Solution: Substituting $x^{n+1}=t$, we have $(n+1)x^n \mathrm{d} x=\mathrm{d} t$. At $x=0$, $t=0$ and at $x=1$, $t=1$. Thus, we get \[ \int_{0}^{1}nx^ne^{x^2}\,{\mathrm{d} }x=\int_{0}^{1}\frac{n}{n+1}e^{t^{\frac{2}{n+1}}}\,{\mathrm{d} }t. \] Now $e^{t^{\frac{2}{n+1}}}$ is bounded in $(0,1)$ for all $n$, and $e^{t^{\frac{2}{n+1}}}\to e$ pointwise in $(0,1)$. Thus, by Dominated Convergence Theorem, we have \begin{align*} \lim_{n\to \infty}\int_{0}^{1}nx^ne^{x^2}\,{\mathrm{d} }x= &\lim_{n\to \infty}\int_{0}^{1}\frac{n}{n+1}e^{t^{\frac{2}{n+1}}}\,{\mathrm{d} }t\\ &=\int_{0}^{1}\lim_{n\to \infty}\frac{n}{n+1}e^{t^{\frac{2}{n+1}}}\,{\mathrm{d} }t\\ &=\int_{0}^{1}1\cdot e \,{\mathrm{d} }t=e \end{align*}