$(1) \lim_{x\to \infty} \frac{1}{f(x)}\int_x^{\infty}f(s)ds=0$ and $(2) \lim_{x\to \infty} \frac{f(x+t)}{f(x)}=0.$ 
Let $f: [0,\infty)\to  [0,\infty)$ be continuous, non-increasing, and integrable. Prove that
$$(1) \lim_{x\to \infty} \frac{1}{f(x)}\int_x^{\infty}f(s)ds=0$$ iff for each $t>0$,
$$(2)  \lim_{x\to \infty} \frac{f(x+t)}{f(x)}=0.$$

This is an exercise from a textbook about real analysis. I think (1) implies (2) is not hard since
$$\frac{1}{f(x)}\int_x^{N}f(s)ds=...?$$ I have no idea about that.
 A: To show that (1) implies (2), note that for any $x,t>0$, $$\frac{1}{f(x)}\int_x^{\infty}f(s)\,ds\geq\frac{1}{f(x)}\int_x^{x+t}f(s)\,ds\geq\frac{1}{f(x)}\int_x^{x+t}f(x+t)\,ds=\frac{tf(x+t)}{f(x)},$$ and let $x\to\infty$.
For the other direction, let $\varepsilon>0$. Then there exists $M>0$ such that, if $x>M$, then $\frac{f(x+\varepsilon/2)}{f(x)}<\frac{1}{2}$. For $x>M$, write $$\frac{1}{f(x)}\int_x^{\infty}f=\frac{1}{f(x)}\int_x^{x+\varepsilon/2}f+\sum_{k=1}^{\infty}\frac{1}{f(x)}\int_{x+k\varepsilon/2}^{x+(k+1)\varepsilon/2}f.$$ The first term is bounded above by $\varepsilon/2$. For the terms in the sum, note that \begin{align*}\frac{1}{f(x)}\int_{x+k\varepsilon/2}^{x+(k+1)\varepsilon/2}f&\leq\frac{f(x+k\varepsilon/2)}{f(x)}\frac{\varepsilon}{2}\\ &=\frac{f(x+k\varepsilon/2)}{f(x+(k-1)\varepsilon/2)}\frac{f(x+(k-1)\varepsilon/2)}{f(x+(k-2)\varepsilon/2)}\cdots \frac{f(x+\varepsilon/2)}{f(x)}\frac{\varepsilon}{2}<\left(\frac{1}{2}\right)^k\frac{\varepsilon}{2},\end{align*} therefore $$\frac{1}{f(x)}\int_x^{\infty}f<\frac{\varepsilon}{2}+\sum_{k=1}^{\infty}2^{-k-1}\varepsilon=\varepsilon.$$
