TITLE: Show the following limit $\lim_{n\rightarrow\infty}\sqrt{n}\int_0^1\frac{f(x)}{1+nx^2}\text{d}x=\frac{\pi}{2}f(0)$
QUESTION [9 upvotes]: Prove that
$$\lim_{n\rightarrow\infty}\sqrt{n}\int_0^1\frac{f(x)}{1+nx^2}\text{d}x=\frac{\pi}{2}f(0),$$
where $f(x)$ is a continuous function on $[0,1].$

REPLY [4 votes]: $$
\sqrt{n}\int_0^1 \frac{f(x)}{1+nx^2}dx
=\int_0^\sqrt{n} \frac{f(u/\sqrt{n})}{1+u^2} du
= \int_0^\infty f_n(x)dx
$$
Here, $f_n(u) = 1_{u\le \sqrt{n}} 
{f(u/\sqrt{n})}/(1+u^2)\to f(0)/(1+u^2)$ (pointwise convergence) and 
$$\int_0^\infty \sup_n|f_n(x)| dx
\le \sup |f|\int_0^\infty  \frac{du}{1+u^2}<\infty
$$(domination), hence the dominated convergence theorem states:
$$
\sqrt{n}\int_0^1 \frac{f(x)}{1+nx^2}dx
\to \int_0^\infty f(0)\frac{du}{1+u^2} = \frac{\pi f(0)}{2}
$$
NB: you only need continuity in 0 and boundedness (and measurability!) of $f$ on $[0,1]$ to conclude.