= Solution
Interpreting the displayed function as
$$
\gamma_n(x)=\frac1\pi\frac{n}{(nx)^2+1},
$$
we have $\gamma_n\geq0$ and, after $u=nx$, $\int_{\mathbb R}\gamma_n=1$. Moreover its mass outside any fixed neighborhood of zero tends to zero. Therefore $(\gamma_n)$ is an <approximate identity>, and for every test function $f$,
$$
\int f(x)\gamma_n(x)\,dx=\frac1\pi\int_{\mathbb R}\frac{f(u/n)}{1+u^2}\,du\longrightarrow f(0).
$$
Thus \b[$\gamma_n\to\delta$ in the sense of distributions].
Back to article page
