Use convergence theorem to prove the limit Suppose $f:\mathbb{R}\to\mathbb{R}$ is $L^1$ and it is continuous at the origin. Let $g_n(x)=\frac{1}{1+x^2}f(\frac{x}{\sqrt{n}})$. The problem is to prove $$\lim_n\int_0^\infty g_ndx=\frac{\pi}{2}f(0)$$ The method I tried so far is using the dominated convergence theorem but failed finding a uniform upper bound for $g_n$. Any suggestions?
 A: EDIT: the fact that the sequence is unbounded for $y = 0$ is not a problem since we don't care about sets of measure $0$.
Clearly $$\lim_{n \to \infty} \frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big) = \frac{1}{1 + x^2}f(0).$$
Moreover we can find $N$ such that if $n \ge N$ then $\Big|f\Big(\frac{x}{\sqrt{n}}\Big) - f(0)\Big| \le 1$, then in particular $$\Big|f\Big(\frac{x}{\sqrt{n}}\Big)\Big| \le 1  + |f(0)|.$$
Then (forgetting about the first $N - 1$ terms of the sequence) we have $$\Big|\frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big) - \frac{1}{1 + x^2}f(0)\Big| \le \frac{1}{1 + x^2}(1 + |f(0)| + |f(0)|) = \frac{1 + 2|f(0)|}{1 + x^2} \in L^{1}.$$
Finally pass to the limit under the integral sign to get the result ;)
Original post (not of any use anymore!):
(Hint: $\frac{x}{\sqrt{n}} = y$, then $dy = \frac{dx}{\sqrt{n}}$. Substituting we get: $$\int_0^{\infty}\frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big)\ dx = \int_0^{\infty}\frac{\sqrt{n}}{1 + ny^2}f(y)\ dy.$$
The sequence in the second integral looks like something that can be bounded by an $L^1$ function. I hope it helps a little bit!)
