How to prove $\lim\limits_{t \to 1^-} \frac{\sqrt{1-t^2}}{2\pi}\int_{S^1}\frac{f(x,y)}{1-tx}ds=f(1,0)$? $f(x,y)$ is a continuous function defined on unit circle $\ S^1 :$ $x^2+y^2=1$, prove
$$\lim\limits_{t \to 1^-} \frac{\sqrt{1-t^2}}{2\pi}\int_{S^1}\frac{f(x,y)}{1-tx}ds=f(1,0)$$

I have tried to solved it by doing a parametric transformation about x and y,that is,
$\ x=\cos\theta$ and $\ y=\sin\theta$ and dealt with it as an improper integrate problem with parameter $\ t$. However, I have been stuck in it. Who can help me complete this proof?
 A: Use the transformation $g(\theta) = f(\cos\theta,\sin\theta)$ for $\theta \in [-\pi,\pi]$, as suggested in the question. 
Then, we need to show that $\displaystyle\lim_{t \to 1^-}\left[\dfrac{\sqrt{1-t^2}}{2\pi}\int_{-\pi}^{\pi}\dfrac{g(\theta)}{1-t\cos\theta}\,d\theta\right] = g(0)$. 
Since $f(x,y)$ is continuous on the unit circle, $g(\theta)$ is continuous on $[-\pi,\pi]$. 
Since $g$ is continuous at $0$, for any $\epsilon > 0$ there is a $\delta > 0$ such that $|g(\theta)-g(0)| < \epsilon$ if $|\theta| < \delta$.
Also, since $g$ is continuous on $[-\pi,\pi]$, $g$ is bounded, i.e. $|g(\theta)-g(0)| \le M$ for some $M > 0$. 
Then, we can break up $\dfrac{\sqrt{1-t^2}}{2\pi}\displaystyle\int_{-\pi}^{\pi}\dfrac{g(\theta)}{1-t\cos\theta}\,d\theta$ as follows: 
$\displaystyle \dfrac{\sqrt{1-t^2}}{2\pi}\left[\int_{-\pi}^{\pi}\dfrac{g(0)}{1-t\cos\theta}\,d\theta + \int_{-\delta}^{\delta}\dfrac{g(\theta)-g(0)}{1-t\cos\theta}\,d\theta + \int\limits_{[-\pi,-\delta]\cup[\delta,\pi]}\dfrac{g(\theta)-g(0)}{1-t\cos\theta}\,d\theta\right]$. 
"Trivially" $\displaystyle\dfrac{\sqrt{1-t^2}}{2\pi}\int_{-\pi}^{\pi}\dfrac{g(0)}{1-t\cos\theta}\,d\theta = \dfrac{g(0)}{\pi}\left[\arctan\left(\sqrt{\dfrac{1+t}{1-t}}\tan\dfrac{\theta}{2}\right)\right]_{-\pi}^{\pi} = g(0)$. 
Also, $\displaystyle\left|\dfrac{\sqrt{1-t^2}}{2\pi}\int_{-\delta}^{\delta}\dfrac{g(\theta)-g(0)}{1-t\cos\theta}\,d\theta\right| \le \dfrac{\sqrt{1-t^2}}{2\pi}\int_{-\delta}^{\delta}\dfrac{|g(\theta)-g(0)|}{|1-t\cos\theta|}\,d\theta$ $\le \displaystyle\dfrac{\sqrt{1-t^2}}{2\pi}\int_{-\delta}^{\delta}\dfrac{\epsilon}{1-t\cos\theta}\,d\theta = \dfrac{\epsilon}{\pi}\left[\arctan\left(\sqrt{\dfrac{1+t}{1-t}}\tan\dfrac{\theta}{2}\right)\right]_{-\delta}^{\delta}$ $= \dfrac{2\epsilon}{\pi}\arctan\left(\sqrt{\dfrac{1+t}{1-t}}\tan\dfrac{\delta}{2}\right) \le \epsilon$ for any $t \in (0,1)$. 
Finally, $\displaystyle\left|\dfrac{\sqrt{1-t^2}}{2\pi}\int_{\delta}^{\pi}\dfrac{g(\theta)-g(0)}{1-t\cos\theta}\,d\theta\right| \le \dfrac{\sqrt{1-t^2}}{2\pi}\int_{\delta}^{\pi}\dfrac{|g(\theta)-g(0)|}{|1-t\cos\theta|}\,d\theta$ $\le \displaystyle\dfrac{\sqrt{1-t^2}}{2\pi}\int_{\delta}^{\pi}\dfrac{M}{1-t\cos\theta}\,d\theta = \dfrac{M}{\pi}\left[\arctan\left(\sqrt{\dfrac{1+t}{1-t}}\tan\dfrac{\theta}{2}\right)\right]_{\delta}^{\pi}$ $= \dfrac{M}{\pi}\text{arccot}\left(\sqrt{\dfrac{1+t}{1-t}}\tan\dfrac{\delta}{2}\right) \to 0$ as $t \to 1^-$ (Similarly for the integral over $[-\pi,-\delta]$). 
Putting all of these pieces back together and letting $\epsilon \to 0$ yields the desired result. 
