x_2)$ and $s=x/r=(\cos\vartheta,\sin\vartheta)$. Using polar coordinates (i.e., $dx_1\,dx_2=r\,dr\,d\vartheta$) we have that
\begin{align*}
\int_{B_1\setminus B_\epsilon}\frac{x\cdot\nabla f}{r^2}\,dx_1\,dx_2 &=\int_\varepsilon^1
\left( \int_{0}^{2\pi}\frac{rs(\vartheta)\cdot\nabla f\big(rs(\vartheta)\big)}{r^2}\,
d\vartheta\right) \,r\,dr \\
&=\int_\varepsilon^1 \int_{0}^{2\pi}s(\vartheta)\cdot\nabla f\big(rs(\vartheta)\big)\,d\vartheta\,dr=\int_{0}^{2\pi}\int_{\varepsilon}^1
\frac{d}{dr} f\big(rs(\vartheta)\big)\,dr\,d\vartheta \\ &=\int_{0}^{2\pi}\big(f\big(s(\vartheta)\big)-f(\varepsilon s(\vartheta))\big)\,
d\vartheta=-\int_{0}^{2\pi}f\big(\varepsilon s(\vartheta)\big)\,d\vartheta,
\end{align*}
as $f\big(s(\vartheta)\big)=0$. Thus
$$
\lim_{\varepsilon\to 0+}\int_{B_1\setminus B_\epsilon}\frac{x\cdot\nabla f}{r^2}\,dx_1\,dx_2=-\lim_{\varepsilon\to 0+}\int_{0}^{2\pi}f\big(\varepsilon s(\vartheta)\big)\,d\vartheta
=-\int_0^{2\pi}f(0)\,d\vartheta=-2\pi\,f(0.)
$$
A: In polar coordinates you have:
$$
 \int_0^{2\pi} \int_\epsilon^{1} \frac {\vec{r}} {r^2} \cdot \left (\frac {\partial f} {\partial r } \hat r \  + \ \frac 1 r \frac {\partial f} {\partial \theta } \hat \theta \right )  r\,dr\, d\theta \,
$$
$$
 = \int_0^{2\pi} \int_\epsilon^{1} \frac {\partial f} {\partial r}\,dr\,d\theta \,
$$
$$
 = \int_0^{2\pi}  \Big ( f (r=1, \theta) - f(r=\epsilon,\theta) \  \Big) d\theta \,
$$
But because $f$ has compact support on $B_1$, $f(r=1,\theta )=0$, and when $\epsilon $
 aproches $0$,  $f$ becomes independent of $\theta$, so you get:
$$
\int_{B_1\setminus B_\epsilon}\frac{x\cdot\nabla f}{r^2}\,dx_1\,dx_2 =-2\pi f(0)
$$
