Why does the Dominated Convergence Theorem fail for evaluating $\lim_{n \to \infty} \ n^2 \int_{0}^{1} xe^{-n^2x^2}dx$? I read in a book that to try to use the dominated convergence theorem for:

$$
\lim_{n \to \infty} \  n^2 \int_{0}^{1} xe^{-n^2x^2}dx
$$

fails. The reasoning was written to be that $nxe^{-n^2x^2} \to 0$ as $n \to \infty$ for all $x\in [0,1]$, so that we cannot interchange integral and limit. However, I fail to see why the Dominated Convergence theorem fails exactly in this case. Would anyone have any ideas? Thanks!
Because $\sup_{n\ge 1} n^2 x e^{-n^2x^2}={1\over x\cdot e}$ for $x\in(0,1]$.
To show that the conclusion of the dominated convergence theorem does not hold, assume it does i.e. assume there exists some dominating function. Then show that there is a contradiction.



Related exercise in Probability w/ Martingales:







Apply the same logic to $g:= \sup_n |n^2xe^{-n^2x^2}|$.

If we suppose g is integrable, then by the dominated convergence theorem, we have

$$0 = \int_{0}^{1} \lim_{n \to \infty} f_n dx  = \lim_{n \to \infty} \int_{0}^{1} f_n dx = 1/2$$

The reason why we get the supremum of the integrand ie $g: =\sup_n |n^2xe^{-n^2x^2}|$ is because DCT assumes an integrable g exists s.t. $|f_n| \le g$. The smallest possible g is  $\sup_n |f_n|$ so if $\sup_n |f_n|$ doesn't work, by monotonicity of integration, any $g \ge \sup_n |f_n|$ won't work and hence DCT does not apply