Is it true that $\lim_{n\to \infty}\int_0^1 f_n(x) dx=\int_0^1\lim_{n\to \infty} f_n(x) dx$? Let $f_n(x)=\dfrac{1}{1+n^2x^2};n\in \Bbb N;x\in \Bbb R$.
Is it true that $\lim_{n\to \infty}\int_0^1 f_n(x) dx=\int_0^1\lim_{n\to \infty} f_n(x) dx$?
$f(x)=\lim_{n\to \infty} f_n(x)=$\begin{cases} 1 & x=0 \\ 0 &x\neq 0\end{cases}
But $\lim_{n\to \infty}\int_0^1 f_n(x) dx=\lim(\tan^{-1} nx)|_0^1=\lim \tan^{-1}(n)=\tan^{-1}(\frac{\pi}{2})\neq \int_0^1 f(x) dx$
Am I right?
Please check.

Yes, it is true. Note that
$$\lim_{n\to \infty}\int_0^1 \dfrac{1}{1+n^2x^2} dx=\lim_{n\to \infty}\left[\frac{\arctan(nx)}{n}\right]_0^1=\lim_{n\to \infty}\frac{\arctan(n)}{n}=0$$
where in the last step we used the fact that $\lim_{n\to \infty}\arctan(n)=\pi/2$.
Moreover
$$\int_0^1 f(x) dx=0$$
where $$\lim_{n\to \infty}\dfrac{1}{1+n^2x^2}=f(x)=\begin{cases} 1 & x=0 \\ 0 &x \not= 0\end{cases}.$$
