How to find the integral $\lim_n \int_{-1}^{\infty}\frac{\sqrt{n}f(x)}{1+n x^2}dx$ where $f$ is continuous and integrable on $\mathbb{R}$ How to find the integral $\lim_n\int_{-1}^{\infty}\frac{\sqrt{n}f(x)}{1+n x^2}dx$ where $f$ is continuous and integrable on $\mathbb{R}$.
I tried to change variable $\int_{-1}^{\infty}\frac{\sqrt{n}f(x)}{1+n x^2}dx=\int_{-\sqrt{n}}^\infty \frac{f(x/\sqrt{n})}{1+x^2}dx$. But $f$ may not be bounded, I guess the limit  is $f(0)\pi$.
 A: Split the integral:
\begin{align}
\int_{-\sqrt n}^{\infty} \frac{f(x/\sqrt n)}{1 + x^2} dx
& = \int_{-\sqrt n}^{\sqrt n} \frac{f(x/\sqrt n)}{1 + x^2} dx + \int_{\sqrt n}^{\infty} \frac{f(x/\sqrt n)}{1 + x^2}dx \\
\lim_{n\to\infty} \int_{-\sqrt n}^{\infty} \frac{f(x/\sqrt n)}{1 + x^2} dx
& = \lim_{n\to\infty} \int_{-\sqrt n}^{\sqrt n} \frac{f(x/\sqrt n)}{1 + x^2} dx + \lim_{n\to\infty} \int_{\sqrt n}^{\infty} \frac{f(x/\sqrt n)}{1 + x^2}dx.
\end{align}
The second integral will go to $0$ because
$$
\int_{\sqrt n}^\infty \left|\frac{f(x / \sqrt n)}{1 + x^2}\right| dx
\le \left(\int_{\sqrt n}^{\infty} |f(x / \sqrt n)| dx\right)
\frac{1}{1 + n}
\le \frac{C\sqrt n}{1 + n}
$$
and the right-hand side goes to $0$ as $n \to \infty$. ($C$ exists because $f$ is integrable.)
Next, set
$$
g_n(x) = \begin{cases}
\frac{f(x/\sqrt n)}{1 + x^2} & ; x \in [-\sqrt n, \sqrt n] \\
0 & ; x \notin [-\sqrt n, \sqrt n]
\end{cases}
$$
The limit in question is $\lim_{n\to\infty} \int_{-\infty}^{\infty} g_n(x) dx$.
Each $|g_n|$ is bounded by $\frac{\sup_{\xi \in[-1, 1]} |f(\xi)|}{1 + x^2}$, so the dominated convergence theorem applies. (We know that $f$ is bounded in $[-1, 1]$ because it's continuous.) The pointwise limit of $g_n$ is $g(x) = \frac{f(0)}{1 + x^2}$, and its integral is $f(0)\pi$, as you speculated.
A: Note that $\int_1^\infty \frac{\sqrt{n}|f(x)|}{1+n x^2}dx \le 
\int_1^\infty \frac{\sqrt{n}|f(x)|}{1+n}dx \le \frac{\sqrt{n}}{1+n} \int |f|$, and hence goes to zero.
We are left with $ I_n =\int_{-1}^{1}\frac{\sqrt{n}f(x)}{1+n x^2}dx $. Using the substitution $u = \arctan \sqrt{n}x$, we have $du = { \sqrt{n} \over 1 + n x^2} dx$ and so the integral becomes $I_n = \int_{- { \arctan \sqrt{n}}}^{\arctan \sqrt{n}} f ({1 \over \sqrt{n}} \tan u) du$.
Now let $g_n(u) = f ({1 \over \sqrt{n}} \tan u) 1_{(-\arctan \sqrt{n},\arctan \sqrt{n}) } (u)$, and note that $g_n(u) \to f(0)1_{(-{ \pi \over 2}, { \pi \over 2} )}(u)$,
since $f$ is continuous at $0$.
Since $f$ is continuous on $[-1,1]$, it is bounded, hence $g_n$ is uniformly bounded and hence the DCT applies.
It follows that $\lim_n I_n = \lim_n \int g_n = \pi f(0)$.
