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Consider a continuous function $f:[0, \infty)\rightarrow\mathbb R$ such that $\lim_{x\rightarrow\infty} f(x)=1$. I would like to prove $$ \lim_{n\rightarrow\infty}\frac{1}{n!}\int_0^\infty f(x)e^{-x}x^n\ dx = 1. $$

I first thought of using Lebesgue's DCT, but soon realized the improper integral may not converge absolutely because $f$ is not guaranteed to be positive. I am at a loss as to how to prove this.

I would be happy if you could help me solve the problem (I also welcome a partial solution, so that I can fill in its gap).

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This is not true. Consider $f(x)=1$ – Norbert Aug 25 '13 at 11:23
@Norbert I fixed a typo. – Semo Ponume Aug 25 '13 at 11:26

You can prove this without the dominated convergence theorem. $f$ is continuous and has a limit at $\infty$, so there exists an $M>0$ such that $|f(x)|<M$ for all $x\geq 0$. Now, let $\varepsilon>0$. Then, there exists $N>0$ such that, for all $x\geq N$, you have that $|f(x)-1|\leq\varepsilon$. So, you have that $$\left|\frac{1}{n!}\int_0^{\infty}f(x)e^{-x}x^n\,dx-1\right|\leq\left|\frac{1}{n!}\int_0^Nf(x)e^{-x}x^n\,dx\right|+\left|\frac{1}{n!}\int_N^{\infty}f(x)e^{-x}x^n\,dx-1\right|\leq$$$$\frac{1}{n!}\int_0^NMe^{-x}N^n\,dx+\left|\frac{1}{n!}\int_N^{\infty}(f(x)-1)e^{-x}x^n\,dx\right|+\frac{1}{n!}\int_0^Ne^{-x}x^n\,dx,$$ where we used that $\int_0^{\infty}e^{-x}x^n\,dx=n!$. So, this is less than or equal to $$\frac{MN^n}{n!}(1-e^{-N})+\frac{1}{n!}\varepsilon\int_0^{\infty}e^{-x}x^n\,dx+\frac{N^n(1-e^{-N})}{n!},$$ which is less than or equal to $\varepsilon$ at the limit as $n$ goes to $\infty$.

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Thank you for your helpful answer. I would be more than happy if you could explain the intuition behind this solution, or how you came up with it. – Semo Ponume Aug 25 '13 at 12:08
What you need to do is to show that the first term goes to zero. For $x$ near infinity, there is no problem, since $f$ is close to $1$. The problem comes from intervals of the form $[0,N]$, where you have no information about $f$, and it could be arbitrarily large there. So, you split the integral in two parts: from $0$ to $N$, and from $N$ to $\infty$. N will be chosen such that $f$ is $\varepsilon$-close to $1$ if $x\geq N$, and the rest follows. – detnvvp Aug 25 '13 at 12:11

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