# A Problem on the Limit of an Integral

While I investigate the property of positive random variables, I encountered the following question, not easy to solve.

Let $$f:\left[0,\infty\right)\rightarrow\mathbb{R}$$ be a continuous positive function. Define $$\displaystyle F(s) := \int_0^s f(x)dx\quad(s\ge0)$$ and assume $$\lim_{s\rightarrow\infty}F(s)=1\quad\text{and}\quad \lim_{s\rightarrow\infty}\frac {s\left(1-F(s)\right)} {\int_0^s xf(x)dx}= 0.$$ Then prove or disprove that $$\lim_{s\rightarrow\infty}\frac{\int_0^s x^2 f(x)dx}{s\int_0^s xf(x)dx}=0.$$

I tried in various way but could not succeed. I would be very happy if someone could answer this question.

What I have tried:

I tried some known inequalities such as Holder's inequality and Chebyshev's inequality but could not find something that matches this problem. One approach i made is following:

Let $$F(s_n) = 1- \frac1{2^n}$$. Then $$\frac {s_n\left(1-F(s_n)\right)} {\int_0^{s_n} xf(x)dx}$$ can be approximated by $$u_n := \frac1{\frac{s_1}{s_n}2^{n-1}+\frac{s_2}{s_n}2^{n-2}+\cdots+\frac{s_{n-1}}{s_n}2+1}$$ and $$\frac{\int_0^{s_n} x^2 f(x)dx}{s_n\int_0^{s_n} xf(x)dx}$$ is approximated by $$\frac{t_n}{u_n} := \frac{\frac{s_1^2}{s_n^2} 2^{n-1}+\frac{s_2^2}{s_n^2}s^{n-2}+\cdots+\frac{s_{n-1}^2}{s_n^2}2+1} {\frac{s_1}{s_n}2^{n-1}+\frac{s_2}{s_n}s^{n-2}+\cdots+\frac{s_{n-1}}{s_n}2+1}$$ Then our new question is whether $$\lim_{n\rightarrow\infty}\frac{t_n}{u_n}=0$$ when $$\lim_{n\rightarrow\infty}u_n = 0$$. I expected that this would give me a key but found this not easy at all compared to the original one. I also posted this another form after modifying some notations on A Problem on the Limit of a Sequence. Two questions have not been solved now.