Given $V \colon (0,\infty) \to (0,\infty)$ monotone increasing, such that for all $x>0$, $\lim_{t \to \infty} \frac{V(tx) - V(t)}{a(t)} = \log x$ for some function $a(t) > 0$, how can we prove that $V$ is slowly-varying? i.e. for all $x>0$, $\lim_{t \to \infty} \frac{V(tx)}{V(t)} = 1$.
Some facts that may be relevant is that we can prove that $a(t)$ must be slowly-varying. And we can write $V(x)-V(1) = \int^x_1 K(u) du - xK(x) + K(1)$ where $K$ is $-1$-varying, i.e. $\frac{K(tx)}{K(t)} \to \frac{1}{x}$. Sadly, the difference of slowly-varying function may not be slowly-varying so proving $V$ is slowly-varying through this representation doesn't seem to work.
Resnick, Sidney I., Extreme values, regular variation, and point processes, Applied Probability, Vol. 4, New York etc.: Springer-Verlag. XII, 320 p.; DM 145.00 (1987). ZBL0633.60001. 1: