Evaluating $\lim_{n\to\infty}\int_0^n(1-(x/n))^ne^{x/2}dx$

Question

$$ \mbox{How to compute}\quad \lim_{n \to \infty}\,\,\int_{0}^{n}\left(1 -{x \over n}\right)^{n} \,\mathrm{e}^{x/2}\,\,\mathrm{d}x\,\,\, ?. $$

No ideas how to start this one. I see that the limit of the $\left[1 - \left(x/n\right)^{n}\right]$ looks close to being $\,\mathrm{e}$, but it should be a plus sign instead of a minus sign$\ldots$

Answer 1

A similar limit is used to prove Euler's limit product formula for the $\Gamma$ function: $$ t! = \lim_{n\to +\infty}\frac{n! n^t}{(t+1)(t+2)\cdot\ldots\cdot(t+n)}\tag{1}.$$ For any $x\in\mathbb{R}^+$ we have $$ \max_{n\geq x} \left(1-x/n\right)^n\leq e^{-x}\tag{2} $$ and $e^{-x}\cdot e^{x/2}\in L^1(\mathbb{R}^+)$, hence by monotone or dominated convergence we have: $$ \lim_{n\to +\infty}\int_{0}^{n}\left(1-\frac{x}{n}\right)^n e^{x/2}\,dx = \int_{0}^{+\infty} e^{-x/2}\,dx = \color{red}{2}.\tag{3}$$

Answer 2

If $0\le x \le n,$ then $(1-x/n)^n \le e^{-x}.$ Proof: This is certainly true at the end points. For $0<x<n,$

$$\tag 1\ln (1-x/n)^n = n \ln (1-x/n) < n(-x/n) = -x.$$

Here we have used the inequality $\ln (1-u) < - u,$ which is valid for $ 0 < u < 1.$ Exponentiate back in $(1)$ to get the desired inequality.

To finish the problem, let $f_n(x) = (1-x/n)^ne^{x/2}\chi_{[0,n]}(x)$ for $x\in [0,\infty).$ Then by the above, $0 \le f_n(x) \le e^{-x/2}, x\ge 0$ You're now good to go for the dominated convergence theorem on $[0,\infty).$