Find $\lim_{n \rightarrow \infty} \int_0^n (1+ \frac{x}{n})^{n+1} \exp(-2x) \, dx$ Find:
$$\lim_{n \rightarrow \infty} \int_0^n \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx$$
The sequence $\left(1+ \frac{x}{n}\right)^{n+1} \exp{(-2x)}$ converges pointwise to $\exp{(-x)}$. So if we could apply Lebesgue's monotone convergence theorem, we have:
\begin{align}
\lim_{n \rightarrow \infty} \int_0^n \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx &=\lim_{n \rightarrow \infty} \int_{\mathbb{R}} I_{(0,n)(x)} \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx \\[10pt]
&= {} \int_{- \infty}^\infty \exp(-x) \, dx= +\infty
\end{align}
Can someone help me to prove that the sequence of integrands is monotone?
 A: Using the indicator function, we can write
$$\int_0^n \left(1+\frac xn\right)^{n+1}e^{-2x}\,dx=\int_0^\infty \left(1+\frac xn\right)^{n+1}e^{-2x}\xi_{[0,n]}(x)\,dx$$
Then, note that for $0\le x\le n$, we have
$$\left(1+\frac xn\right)^{n+1}=\left(1+\frac xn\right)\left(1+\frac xn\right)^{n}\le 2e^{x}$$

EDIT:  The OP requested a source for the preceding inequality.
  In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the exponential function satisfies the inequality $$e^x\ge \left(1+\frac xn\right)^{n}$$for $x>-n$.

We can assert, therefore, that 
$$\left(1+\frac xn\right)^{n+1}e^{-2x}\xi_{[0,n]}(x)\le 2e^{-x}$$
Applying the Dominated convergence theorem, we have
$$\begin{align}
\lim_{n\to \infty} \int_0^n \left(1+\frac xn\right)^{n+1}e^{-2x}\,dx&=\int_0^\infty \lim_{n\to \infty}\left(\left(1+\frac xn\right)^{n+1}\xi_{[0,n]}(x)\right)\,e^{-2x}\,dx\\\\
&=\int_0^\infty e^{x}\,(1)\,e^{-2x}\,dx\\\\
&=1
\end{align}$$ 
A: Seems easier to use dominated convergence. Hint: $1+t\le e^t$, so $(1+x/n)^{n+1}\le\dots$; for $n\ge 2$ this is less than or equal to $\dots$.
A: Firstly, you have a mistake in your last line:
\begin{align}\lim_{n \rightarrow \infty} \int_{\mathbb{R}} I_{(0,n)(x)} \left(1+ \frac{x}{n}\right)^{n+1} \exp(-2x) \, dx = \int_{0}^\infty \exp(-x) \, dx= 1 \end{align}
and not $+\infty$ as you have it (because you integrated from $-\infty$ and not from $0$). You could also change variable first $y=\frac{x}{n}$: $$\int_{0}^{n} (1+ \frac{x}{n})^{n+1} \exp{(-2x)} dx=\int_{0}^{1} (1+y)^{n+1} \exp{(-2ny)} n\,dy$$ and now everything happens in $[0,1]$. 
