Evaluating $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^n}{ne^{-x}}dx$

Question

Question: Find $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^n}{ne^{-x}}dx$.

My thoughts: First, I'd like to bring the limit inside the integral, because $\lim_{n\rightarrow\infty}\frac{(1-\frac{x}{n})^n}{ne^{-x}}=\frac{e^{-x}}{ne^{-x}}\rightarrow0$ and $n\rightarrow\infty$, and so the value of the integral would be $0$. However, I am a bit stuck on justifying pulling the limit inside the integral. I was hoping to be able to use the Dominated Covergence Theorem, so I need to find an integral majorant. The way that I have always gone about doing that (when the answer isn't obvious to me) is to take the derivative of the denominator with respect to $n$ and set it equal to $0$ to minimize it, then get $n$ in terms of $x$. Next, find the minimum over $n$ of my denominator (now in terms of $x$), and then find the supremum of the fraction, and see when that integral converges. However, for this one, I am a bit stuck..... maybe DCT isn't best here?

Answer 1

Easier: Move everything to $[0,1]$, so $$ \int_0^n\frac{(1-(x/n))^n}{ne^{-x}}\,\mathrm{d}x=\int_0^1\frac{(1-t)^n}{e^{-nt}}\,\mathrm{d}t=\int_0^1[(1-t)e^t]^n\,\mathrm{d}t $$ But $0\leq (1-t)e^t\leq 1$ for $t\in[0,1]$, so DCT gives the limit $0$.