Prove $\lim_{n\rightarrow \infty} n\int_0^{+\infty}{\frac{g\left( x \right)}{n^2+x^2}\text{d}x=\frac{\pi}{2}c}$ Let $g\in C[0,+\infty)$, and $\displaystyle \lim_{x\to +\infty }g(x)=c\in \mathbb R$. Prove that
$$
\lim_{n \to \infty}\left[n\int_{0}^{\infty}
\frac{g\left(x\right)}{n^{2} + x^{2}}\,\mathrm{d}x\right] =
\frac{\pi}{2}\,c
$$
I think the integral is related to the following
$$
\int_{0}^{\infty}
{\frac{c}{1 + x^{2}}\,\mathrm{d}x} = \frac{\pi}{2}\,c
$$
But I failed to find the precise relationship. Can anyone help $?$.
 A: Here's a proof that doesn't use the DCT:
Consider the quantity
$$Q=\Bigg|n\int_0^{\infty}\frac{g(x)}{n^2+x^2}dx-\frac{\pi c}{2}\Bigg|$$
This can be cleverly rewritten as follows, after substituting $x=nt$ and $\frac{\pi}{2}=\int_{0}^{\infty}\frac{dt}{1+t^2}$
$$Q=\Bigg|\int_{0}^{\infty}\frac{g(nt)-c}{1+t^2}\Bigg|$$
Now we know that there exists $x_0$ s.t $|g(nt)-c|<\frac{2\epsilon}{\pi}, n>x_0/t$, since the limit of $g$ exists. To apply this inequality successfully, we bound the quantity of interest as follows
$$Q\leq\int_{0}^{M}\frac{|g(nt)-c|}{1+t^2}+
\int_{M}^{\infty}\frac{|g(nt)-c|}{1+t^2}$$
where we arbitrarily split the integral in two pieces. The inequality above can only be applied on the second piece, with $n>\frac{x_0}{M}$. We put a bound on the first piece by using the boundedness of $g$ (inferred by it's continuity and the finite limit at infinity) which implies that $|g(nt)-c|<\Delta ~\forall~ n,t$.
All in all we have that
$$Q<\Delta\int_{0}^M\frac{dt}{1+t^2}+\frac{2\epsilon}{\pi}\int_{M}^\infty\frac{dt}{1+t^2}=\left(\Delta-\frac{2\epsilon}{\pi}\right)\arctan M +\epsilon$$
$Q$ doesn't depend on M so we can conclude that
$$Q<\inf_{M}\left(\Delta-\frac{2\epsilon}{\pi}\right)\arctan M +\epsilon=\epsilon$$
where we made the choice that $\epsilon<\frac{\pi\Delta}{2}$ has to be small enough. Hence we proved by the definition of the limit itself that
$$\lim_{n\to\infty}n\int_{0}^{\infty}\frac{g(x)}{n^2+x^2}dx=\frac{\pi c}{2}$$
A: Substitute $u=x/n$:
$$\int_0^\infty\frac{g(x)}{n^2+x^2}\,dx=\frac{1}{n^2}\int_0^\infty\frac{g(x)}{1+(x/n)^2}\,dx=\frac{1}{n}\int_0^\infty\frac{g(nu)}{1+u^2}\,du.$$
Now since $g$ is continuous on $[0,\infty)$ and has a finite limit at $\infty$, $g$ must be bounded. Therefore, by the dominated convergence theorem:
$$\lim_{n\to\infty}n\int_0^\infty\frac{g(x)}{n^2+x^2}\,dx=\lim_{n\to\infty}\int_0^\infty\frac{g(nu)}{1+u^2}\,du=\int_0^\infty\frac{1}{1+u^2}\lim_{n\to\infty}g(nu)\,du,$$
and now you can use the integral relationship that you mentioned.
