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

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?

• You may calculate the integral first Jul 22, 2020 at 4:53

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 $$0$$.