$\lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^n\text{e}^{\frac{x}{2}}\text{d}x$ Evaluating this limit How to evaluate :
$$\lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^n\text{e}^{\frac{x}{2}}\text{d}x$$
 A: Use the fact that
$$\left(1 - \dfrac{x}n \right)^n \leq e^{-x}$$
i.e.
$$\left(1 - \dfrac{x}n \right)^n e^{x/2} \leq e^{-x/2}$$
and
$$\lim_{n \to \infty}\left(1 - \dfrac{x}n \right)^n = e^{-x}$$
Consider the sequence $$f_n(x) = \begin{cases} \left(1 - \dfrac{x}{n} \right)^n e^{x/2} & x \in [0,n]\\ 0 & x > n\end{cases}$$ which is dominated by $g(x) = e^{-x/2}$. Now apply dominated convergence theorem to get that
$$\lim_{n \to \infty} \int_0^n f_n(x) dx = \lim_{n \to \infty} \int_0^{\infty} f_n(x) dx = \int_0^{\infty} \lim_{n \to \infty} f_n(x) dx = \int_0^{\infty} e^{-x/2} dx = 2$$
A: 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 $n \to \infty$.
