<s>[INST] Given $\frac{A}{4B}+\frac{2a^3A}{B^3}-\frac{a^2A}{B^3}-\frac{2a^2A^2}{B^3}=0$, what does $\frac{a}{B}$ tend to as $\frac{A}{B}\to \infty$? In this problem, all of $A, a, B$ can vary.
My attempt was to let $\frac{A}{B}=C$ and $\frac{a}{B}=c$, which gives $ \frac{C}{4}+2Ac^3-c^2C-2Ac^2C=0$.
Now $C\to \infty $ means either $A\to \infty $ or $B\to 0 $. At this point I'm not sure at all how to proceed.
 [/INST] my attempt, starting from your attempt:
$$\frac{C}{4} + 2Ac^3 - c^2C-2Ac^2C = 0$$
divide by $c^2$:
$$\frac{C}{4c^2} + 2Ac - C-2AC = 0$$
$$2Ac = -\frac{C}{4c^2} + C + 2AC$$
$$2Ac = C(-\frac{1}{4c^2} + 1 + 2A)$$
and isolate $C$:
$$\frac{2Ac}{-\frac{1}{4c^2} + 1 + 2A} = C$$
as we know, $C$ approaches infinity. Therefore, the LHS of the equation should too. Assuming $2Ac$ does not approach infinity, the following can be deduced:
$$-\frac{1}{4c^2} + 1 + 2A \rightarrow 0$$
$$-1 + 4c^2 + 8Ac^2 \rightarrow 0$$
$$ c^2(4 + 8A) = 1$$
and therefore:
$$c = \sqrt{\frac{1}{4+8A}}$$
</s>