Why doesn't $f_n(x)=(n^2x^2)/\left(1+x^2\right)^n$ verify $\lim_{n\to\infty}\int f_n = \int\lim_{n\to\infty} f_n$?

Question

The question is: If $f_n(x)=\dfrac{n^2x^2}{\left(1+x^2\right)^n}$, with $x\in [0,1]$, then $$\lim_{n\to\infty}\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{1}}_{\cssId{lower-bound-mathjax}{0}} \dfrac{n^2x^2}{\left(1+x^2\right)^n}\,\cssId{int-var-mathjax}{\mathrm{d}x} \neq \displaystyle\int\limits^{\cssId{upper-bound-mathjax}{1}}_{\cssId{lower-bound-mathjax}{0}}\lim_{n\to\infty} \dfrac{n^2x^2}{\left(1+x^2\right)^n}\,\cssId{int-var-mathjax}{\mathrm{d}x}$$

The problem gives a hint: try to find a lower bound for $f_n$ in $[0,\frac{1}{\sqrt n}]$.

Thanks.

Answer 1

If $0\leqslant x\leqslant\frac1{\sqrt n}$, then$$(1+x^2)^n\leqslant\left(1+\frac1n\right)^n\leqslant3$$and therefore\begin{align}\int_0^1\frac{n^2x^2}{(1+x^2)^n}\,\mathrm dx&\geqslant\int_0^{1/\sqrt n}\frac{n^2x^2}{(1+x^2)^n}\,\mathrm dx\\&\geqslant\frac13\int_0^{1/\sqrt n}n^2x^2\,\mathrm dx\\&=\frac{\sqrt n}9.\end{align}So,$$\lim_{n\to\infty}\int_0^1\frac{n^2x^2}{(1+x^2)^n}\,\mathrm dx=\infty.$$But, for each $x\in(0,1]$,$$\lim_{n\to\infty}\frac{n^2x^2}{(1+x^2)^n}=0.$$

Answer 2

We can try to find a similar limit. We have that

$$\int_0^1 \frac{n^2x^2}{(1+x^2)^n}\:dx \geq \int_0^1 \frac{n^{\frac{3}{2}}x^2}{(1+x^2)^n}\:dx$$

for all $n \in \Bbb{N}$. For the integral on the right, use the substitution $u = \sqrt{n}x$:

$$\lim_{n\to\infty}\int_0^{\sqrt{n}}\frac{u^2}{\left(1+\frac{u^2}{n}\right)^n}\:du \longrightarrow \int_0^\infty u^2e^{-u^2}\:du = \frac{\sqrt{\pi}}{4}$$

by dominated convergence. Thus we have that while the right hand side goes to $0$, the left hand side is larger than a positive number (in fact it even grows without bound).