How find this limit $\lim_{n\to\infty}\left(\sqrt{n}\int_{0}^{1}(e^x(1-x))^ndx\right)$ Question:
Find this limit
$$\lim_{n\to\infty}\left(\sqrt{n}\int_{0}^{1}(e^x(1-x))^ndx\right)$$
my idea: since $$e^{x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$
so
$$(1-x)e^x=\sum_{k=0}^{\infty}\dfrac{(1-x)x^k}{k!}$$
so
$$\lim_{n\to\infty}\left(\sqrt{n}\int_{0}^{1}\sum_{k=0}^{\infty}\dfrac{x^k(1-x)}{k!}dx\right)$$
then I fell very ugly

It is better to exploit
$$\log(1-x) = -x-\frac{x^2}{2}+O(x^3)$$ in order to have:
$$\begin{eqnarray*}\int_{0}^{1}((1-x)e^x)^n\,dx &=& \int_{0}^{1}\exp\left(-\frac{n x^2}{2}+n\,O(x^3)\right)\,dx\\ &=&\frac{1}{\sqrt{n}}\int_{0}^{\sqrt{n}}\exp\left(-\frac{x^2}{2}+\frac{O(x^3)}{\sqrt{n}}\right)\,dx\end{eqnarray*}$$
hence the dominated convergence theorem easily gives:
$$\lim_{n\to +\infty}\left(\sqrt{n}\int_{0}^{1}((1-x)e^x)^n\,dx\right)=\int_{0}^{+\infty}e^{-x^2/2}\,dx = \color{red}{\sqrt{\frac{\pi}{2}}}.$$
