On the evaluation of a limit of a definite integral 
Why is it that
  $$
\lim_{\epsilon\to 0} \, \frac{2}{\pi} \int_0^\epsilon \frac{f(x)}{\sqrt{\epsilon^2-x^2}} \, \mathrm{d}x = f(0) \, ?
$$

In the particular case when $f(x) = c$ is a constant, the identity follows forthwith. Is it possible to show that this is true for an arbitrary real valued function $f(x)$? 
Thanks 
Best
fede
 A: Note that
$$\left|\frac{2}{\pi} \int_0^\epsilon \frac{f(x)}{\sqrt{\epsilon^2-x^2}} \, \mathrm{d}x-f(0)\right|=\frac{2}{\pi} \left|\int_0^\epsilon \frac{f(x)-f(0)}{\sqrt{\epsilon^2-x^2}} \, \mathrm{d}x\right|\leq 
\frac{2}{\pi} \int_0^\epsilon \frac{|f(x)-f(0)|}{\sqrt{\epsilon^2-x^2}} \, \mathrm{d}x.$$
Then by the continuity of $f$ at $0$ (we need this property otherwise it is not true), for $\epsilon'>0$ there is $\delta>0$ such that $|f(x)-f(0)|<\epsilon'$ for $0\leq x<\delta$. Hence for $0<\epsilon<\delta$,
$$\left|\frac{2}{\pi} \int_0^\epsilon \frac{f(x)}{\sqrt{\epsilon^2-x^2}} \, \mathrm{d}x-f(0)\right|\leq 
\frac{2}{\pi} \int_0^\epsilon \frac{\epsilon'}{\sqrt{\epsilon^2-x^2}} \, \mathrm{d}x=\epsilon'$$
that is
$$\lim_{\epsilon\to 0} \, \frac{2}{\pi} \int_0^\epsilon \frac{f(x)}{\sqrt{\epsilon^2-x^2}} \, \mathrm{d}x = f(0).$$
A: For any $\varepsilon>0$
$$ \frac{2}{\pi}\int_{0}^{\varepsilon}\frac{f(x)\,dx}{\sqrt{\varepsilon^2-x^2}}\stackrel{x\mapsto \varepsilon z}{=}\int_{0}^{1}f(\varepsilon z)\frac{2}{\pi\sqrt{1-z^2}}\,dz \tag{1}$$
and we have $\int_{0}^{1}\frac{2\,dz}{\pi\sqrt{1-z^2}}=1$. In particular, if $\lim_{u\to 0^+} f(u)$ exists then
$$ \lim_{\varepsilon\to 0^+}\frac{2}{\pi}\int_{0}^{\varepsilon}\frac{f(x)\,dx}{\sqrt{\varepsilon^2-x^2}} = \lim_{u\to 0^+} f(u) \tag{2}$$
by the dominated convergence theorem.
