I am cross-posting the question below, which I asked in Mathematics StackExchange a week ago and did not receive answers there. Thank you for your help!

It is well known that, if $x\mapsto f(x)$ is a slowly-varying function, then $$ \lim_{x\to\infty}\sup_{\lambda\in K}\left|\frac{f(\lambda x)}{f(x)}-1\right| = 0, $$ where $K\subset (0,\infty)$ is a compact set.

Of course, it is quite important that $K$ does not contain 0 in the above result. However, I am wondering if this uniform convergence (or, in fact, some other property of slowly-varying functions such as their explicit representation in terms of the exponential function, see below) can be exploited to prove something like the following:

Let $x\mapsto\lambda(x)$ be a function that goes to zero "sufficiently slowly" as $x\to\infty$ and let $x\mapsto f(x)$ be a slowly-varying function, then $$ \lim_{x\to\infty}\frac{f(\lambda(x) x)}{f(x)} = 1. $$

Can this be proven under some assumptions on $\lambda(\cdot)$?

One might be tempted to find some conditions on $\lambda(\cdot)$ via the representation $$ f(x) = \exp\left(\eta(x) + \int_c^{x}\frac{\varepsilon(t)}{t}\mathrm d t\right), $$ where $c>0$ is some arbitrary constant, $x\mapsto\eta(x)$ is measurable, bounded, and admits a limit as $x\to\infty$, and $t\to\varepsilon(t)$ is measurable, bounded, and converges to zero as $t\to\infty$. We get $$ \frac{f(\lambda(x)x)}{f(x)} = \exp(\eta(\lambda(x)x)-\eta(x))\times\exp\left(\int_{\lambda(x)x}^{x}\frac{\varepsilon(t)}{t}\mathrm d t\right). $$ If $\lambda(x)x\to\infty$ as $x\to\infty$, then the first term on the right-hand side converges to zero. However, bounding the second term on the right-hand side is difficult if one lacks any sort of estimate or bound on $\varepsilon(\cdot)$. It seems to me that the best one can do is exploit that $\varepsilon(t)\leq C$ for some $C>0$ and all $t>c$, and get \begin{align*} \exp\left(\int_{\lambda(x)x}^{x}\frac{\varepsilon(t)}{t}\mathrm d t\right) &\leq \exp\left(\int_{\lambda(x)x}^{x}\frac{C}{t}\mathrm d t\right)\\ % &=\exp\left(C(\log(x)-\log(\lambda(x)x)\right)\\ % &=\exp(-C\log(\lambda(x)), \end{align*} but this is insufficient since $\lambda(x)\to0$ as $x\to\infty$. So it seems that one has to exploit the fact that $\varepsilon(t)\to0$ as $t\to\infty$, but without more explicit estimates I would not know how.