A function, $L$ is slowly varying if for all $b>0$:
$$ \lim_{x \to \infty} \frac{L(bx)}{L(x)}=1. $$
If two functions, $L_1, L_2$ are slowly varying, and we have further that $L_2(x) \to \infty$ as $x \to \infty$, I am stuck on proving that $L_1(L_2) $ is slowly varying as well (the composite).
It seems obvious since $L_1$ is slowly varying and so letting the argument of the numerator and denominator go to infinity will surely give us a result of 1, but I'm not sure how to show this analytically?