Is the ''proof" for a proposition about slowly varying function correct?

Question

The positive function $f: [0,+\infty)\to\mathbb{R}^+$ is called slowly varying function if $$\lim_{x\to+\infty}\frac{f(cx)}{f(x)}=1 \text{ for any $c>0$}. $$ Now there is a ''proof" for this proposition: Let $f$ be a slowly varying function, for every $r>0$ , $$\int_x^{\infty}t^{-r-1}f(t)\text{d}t\sim \frac{1}{r}x^{-r}f(x) \text{ as }x\to +\infty.$$ Proof. For every $c>0$ and $\varepsilon>0$, there exists a positive number $X$ such taht \begin{equation}\label{1} 1-\varepsilon<\frac{f(cx)}{f(x)}<1+\varepsilon \text{ as } x>X. \end{equation} We take $c=t/x$ and have $$(1-\varepsilon)f(x)<f(t)<(1+\varepsilon)f(x).$$ So $$r(1-\varepsilon)f(x)\int_x^{\infty}t^{-r-1}\text{d}t<r\int_x^{\infty}t^{-r-1}f(t)\text{d}t<r(1+\varepsilon)f(x)\int_x^{\infty}t^{-r-1}\text{d}t.$$ This means that $$(1-\varepsilon)x^{-r}f(x)<r\int_x^{\infty}t^{-r-1}f(t)\text{d}t<(1+\varepsilon)x^{-r}f(x).$$ For the arbitrariness of $\varepsilon$ we can assert the validity of this proposition. This is the end of the proof. But I can't sure the above ''proof" is right. I belive that the $\varepsilon$ is related to $c$. So we can not take $c=t/x$ since $t>x$, both $t$ and $x$ are variables. Is my view correct? How to prove this proposition?