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?

This may be more of a sketch, but should give you the idea.
$\displaystyle |\sqrt{n}\int_0^1 \frac{f(x)}{1+nx^2}\,dx-\frac{\pi}{2}{f(0)}|=
|\sqrt{n}\int_0^1 \frac{f(x)-\frac{\pi}{2\arctan(\sqrt{n})}f(0)}{1+nx^2}\,dx|\le|\sqrt{n}\int_0^\delta \frac{f(x)-\frac{\pi}{2\arctan(\sqrt{n})}f(0)}{1+nx^2}\,dx|+|\sqrt{n}\int_\delta^1 \frac{f(x)-\frac{\pi}{2\arctan(\sqrt{n})}f(0)}{1+nx^2}\,dx|\,.$ 
The second term goes to zero uniformly for $\delta >0$, and for the first term
$\displaystyle |\sqrt{n}\int_0^\delta \frac{f(x)-\frac{\pi}{2\arctan(\sqrt{n})}f(0)}{1+nx^2}\,dx|\le\sqrt{n}M(\delta,n)\int_0^\delta \frac{1}{1+nx^2}\,dx=M(\delta,n)\arctan(\sqrt{n}\delta)<\epsilon$ 
for any $\epsilon>0$ for appropriate $\delta, n$.
