TITLE: $\lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^n\text{e}^{\frac{x}{2}}\text{d}x$ Evaluating this limit
QUESTION [5 upvotes]: How to evaluate :
$$\lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^n\text{e}^{\frac{x}{2}}\text{d}x$$

REPLY [6 votes]: Here's a proof using just the squeeze theorem (as requested by Ryan in the comments on Marvis' answer).
For $x = o(n)$ and $n$ large enough we have
$$
n \log\left(1-\frac{x}{n}\right) = -\sum_{k=1}^{\infty} \frac{x^k}{k n^{k-1}} \geq - x - \frac{x^2}{n},
$$
so that
$$
\left(1-\frac{x}{n}\right)^n \geq e^{-x-x^2/n}.
$$
This gives
$$
\begin{align*}
\int_0^n \left(1-\frac{x}{n}\right)^n e^{x/2} \,dx &\geq \int_0^{n^{1/4}} \left(1-\frac{x}{n}\right)^n e^{x/2}\,dx \\
&\geq \int_0^{n^{1/4}} e^{-x/2-x^2/n}\,dx \\
&\geq e^{-1/\sqrt{n}} \int_0^{n^{1/4}} e^{-x/2}\,dx.
\end{align*}
$$
The last expression converges to
$$
\int_0^\infty e^{-x/2}\,dx = 2
$$
as $n \to \infty$.  Marvis' answer shows that
$$
\int_0^n \left(1-\frac{x}{n}\right)^n e^{x/2} \,dx \leq \int_0^\infty e^{-x/2}\,dx = 2,
$$
so we conclude from the squeeze theorem that
$$
\int_0^n \left(1-\frac{x}{n}\right)^n e^{x/2}\,dx \longrightarrow 2
$$
as $n \to \infty$.