If $f$ is a function of slow increase, then $f$ is slowly varying?

Following up on the question Slow variation vs slow increase. Consider the definitions below; assume that all functions are defined on the interval $x\in[a,\infty)$ for some $a\in{\mathbb R}$. (The value of $a$ is assumed to exist, but need not be the same for every function.)

Definition (Karamata): A function $L(x)>0$ is called slowly varying if for any $\lambda>0$ we have $$\lim_{x\to\infty}{L(\lambda x)\over L(x)} = 1. \tag{1}$$

Definition (Jakimczuk): Suppose that $f(x)>0, \ \lim\limits_{x\to\infty}f(x)=\infty$, and there exists a continuous derivative $f'(x)>0$. The function $f(x)$ is of slow increase if $$\lim_{x\to\infty} {f'(x) \over f(x)/x} = 0. \tag{2}$$

Let ${\mathbb L}$ be the set of all slowly varying functions, and let ${\mathbb J}$ be the set of all functions of slow increase.

Prove or disprove: $\ \$ If $f\in{\mathbb J}$, then $f\in{\mathbb L}$.

(We know that if $f\in{\mathbb J}$ and $f'(x)$ is decreasing, then $f\in{\mathbb L}$; see Theorem 9 of Jakimczuk's paper.)

Claim : $\mathbb{J}\subset \mathbb{L}$. Consider $f \in \mathbb{J}$ and $\lambda \ge 1$ (what follows can be adapted for $\lambda \le 1$).
Fix $\varepsilon>0$ : there exists $x_0>0$ such that $\forall x \ge x_0,\ 0 < \frac{f'(x)}{f(x)} \le \frac{\varepsilon}{x}$. Hence for $x \ge x_0$, $0 < \displaystyle{\int_x^{\lambda x}} \frac{f'(\xi)}{f(\xi)}d\xi \le \varepsilon \cdot \displaystyle{\int_x^{\lambda x}} \frac{d\xi}{\xi}$, and thus $0 < \mathrm{ln}\big(\frac{f(\lambda x)}{f(x)}\big) \le \varepsilon \cdot \mathrm{ln}(\lambda)$, so $1 < \frac{f(\lambda(x))}{f(x)} \le \lambda^{\varepsilon} \underset{\varepsilon \to 0}{\longrightarrow} 1$, and finally $\frac{f(\lambda x)}{f(x)} \underset{x \to +\infty}{\longrightarrow} 1$, so $f \in \mathbb{L}$.
• Thanks a lot! Very nice proof; and it works not only for $f\in{\mathbb J}$ but also for a wider set of functions $f$ with $$\lim\limits_{x\to\infty}f(x)=K\le+\infty.$$