$$\lim_{n \to \infty}\int_0^n(1-\frac{x}{n})^ne^{-x}dx$$
I am going to use the Dominated Convergence Theorem to solve this.
$$\lim_{n \to \infty}\int_0^{\infty}\chi_{[0, n]}(1-\frac{x}{n})^ne^{-x}dx$$
Step 1 - Find a dominating function $g$
Let $h_n = (1-\frac{x}{n})^n$
$\chi_{[0, n]}$ is telling us that $h_n = 0$ whenever $x > n$. This means that $h_n(0) = 1$ and $h_n(n) = 0$ for all $n$. So $h_n$ is a sequence of measurable functions that is bounded above by $1$.
Let $f_n = (1-\frac{x}{n})^n \cdot e^{-x} = h_n \cdot e^{-x}$. This also has a maximum value of $1$. So $f_n$ is dominated by the constant function $g(x) = 1$.
Step 2 - Find a function $f$ such that $f_n \to f$ pointwise almost everywhere
Now consider $\lim_{n \to \infty}(1-\frac{x}{n})^ne^{-x}$
$e^{-x}\cdot e^{-x} = e^{-2x}$
So we have that $f_n \to f = e^{-2x}$ $f$ is integrable and it's integral is \begin{align*} &= \lim_{k \to \infty}\int_0^{k}e^{-2x}dx\\ &= \lim_{k \to \infty}-\frac{1}{2}e^{-2x}\mid_0^k\\ &= \lim_{k \to \infty}[-\frac{1}{2}e^{-2\cdot k} - (-\frac{1}{2} e^{2\cdot0})]\\ &=-\frac{1}{2}(0) + \frac{1}{2}\\ &= \frac{1}{2} \end{align*}
So we have a sequence of functions $f_n$ such that $\mid f_n \mid \le g$ and $f_n \to f$, with $f$ integrable. Hence, by the Dominated Convergence Theorem we have that $$\lim_{n \to \infty}\int_0^{\infty} f_n = \int_0^{\infty} f = \frac{1}{2}$$
Have I made any incorrect assumptions in this solution?
