<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] You can rewrite the equation to get a function $c=f(C)$ and then calculate $\lim_{C\to\infty}f(C)$. It requires you to solve a cubic equation. However, there is an easier solution:
You rewrite the equation to this form:$$
C=\frac{-2Ac^3}{\frac14-c^2-2Ac^2}=\frac{2Ac^3}{(2A+1)c^2-\frac14}
$$
If $2A+1>0$ then:
$$
C=\frac{2Ac^3}{\left(c\sqrt{2A+1}-\frac12\right)\left(c\sqrt{2A+1}+\frac12\right)}
$$
You need to make something with $c$ to make $C$ go to infinity. Let's assume that $A$ is positive. Now what do you have to do to make $C=\infty$ if you can change only $c$? If $c$ approaches $\infty$, $\frac{1}{2\sqrt{2A+1}}$ from right or $-\frac{1}{2\sqrt{2A+1}}$ from right, $C$ goes to infinity. And vice-versa, if $C$ goes to infinity, $c$ approaches $\infty$, $\frac{1}{2\sqrt{2A+1}}$ from right or $-\frac{1}{2\sqrt{2A+1}}$ from right. And if $A$ is negative, we just flip the sign.
If $2A+1\leq0$ then $(2A+1)c^2-\frac14$ is always nonpositive. It also means that $A<0$. In this case, $c$ must go to $\infty$.
We have up to three results, so what now? It is because the equation gives up to three sollutions for each $C$. So the sollution varies on which of these solutions you pick.
To sum it up:
$$
\lim_{\frac{A}{B}\to\infty}=
\begin{cases}
\infty \text{ or} \pm\frac{1}{2\sqrt{2A+1}},  & \text{if $\liminf A>0$} \\
-\infty \text{ or} \pm\frac{1}{2\sqrt{2A+1}},  & \text{if $\liminf A>-\frac12$ and $\limsup A<0$} \\
\infty, & \text{if $\limsup A\leq-\frac12$}
\end{cases}
$$
Note that $\frac{A}{B}$ can't go to $\infty$ if $A=0$.
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