To calculate a limit by Dominated Convergence. $$\lim\limits_{n\rightarrow \infty}\int^{n}_{0}\left(1+\frac{x}{n}\right)^{n}e^{-2x}dx$$
Let $f_{n}, f:[0,\infty]\longrightarrow [0,\infty]$, defined by $f_{n}(x):=\left(1+\frac{x}{n}\right)^{n}e^{-2x}1_{[0,n]}(x)$ and $f(x):=e^{-x}$. Note that $f_{n}$ is a measurable function, because is continuous. Furthermore, since $\left(1+\frac{x}{n}\right)^{n}$ is increasing and 
$$\lim\limits_{n\rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}=\lim\limits_{y\rightarrow 0} \left(1+y\right)^{\frac{x}{y}}=e^{x}$$
this means that $f_{n}(x)\rightarrow e^{x}e^{-2x}=e^{-x}=f(x)$ and
 taking $g(x)=e^{-x}$ we have $g$ integrable and $|f_{n}(x)|\leq e^{x}e^{-2x}=e^{x}=g(x)$. Therefore, by Dominated Convergence Theorem, we have
$$\lim\limits_{n\rightarrow \infty}\int f_{n}(x)=\int f(x)$$
i.e., 
$$\lim\limits_{n\rightarrow \infty}\int^{n}_{0}\left(1+\frac{x}{n}\right)^{n}e^{-2x}dx=\int^{\infty}_{0}e^{-x}=\lim\limits_{t\rightarrow \infty}\int^{t}_{0}e^{-t}=\lim\limits_{t\rightarrow \infty}(-e^{-t}+1)=1$$
I would like to know similar examples. Thank you.
 A: You must define your $f_n$ by $$f_n(x)=\left(1+\frac{x}{n}\right)e^{-2x}\mathrm{1}_{[0,n]}(x)$$
in your example. There are a lot of examples using this technique, such as
$$ \Gamma(x)=\lim\limits_{n\rightarrow+\infty}\int_0^n{\left(1-\frac{t}{n}\right)^nt^{x-1}dt} $$
which leads you to the formula
$$ \Gamma(x)=\lim\limits_{n\rightarrow+\infty}\frac{n!n^x}{x(x+1)\ldots(x+n)} $$
A trickiest exercise could be : find the following limit
$$ \lim\limits_{n\rightarrow+\infty}\int_0^n{\left(\cos\left(\frac{x}{n}\right)\right)^{n^2}dx} $$
You can also prove D'Alembert-Gauss' theorem by dominated convergence : let $f:[0,2\pi]\longrightarrow\mathbb{C}^*$, we define
$$ I(f):=\frac{1}{2i\pi}\int_0^{2\pi}\frac{f'(t)}{f(t)}dt $$
Let $P\in\mathbb{C}[X]$ a non-constant polynom, let us suppose that $\forall z\in\mathbb{C},\,P(z)\neq 0$, and let $f_r(t)=P(re^{it})$ for all $r\in\mathbb{R}^+$. Since $f_r(t)\neq 0$ for all $t\in[0,2\pi]$ we can calculate $I(f_r)$. One can show that $I(f_r)=0$ because $P$ has no roots, but
$$ I(f_r)=\frac{1}{2\pi}\int_0^{2\pi}\frac{rP'(re^{it})}{P(re^{it})}dt\underset{r\rightarrow+\infty}{\longrightarrow}\deg P\geqslant 1 $$
by dominated convergence, which is not, thus $P$ has a root.
