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?
A good property of the slowly varying functions is that the convergence in the definition is actually uniform on compact sets, that is, for all $0\lt a<b$, $$ \lim_{x\to +\infty}\sup_{a\leqslant c\leqslant b}\left\lvert \frac{L(cx)}{L(x)}-1\right\rvert=0. $$ Fix $c\gt 0$ and let $x_0$ be such that for all $x\geqslant x_0$, $1/2\leqslant L_2(cx)/L_2(x)\leqslant 3/2$ and $L_2(x)$. Then $$ \left\lvert \frac{L_1\left(L_2(cx)\right)}{L_1\left(L_2(x)\right)}-1\right\rvert =\left\lvert \frac{L_1\left(\frac{L_2(cx)}{L_2(x)}L_2(x)\right)}{L_1\left(L_2(x)\right)}-1\right\rvert\leqslant \sup_{1/2\leqslant R\leqslant 3/2}\left\lvert \frac{L_1\left(RL_2(x)\right)}{L_1\left(L_2(x)\right)}-1\right\rvert. $$ Since $L_2(x)\to +\infty$, the last expression goes to $0$ as $x$ goes to infinity.