Integral $ \lim_{n\rightarrow\infty}\sqrt{n}\int\limits_0^1 \frac {f(x)dx}{1 + nx^2} = \frac{\pi}{2}f(0) $ 
Show that for $ f(x) $ a continuous function on $ [0,1] $ we have
  \begin{equation}
\lim_{n\rightarrow\infty}\sqrt{n}\int\limits_0^1 \frac {f(x)dx}{1 + nx^2} = \frac{\pi}{2}f(0)
\end{equation}


It is obvious that
\begin{equation}
\sqrt{n}\int\limits_0^1 \frac {f(x)dx}{1 + nx^2} = \int\limits_0^1 f(x) d [\arctan(\sqrt{n}x)]
\end{equation}
and for any $ x \in (0, 1] $
\begin{equation}
\lim_{n\rightarrow\infty}{\arctan(\sqrt{n}x)} = \frac{\pi}{2},
\end{equation}
so the initial statement looks very reasonable. But we can't even integrate by parts because $ f(x) $ is in general non-smooth! Can anybody help please?

Here is another approach, via hint: it's easy to verify for polynomials (just check that it works on each of the terms $c,x,x^2,\ldots$). Then, show that the distance between your limit and the desired constant is less than $\varepsilon$ for every $\varepsilon>0$. To do this, take a polynomial such that $\|p-f\|_\infty<\frac{\varepsilon}{3}$ and do the obvious estimations.
