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Evaluating $lim_{nrightarrowinfty}int_0^nfrac{(1-frac{x}{n})^n}{ne^{-x}}dx$

Mathematics Asked on November 18, 2021

Question: Find $lim_{nrightarrowinfty}int_0^nfrac{(1-frac{x}{n})^n}{ne^{-x}}dx$.

My thoughts: First, I’d like to bring the limit inside the integral, because $lim_{nrightarrowinfty}frac{(1-frac{x}{n})^n}{ne^{-x}}=frac{e^{-x}}{ne^{-x}}rightarrow0$ and $nrightarrowinfty$, 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?

One Answer

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

Answered by user10354138 on November 18, 2021

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