How prove this integral limit is exsit $\lim_{\varepsilon\to 0^{+} }f(x,y)dxdy$ Question:
This problem  is the 2013 Beijing university mathematics examination the last question,and I consider sometimes,and I can't,
let  $D$ is with smooth boundary bounded region in plane,and the function 
$f(x,y)$  is Continuous differentiable on $\overline{D}$,and $\forall P_{0}=(x_{0},y_{0})\in D$, show that this limit

$$A=\lim_{\varepsilon\to 0^{+}}\int\int_{D/B_{\varepsilon}(P_{0})}\dfrac{\dfrac{\partial f}{\partial y}(x,y)(y-y_{0})+\dfrac{\partial f}{\partial x}(x,y)(x-x_{0})}{(x-x_{0})^2+(y-y_{0})^2}dxdy$$
  is exsit,where $B_{\varepsilon}(P_{0})=\{(x,y)|(x-x_{0})^2+(y-y_{0})^2\le\varepsilon^2,,x_{0},y_{0}\in D\}$
(2):show that
  $$f(x_{0},y_{0})=\dfrac{1}{2\pi}\left(\int_{\partial D}\dfrac{f(x,y)}{(x-x_{0})^2+(y-y_{0})^2}((x-x_{0})dy-(y-y_{0})dx)-A\right)$$

First,I want use this Taylor lemma:
$$f(x,y)=f(x_{0},y_{0})+f'_{x}(x_{0},y_{0})(x-x_{0})+f'_{y}(x_{0},y_{0})(y-y_{0})+\cdots $$
But I can't any work .Thank you for you help!
 A: 1) Without loss of generality we can consider the region D as following:
$$
D = \{(x,y): (x-x_0)^2+(y-y_0)^2 \leqslant R^2 \},
$$
where R > 0. It does not change anything dramatically, one can easily prove for a random domain D. It does not really affect the proof.
Let us now consider the calculation of A, the integral becomes simple if we use the polar coordinates with $(x_0,y_0)$ as the center:
$$
x = x_0 + \sigma \cos \phi\\
y = y_0 + \sigma \sin \phi
$$
For the region $D\backslash B_\varepsilon (P_0)$ it means that $\phi \in [ 0, 2\pi)$ and $\sigma \in [\varepsilon, R]$. Now lets make the substitution:
$$
A = \lim\limits_{\varepsilon \to +0} \int\limits_{D\backslash B_\varepsilon (P_0)} \frac{\frac{\partial f}{\partial x}\cdot \sigma \cos \phi + \frac{\partial f}{\partial y} \cdot \sigma \sin \phi}{\sigma^2} \sigma d\sigma d\phi = \lim\limits_{\varepsilon \to +0} \int\limits_{D\backslash B_\varepsilon (P_0)} [ \frac{\partial f}{\partial x}\cdot \cos \phi + \frac{\partial f}{\partial y} \cdot \sin \phi ] d\sigma d\phi
$$
Here we keep in mind that the substitution also took place in partial derivatives but I omitted it to avoid text overloading.
It is clear that the limit exists as the function $[ \frac{\partial f}{\partial x}\cdot \cos \phi + \frac{\partial f}{\partial y} \cdot \sin \phi ]$ is continuously differentiable, therefore, integrable over any compact, such is $D\backslash B_\varepsilon (P_0)$ for any $\varepsilon$.
Ergo we have just proven the first part. Let's move to the second one.
2) I suggest the following notation in this part: $K_\varepsilon$ as the boundary of $B_\varepsilon(P_0)$, A is some random point on the boundary $\partial D$, B is a point on the boundary $\partial K_\varepsilon$ such that A, B and $P_0$ belong to the same line. Let's consider a contour:
$$
\Gamma = \partial D + AB + \partial K^-_\varepsilon + BA
$$
(Remember the contours used in theory of complex variables?) Let's use Cauchy's integral theorem for domain bound with $\Gamma$, thus we obtain:
$$
\oint\limits_\Gamma = \oint\limits_{\partial D} + \oint\limits_{\partial K^-_\varepsilon}
$$
First let's calculate:
$$
\lim\limits_{\varepsilon \to +0}\oint\limits_{\partial K^-_\varepsilon} \frac{f(x,y)}{(x-x_0)^2+(y-y_0)^2}[(x-x_0)dy - (y-y_0) dx] =\\ -\lim\limits_{\varepsilon \to +0} \int\limits_0^{2\pi} \frac{f(x_0+\varepsilon \cos \phi, y_0 + \varepsilon \sin \phi)}{\varepsilon^2} [\varepsilon^2 \cos^2\phi + \varepsilon^2 \sin^2\phi]d\phi = \\-\lim\limits_{\varepsilon \to +0} \int\limits_0^{2\pi} f(x_0+\varepsilon \cos \phi, y_0 + \varepsilon \sin \phi)d\phi = -2\pi f(x_0,y_0).
$$
Now let's move to the integral over $\Gamma$ curve (using Green's theorem):
$$
\lim\limits_{\varepsilon \to +0}\oint\limits_{\Gamma} \frac{f(x,y)}{(x-x_0)^2+(y-y_0)^2}[(x-x_0)dy - (y-y_0) dx] = \\ 
\lim\limits_{\varepsilon \to +0} \int\limits_{D\backslash B_\varepsilon (P_0)} \frac{\frac{\partial f}{\partial x}(x-x_0) + \frac{\partial f}{\partial y}(y-y_0)}{(x-x_0)^2+(y-y_0)^2} dx dy = A
$$
Subsitituting the calculated integrals into Cauchy's theorem one gets:
$$
f(x_0, y_0) = \frac{1}{2\pi} \oint\limits_{\partial D} \frac{f(x,y)}{(x-x_0)^2+(y-y_0)^2}[(x-x_0)dy - (y-y_0) dx] - \frac{A}{2\pi}.
$$
I hope the proof was understandable.
