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Using Lebesgue theorem solve following limit $$ \lim_{n\to\infty}\int_{0}^{n}\big(1+\frac{x}{n}\big)^{n+1}e^{-2x}\,dx $$

I cannot find limiting function $(1+\frac{x}{n})^ne^{-2x}$. I do not know if I miss something obvious or not.

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    $(1 + \frac{x}{n})^n < e^x$ for every $n$, so your function is $\le(1+x)e^{-x} \chi_{[0,n]}$ on the non-negative reals. Prove that this is a suitable dominating function! – 2017-01-22
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    It's not true for all $n$. Only when $n$ is big enough. And when it is, depends on $x$. Or wait. You may be right. – 2017-01-22

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Let ${\{f_n\}}_{n = 1}^{\infty}$ be the functions sequence $$ f_n(x) = \left(1 + \frac{x}{n}\right) {\left(1 + \frac{x}{n}\right)}^n e^{- 2 x} {\chi}_{(0 , n)}(x) \quad \mbox{ for all } \quad x \in \mathbb{R} \quad \mbox{ and for all } \quad n = 1 , 2 , \ldots\mbox{,} $$ so we have to study the limit $$ \lim_{n \to \infty} \int_0^n {\left(1 + \frac{x}{n}\right)}^{n + 1} e^{- 2 x} dx = \lim_{n \to \infty} \int_{(0 , \infty)} f_n(x) dx\mbox{.} $$ Well, on the one hand, $f_n$ is non-negative for all $n = 1 , 2 , \ldots$ and if you choose $n \in \mathbb{N}$ and $x \in (0 , \infty)$, you have two cases: $x \geq n$ and $x < n$. If $x \geq n$, $f_n = 0$ because ${\chi}_{(0 , n)}(x) = 0$. If $x < n$ you have that $$ 1 + \frac{x}{n} \leq 2\mbox{.} \tag{1} $$ On the other hand, we know the sequence ${\left\{{\left(1 + \frac{x}{n}\right)}^n\right\}}_{n = 1}^{\infty}$ is growing to $e^x$. In other words, $$ {\left(1 + \frac{x}{n}\right)}^n \leq e^x\mbox{.} \tag{2} $$ Then, by $(1)$ and $(2)$, $$ f_n(x) = \left(1 + \frac{x}{n}\right) {\left(1 + \frac{x}{n}\right)}^n e^{- 2 x} {\chi}_{(0 , n)}(x) \leq 2 e^x e^{- 2 x} = 2 e^{- x} = g(x) $$ for all $x \in (0 , \infty)$ and for all $n = 1 , 2 , \ldots$. Further, if you fix $x \in (0 , \infty)$, $$ \lim_{n \to \infty} \left(1 + \frac{x}{n}\right) = 1 \quad \mbox{ and } \quad \lim_{n \to \infty} {\left(1 + \frac{x}{n}\right)}^n = e^x\mbox{,} $$ so you can use the properties about limits (for sequences of real numbers) to obtain that $$ \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right) {\left(1 + \frac{x}{n}\right)}^n e^{- 2 x} {\chi}_{(0 , n)}(x) = 1 \cdot e^x e^{- 2 x} \cdot 1 = e^{- x} = f(x) $$ for all $x \in (0 , \infty)$. Obviosuly, $g$ is integrable on $(0 , \infty)$, so you can use Dominated Convergence Theorem to say that $$ \lim_{n \to \infty} \int_0^n {\left(1 + \frac{x}{n}\right)}^{n + 1} e^{- 2 x} dx = \lim_{n \to \infty} \int_{(0 , \infty)} f_n(x) dx = \int_{(0 , \infty)} f(x) dx = \int_{(0 , \infty)} e^{- x} dx = 1\mbox{.} $$

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Hint. Observe that, for $x \in [0,\infty)$, $$ \lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n=e^x $$ and that, for $x \in [0,\infty)$, $$ 1_{[0,n]}(x)\left(1+\frac{x}{n}\right)^ne^{-2x}\le e^{-x}, \quad n\ge1. $$