$\lim_{x\to\infty}\int_x^{x+1}f(y)dy=0$ implies $\lim_{x\to\infty}\frac{\int_0^{x}f(y)dy}{x}=0$ Let $f$ be a non negative continuous function on $[0,\infty)$ such that $$\lim_{x\to\infty}\int_x^{x+1}f(y)dy=0.$$
How do we prove that
$$\lim_{x\to\infty}\frac{\int_0^{x}f(y)dy}{x}=0.$$
If we see this question using the primitive function, do we have the following result for a continuous function $F$:
$$\lim_{x\to\infty}F(x+1)-F(x)=0$$
implies that
$$\lim_{x\to\infty}\frac{F(x)}{x}=0.$$
 A: Hint:


*

*Since $\lim\limits_{x\to\infty}\int_x^{x+1}f(t)dt$ then, by the definition of the limit and since $f$ is non-negative, for any $\epsilon>0$ there exist an $x_*$ s.t. if $x>x_*$ then $ 0 \leq \int_x^{x+1}f(t)dt < \frac{\epsilon}{2}$. 

*We have $\frac{\int_0^xf(t)dt}{x} = \frac{\int_0^{x_*}f(t)dt}{x} + \frac{\int_{x_*}^xf(t)dt}{x} $

*The last term in 2) is by 1) bounded by $\frac{(x-x_*)\epsilon}{2x} < \frac{\epsilon}{2}$ since $\int_{x_*}^xf(t)dt = \int_{x_*}^{x_*+1}f(t)dt +\int_{x_*+1}^{x_*+2}f(t)dt + \ldots + \int_{x_*+\lfloor x-x_*\rfloor}^{x}f(t)dt$.

*The first term in 2) can be made as small as possible ($<\frac{\epsilon}{2}$) by taking $x$ large enough.
A: Continuing from your last step in question.
see that $F(x)$ is a differentiable function. 
Then $F(x+1)-F(x)=F'(\eta_x)(x+1-x)=F'(\eta_x)$ where $x\leq \eta_x \leq x+1$.(By mean value theorem)
since $\lim_{x\rightarrow \infty}(F(x+1)-F(x))=0\implies \lim_{x\rightarrow \infty}F'(\eta_x)=0$
Now 
$$
\lim_{x\rightarrow \infty}(\frac{F(x)}{x})=\lim_{x\rightarrow \infty}F'(x)=\lim_{\rightarrow \infty}F'(\eta_x)=0
$$
Use L'Hospital's rule in first inequality
